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Hi zetafunc ,
Will you please give an example ?
Hi thickhead ,
Once you have got a valid pair of i and j , then you may add
n * 182 * 165 to both sides of 182 * i and 165 * j and obtain a bunch of i and j as you had stated in # 3 . While to express a certain prime as a * 17 + b * 5 may need trial and error .
Hi thickhead ,
Your answer is correct but the values i = 967 and j = 1061 are too large .
I have got an answer for i = 142 and j = 151 .
By the method you stated in # 3 , I tried 929 = a * 17 + b * 5 and got
929 = 2 * 17 + 179 * 5 ( by trial and error ) , thus
929 = 2 * (182 - 165 ) + 179 * ( 10 * 182 - 11 * 165 )
= 2 * 182 + 179 * 10 * 182 - ( 2 * 165 + 179 * 11 * 165 )
= 1792 * 182 - 1971 * 165
Thus an answer is i = 1792 and j = 1971 .
Since 1792 = 10 * 165 + 142 while 1971 = 10 * 182 + 151 ,
thus 929 = ( 10 * 165 + 142 ) * 182 - ( 10 * 182 + 151 ) * 165
= 10 * 165 * 182 + 142 * 182 - ( 10 * 182 * 165 + 151 * 165 )
= 142 * 182 - 151 * 165
Thus another answer is i = 142 and j = 151
Your answer can also be simplified as 967 = 5 * 165 + 142 while
1061 = 5 * 182 + 151 .
Hi zetafunc ,
Thanks for your opinion !
Hi thickhead ,
Thanks much for your reply !
Originally I wonder why I can't get the 6 primes from (1) to (6) using the
expression 182 * i - 165 * j , now I have found that it's because I omitted
i for 53 after (108) at the algorithm .
As 182 * 53 = 9646 - 165 * 55 = 9075 ⇒ d = 571 (P) for i=53 and j=55
Let 182 * i - 165 * j = -571 ⇒ 182 ( 53 + i ) - 165 ( 55 + j ) = 0
⇒ 182 ( 53 + i ) =165 ( 55 + j ) ⇒ i = 165 - 53 = 112 and j = 182 - 55 = 127
Thus the prime -571 will occur in its symmetric place with i = 112 and j = 127 .
However , the method at shown in # 3 seems not work since i and j should be
relatively prime .
For equations of the form 182 * i - 165 * j = p
where i and j are integers and p is prime .
( 182 = 2*7*13 while 165 = 3*5*11 . )
i is relatively prime with 165 and j while j is relatively prime
with 182 and i . Solve i and j for
(1) 182 * i - 165 * j = 89
(2) 182 * i - 165 * j = 241
(3) 182 * i - 165 * j = 419
(4) 182 * i - 165 * j = 571
(5) 182 * i - 165 * j = 929
(6) 182 * i - 165 * j = 997
Composite no. < 1000 containing at least 1 prime factor from 17 to 31 .(The other prime factor may > 31 , but will < 1000 / 17 = 58 . 8 , i.e. 53 ) ( 27 ones ) :
( A ) Arranged in products of prime factors
* | 17 | 19 | 23 | 29 | 31 || 37 | 41 | 43 | 47 | 53 |
17 | 289 | 323 | 391 | 493 | 527 || 629 | 697 | 731 | 799 | 901 |
19 | / | 361 | 437 | 551 | 589 || 703 | 779 | 817 | 893 | / |
23 | / | / | 529 | 667 | 713 || 851 | 943 | 989 | / | / |
29 | / | / | / | 841 | 899 || / | / | / | / | / |
31 | / | / | / | / | 961 || / | / | / | / | / |
( B ) Rearranged in ascending order
2xx : 289
3xx : 323 , 361 , 391
4xx : 437 , 493
5xx : 527 , 529 , 551 , 589
6xx : 629 , 667 , 697
7xx : 703 , 713 , 731 , 779 , 799
8xx : 817 , 841 , 851 , 893 , 899
9xx : 901 , 943 , 961 , 989
Since the nos. generated are found to be symmetric from (175) upwards with (176) downwards . It can be induced that the value in ( r ) is
identical with that in (351 - r ) by * ( -1 ) for r ≧ 7 . E.g. the values obtained in
(173) is 239 and (178) - 239 .
Similarly we can predict that the value obtained in (7) , i.e.
182 * 1 - 165 * 1 = 17 will occur in ( 351 - 7 ) = ( 344 ) as - 17 .
Let the expression in (344) be expressed as 182 i - 165j = -17
where i is relatively prime with 165 * j while j is relatively prime
with 182 * i .
Since 182i - 165j = -17 ⇒ 165 j - 182i = 182 - 165
⇒ 165 j + 165 = 182 i + 182
⇒ 165 ( j + 1) = 182 ( i + 1)
A solution for the equation is j+1 = i+1 = 0 ⇒ j = -1 and i = -1 ,
i.e. the expression in (7) : 182 * 1 - 165 * 1 = 17 .
Another solution is i + 1 = 165 and j + 1 = 182 ,
⇒ i = 164 and j = 181 .
Thus the expression in (344) may be expressed as
182 * 164 - 165 * 181= 29848 - 29865= -17
A third solution for the equation may be i + 1 = 2 * 165 = 330
and j+1 = 2 * 182 = 364 ⇒ i = 329 and j = 363 .
i.e. 182 * 329 = 59878 - 165 * 363 = 59895 = -17 .
( Continued from # 17 )
(104) 182 * 52 = 9464 165 * 53 = 8745 719 (P)
(105) 182 * 52 = 9464 165 * 55 = 9075 389 (P)
(106) 182 * 52 = 9464 165 * 57 = 9405 59 (P)
(107) 182 * 52 = 9464 165 * 59 = 9735 - 271 (P)
(108) 182 * 52 = 9464 165 * 61 = 10065 -601(P)
(109) 182 * 56 = 10192 165 * 57 = 9405 787 (P)
(110) 182 * 56 = 10192 165 * 59 = 9735 457 (P)
(111) 182 * 56 = 10192 165 * 61 = 10065 127 (P)
(112) 182 * 56 = 10192 165 * 67 = 11055 - 863 (P)
(113) 182 * 58 = 10556 165 * 59 = 9735 821 (P)
(114) 182 * 58 = 10556 165 * 61 = 10065 491 (P)
(115) 182 * 58 = 10556 165 * 67 = 11055 - 499 (P)
(116) 182 * 58 = 10556 165 * 69 = 11385 - 829 (P)
(117) 182 * 59 = 10738 165 * 61 = 10065 673 (P)
(118) 182 * 59 = 10738 165 * 67 = 11055 - 317 (P)
(119) 182 * 59 = 10738 165 * 69 = 11385 - 647 (P)
(120) 182 * 59 = 10738 165 * 71 = 11715 - 977 (P)
(121) 182 * 61 = 11102 165 * 67 = 11055 47 (P)
(122) 182 * 61 = 11102 165 * 69 = 11385 - 283 (P)
(123) 182 * 61 = 11102 165 * 71 = 11715 - 613 (P)
(124) 182 * 61 = 11102 165 * 73 = 12045 - 943 ( = 23 * 41 )
(125) 182 * 62 = 11284 165 * 67 = 11055 229 (P)
(126) 182 * 62 = 11284 165 * 69 = 11385 - 101 (P)
(127) 182 * 62 = 11284 165 * 71 = 11715 - 431(P)
(128) 182 * 62 = 11284 165 * 73 = 12045 - 761(P)
(129) 182 * 64 = 11648 165 * 67 = 11055 593 (P)
(130) 182 * 64 = 11648 165 * 69 = 11385 263(P)
(131) 182 * 64 = 11648 165 * 71 = 11715 - 67 (P)
(132) 182 * 64 = 11648 165 * 73 = 12045 - 397 (P)
(133) 182 * 64 = 11648 165 * 75 = 12375 - 727 (P)
(134) 182 * 67 = 12194 165 * 69 = 11385 809 (P)
(135) 182 * 67 = 12194 165 * 71 = 11715 479 ( P)
(136) 182 * 67 = 12194 165 * 73 = 12045 149 (P)
(137) 182 * 67 = 12194 165 * 75 = 12375 - 181 (P)
(138) 182 * 67 = 12194 165 * 79 = 13035 - 841 (= 29 * 29 )
(139) 182 * 68 = 12376 165 * 69 = 11385 991 (P)
(140) 182 * 68 = 12376 165 * 71 = 11715 661 (P)
(141) 182 * 68 = 12376 165 * 73 = 12045 331 (P)
(142) 182 * 68 = 12376 165 * 75 = 12375 1
(143) 182 * 68 = 12376 165 * 79 = 13035 - 659 (P)
(144) 182 * 68 = 12376 165 * 81 = 13365 - 989 (=23*43)
(145) 182 * 71 = 12922 165 * 73 = 12045 877 (P)
(146) 182 * 71 = 12922 165 * 75 = 12375 547 (P)
(147) 182 * 71 = 12922 165 * 79 = 13035 - 113 (P)
(148) 182 * 71 = 12922 165 * 81 = 13365 - 443 (P)
(149 ) 182 * 71 = 12922 165 * 83 = 13695 - 773 (P)
(150) 182 * 73 = 13286 165 * 75 = 12375 911 (P)
(151) 182 * 73 = 13286 165 * 79 = 13035 251 (P)
(152) 182 * 73 = 13286 165 * 81 = 13365 - 79 (P)
(153) 182 * 73 = 13286 165 * 83 = 13695 - 409 (P)
(154) 182 * 73 = 13286 165* 85 = 14025 - 739 (P)
(155) 182 * 74 = 13468 165 * 79 = 13035 433 (P)
(156) 182 * 74 = 13468 165 * 81 = 13365 103 (P)
(157) 182 * 74 = 13468 165 * 83 = 13695 - 227 (P)
(158) 182 * 74 = 13468 165 * 85 = 14025 - 557 (P)
(159) 182 * 74 = 13468 165 * 87 = 14355 - 887 (P)
(160) 182 * 76 = 13832 165 * 79 = 13035 797 (P)
(161) 182 * 76 = 13832 165 * 81 = 13365 467 (P)
(162) 182 * 76 = 13832 165 * 83 = 13695 137 (P)
(163) 182 * 76 = 13832 165 * 85 = 14025 - 193 (P)
(164) 182 * 76 = 13832 165 * 87 = 14355 - 523 (P)
(165) 182 * 76 = 13832 165 * 89 = 14685 - 853 (P)
(166) 182 * 79 = 14378 165 * 83 = 13695 683 (P)
(167) 182 * 79 = 14378 165 * 85 = 14025 353 (P)
(168) 182 * 79 = 14378 165 * 87 = 14355 23 (P)
(169) 182 * 79 = 14378 165 * 89 = 14685 - 307 (P)
(170) 182 * 79 = 14378 165 * 93 = 15345 - 967 (P)
(171) 182 * 82 = 14924 165 * 85 = 14025 899 (= 29 * 31)
(172) 182 * 82 = 14924 165 * 87 = 14355 569 (P)
(173) 182 * 82 = 14924 165 * 89 = 14685 239 (P)
(174) 182 * 82 = 14924 165 * 93 = 15345 - 421(P)
(175) 182 * 82 = 14924 165 * 95 = 15675 - 751(P)
(176) 182 * 83 = 15106 165 * 87 = 14355 751 ( Re)
(177) 182 * 83 = 15106 165 * 89 = 14685 421
(178) 182 * 83 = 15106 165 * 93 = 15345 - 239
(179) 182 * 83 = 15106 165 * 95 = 15675 - 569
(180) 182 * 83 = 15106 165 * 97 = 16005 - 899 (= 29 * 31 )
(181) 182 * 86 = 15652 165 * 89 = 14685 967
(182) 182 * 86 = 15652 165 * 93 = 15345 307
(183) 182 * 86 = 15652 165 * 95 = 15675 - 23
(184) 182 * 86 = 15652 165 * 97 = 16005 - 353
(185) 182 * 86 = 15652 165 * 99 = 16335 - 683
(186) 182 * 89 = 16198 165 * 93 = 15345 853
(187) 182 * 89 = 16198 165 * 95 = 15675 523
(188) 182 * 89 = 16198 165 * 97 = 16005 193
(189) 182 * 89 = 16198 165 * 99 = 16335 - 137
(190) 182 * 89 = 16198 165 * 101 = 16665 - 467
(191) 182 * 89 = 16198 165 * 103 = 16995 - 797
(192) 182 * 91 = 16562 165 * 95 = 15675 887
(193) 182 * 91 = 16562 165 * 97 = 16005 557
(194) 182 * 91 = 16562 165 * 99 = 16335 227
(195) 182 * 91 = 16562 165 * 101 = 16665 - 103
(196) 182 * 91 = 16562 165 * 103 = 16995 - 433
(197) 182 * 92 = 16744 165 * 97 = 16005 739
(198) 182 * 92 = 16744 165 * 99 = 16335 409
(199) 182 * 92 = 16744 165 * 101 = 16665 79
(200) 182 * 92 = 16744 165 * 103 = 16995 - 251
(201) 182 * 92 = 16744 165 * 107 = 17655 - 911
(202) 182 * 94 = 17108 165 * 99 = 16335 773
(203) 182 * 94 = 17108 165 * 101 = 16665 443
(204) 182 * 94 = 17108 165 * 103 = 16995 113
(205) 182 * 94 = 17108 165 * 107 = 17655 - 547
(206) 182 * 94 = 17108 165 * 109 = 17985 - 877
(207) 182 * 97 = 17654 165 * 101 = 16665 989 ( = 23 * 43 )
(208) 182 * 97 = 17654 165 * 103 = 16995 659
(209) 182 * 97 = 17654 165 * 107 = 17655 - 1
(210) 182 * 97 = 17654 165 * 109 = 17985 - 331
(211) 182 * 97 = 17654 165 * 111 = 18315 - 661
(212) 182 * 97 = 17654 165 * 113 = 18645 - 991
(213) 182 * 98 = 17836 165 * 103 = 16995 841 ( = 29 * 29 )
(214) 182 * 98 = 17836 165 * 107 = 17655 181
(215) 182 * 98 = 17836 165 * 109 = 17985 - 149
(216) 182 * 98 = 17836 165 * 111 = 18315 - 479
(217) 182 * 98 = 17836 165 * 113 = 18645 - 809
(218) 182 * 101 = 18382 165 * 107 = 17655 727
(219) 182 * 101 = 18382 165 * 109 = 17985 397
(220) 182 * 101 = 18382 165 * 111 = 18315 67
(221) 182 * 101 = 18382 165 * 113 = 18645 - 263
(222) 182 * 101 = 18382 165 * 115 = 18975 - 593
The prime 751 obtained in (175 ) repeated in ( 176 ) , but the former
= 165 * 95 - 182 * 82 and the latter = 182 * 83 - 165 * 87 , while
165 * 95 - 182 * 82 = 182 * 83 - 165 * 87
<=> 165 * 95 + 165 * 87 = 182 * 83 + 182 * 82
<=> 165 * ( 95 + 87 ) = 182 * ( 83 + 82 )
<=> 165 * 182 = 182 * 165 .
It is quite strange that then the prime 421 obtained in (174 ) was
repeated in ( 177 ) by * -1 , and so on for every nos . obtained before , i.e. ,
the list is symmetric ( for absolute value ) upwards at (175) and downwards at (176) .
( It seems that if the list is not symmetric at certain place then there must
be something wrong . )
Numbers with value < 1000 generated mainly from set of primes < 17 .
Let S = { 2 , 3 , 5 , 7 , 11 , 13 } and take A = { 2 , 7 , 13 } and B = { 3 , 5 , 11 } , thus π A = 182 and πB = 165 while L1 = 17 ^2 = 289 .
Multiply 182 by i , and 165 by j where i and j are positive integers . If i is relatively prime with 165 and j , while j is relatively prime with 182 and i , then 182 * i is relatively prime with 165 * j .
Let n be 182 * i ± 165 * j for various values of i and j , clearly n is not divisible by primes < 17 , nor every prime factors of i and j .
If n generated ( as sum or difference ) < 289 , then n = 1 or is a prime .
If n > 289 , then usually we have to check further the divisibility of n by
consecutive primes from 17 up to the prime just < √n ( may be by actual division ) , excluding those prime factors of i or j . If every such prime does not divide n , then n is a prime .
In the following we shall list all such n's with value < 1000 .
( A ) sum
(1) 182 * 1 + 165 * 1 = 347 ( P )
(2) 182 * 1 + 165 * 3 = 677 ( P )
(3) 182 * 1 + 165 * 5 = 1007
(4) 182 * 2 + 165 * 1 = 529 ( = 23 * 23 )
(5) 182 * 2 + 165 * 3 = 859 ( P )
(6) 182 * 4 + 165 * 1 = 893 ( = 19 * 47 )
( B ) difference
(7) 182 * 1 = 182 - 165 * 1 = 165 = 17 ( P )
(8) 182 * 1 = 182 - 165 * 3 = 495 = -313 ( P )
(9) 182 * 1 = 182 - 165 * 5 = 825 = - 643 (P)
(10) 182 * 2 = 364 - 165 * 1 = 165 = 199 (P)
(11) 182 * 2 = 364 - 165 * 3 = 495 = -131 (P)
(12) 182 * 2 = 364 - 165 * 5 = 825 = -461 (P)
(13) 182 * 4 = 728 - 165 * 1 = 165 = 563 ( P)
(14) 182 * 4 = 728 - 165 * 3 = 495 = 233 ( P )
(15) 182 * 4 = 728 - 165 * 5 = 825 = -97 ( P )
(16) 182 * 4 = 728 - 165 * 9 = 1485 = - 757 ( P )
(17) 182 * 7 = 1274 - 165 * 3 = 495 = 779 ( = 19 * 41)
(18) 182 * 7 = 1274 - 165 * 5 = 825 = 449 ( P )
(19) 182 * 7 = 1274 - 165 * 9 = 1485 = - 211( P )
(20) 182 * 7 = 1274 - 165 *11 = 1815 = - 541(P)
(21) 182 * 8 = 1456 - 165 * 3 = 495 = 961( = 31 * 31)
(22) 182 * 8 = 1456 - 165 * 5 = 825 = 631 ( P )
(23) 182 * 8 = 1456 - 165 * 9 = 1485 = - 29 ( P)
(24) 182 * 8 = 1456 - 165 *11 = 1815 = - 359 (P)
(25) 182 * 13 = 2366 - 165 * 9 = 1485 = 881 (P)
(26) 182 * 13 = 2366 - 165 *11 = 1815 = 551( = 19 * 29)
(27) 182 * 13 = 2366 - 165 * 15 = 2475 = - 109 (P)
(28) 182 * 13 = 2366 - 165 * 17 = 2805 = - 439(P)
(29) 182 * 13 = 2366 - 165 * 19 = 3135 = - 769(P)
(30) 182 * 14 = 2548 - 165 *11 = 1815 = 733(P)
(31) 182 * 14 = 2548 - 165 * 15 = 2475 = - 73(P)
(32) 182 * 14 = 2548 - 165 * 17 = 2805 = - 257(P)
(33) 182 * 14 = 2548 - 165 * 19 = 3135 = - 587(P)
(34) 182 * 16 = 2912 - 165 * 15 = 2475 = 437(=19 * 23)
(35) 182 * 16 = 2912 - 165 *17 = 2805 = 107(P)
(36) 182 * 16 = 2912 - 165 * 19 = 3135 = - 223(P)
(37) 182 * 16 = 2912 - 165 * 23 = 3795 = - 883(P)
(38) 182 * 17 = 3094 - 165 * 15 = 2475 = 619(P)
(39) 182 * 17 = 3094 - 165 * 19 = 3135 = - 41(P)
(40) 182 * 17 = 3094 - 165 * 23 = 3795 = 701(P)
(41) 182 * 19 = 3458 - 165 * 15 = 2475 = 983 (P)
(42) 182 * 19 = 3458 - 165 *17 = 2805 = 653(P)
(43) 182 * 19 = 3458 - 165 * 23 = 3795 = - 337(P)
(44) 182 * 19 = 3458 - 165 * 25 = 4125 = - 667 ( = 23 * 29)
(45) 182 * 23 = 4186 - 165 * 25 = 4125 = 61 (P)
(46) 182 * 23 = 4186 - 165 * 27 = 4455 = - 269(P)
(47) 182 * 23 = 4186 - 165 *29 = 4785 = - 599(P)
(48) 182 * 26 = 4732 - 165 * 23 = 3795 = 937(P)
(49) 182 * 26 = 4732 - 165 * 25 = 4125 = 607(P)
(50) 182 * 26 = 4732 - 165 * 27 = 4455 = 277(P)
(51) 182 * 26 = 4732 - 165 * 29 = 4785 = - 53(P)
(52) 182 * 26 = 4732 - 165 * 31 = 5115 = - 383(P)
(53) 182 * 26 = 4732 - 165 * 33 = 3795 = 713 ( = 23 * 31 )
(54) 182 * 28 = 5096 - 165 * 25 = 4125 = 971 (P)
(55) 182 * 28 = 5096 - 165 * 27 = 4455 = 641 (P)
(56) 182 * 28 = 5096 - 165 * 29 = 4785 = 311 (P)
(57)182 * 28 = 5096 - 165 * 31 = 5115 = - 19 (P)
(58)182 * 28 = 5096 - 165 * 33 = 5445 = - 349 (P)
(59) 182 * 29 = 5278 - 165 * 27 = 4455 = 823 (P)
(60) 182 * 29 = 5278 - 165 * 31 = 5115 = 163 (P)
(61)182 * 29 = 5278 - 165 * 33 = 5445 = - 167 (P)
(62)182 * 29 = 5278 - 165 * 37 = 6105 = - 827 (P)
(63) 182 * 31 = 5642 - 165 * 29 = 4785 = 857 (P)
(64) 182 * 31 = 5642 - 165 * 33 = 5445 = 197 (P)
(65) 182 * 31 = 5642 - 165 * 37 = 6105 = - 463 (P)
(66) 182 * 32 = 5824 - 165 * 31 = 5115 = 709 (P)
(67) 182 * 32 = 5824 - 165 * 33 = 5445 = 379 (P)
(68) 182 * 32 = 5824 - 165 * 37 = 6105 = - 281 (P)
(69) 182 * 32 = 5824 - 165 * 41 = 6765 = - 941 (P)
(70)182 * 34= 6188 - 165 * 33 = 5445 = 743 (P)
(71)182 * 34= 6188 - 165 * 37 = 6105 = 83 (P)
(72)182 * 34= 6188 - 165 * 41 = 6765 = - 577 (P)
(73)182 * 34= 6188 - 165 * 43 = 7095 = - 907 (P)
(74) 182 * 37 = 6734 - 165 * 41 = 6765 = - 31 (P)
(75) 182 * 37 = 6734 - 165 * 43 = 7095 = - 361 ( = 19 * 19 )
(76) 182 * 37 = 6734 - 165 * 45 = 7425 = - 691 (P)
(77) 182 * 38 = 6916 - 165 * 37 = 6105 = 811 (P)
(78) 182 * 38 = 6916 - 165 * 41 = 6765 = 151 (P)
(79) 182 * 38 = 6916 - 165 * 43 = 7095 = - 179 (P)
(80) 182 * 38 = 6916 - 165 * 45 = 7425 = - 509 (P)
(81) 182 * 38 = 6916 - 165 * 47 = 7755 = - 839 (P)
(82) 182 * 41 = 7462 - 165 * 43 = 7095 = 367 (P)
(83) 182 * 41 = 7462 - 165 * 45 = 7425 = 37 (P)
(84) 182 * 41 = 7462 - 165 * 47 = 7755 = - 293 (P)
(85) 182 * 41 = 7462 - 165 * 51 = 8415 = - 953 (P)
(86) 182 * 43 = 7826 - 165 * 45 = 7425 = 401 (P)
(87) 182 * 43 = 7826 - 165 * 47 = 7755 = - 71 (P)
(88) 182 * 43 = 7826 - 165 * 51 = 8415 = - 589 ( = 19 * 31 )
(89) 182 * 43 = 7826 - 165 * 53 = 8745 = - 919 (P)
(90) 182 * 46 = 8372 - 165 * 45 = 7425 = 947 (P)
(91) 182 * 46 = 8372 - 165 * 47 = 7755 = 617 (P)
(92) 182 * 46 = 8372 - 165 * 51 = 8415 = -43 (P)
(93) 182 * 46 = 8372 - 165 * 53 = 8745 = - 373 (P)
(94) 182 * 46 = 8372 - 165 * 55 = 9075 = - 703 ( = 19 * 37 )
(95) 182 * 47 = 8554 - 165 * 51 = 8415 = 139 (P)
(96) 182 * 47 = 8554 - 165 * 53 = 8745 = - 191 (P)
(97) 182 * 47 = 8554 - 165 * 55 = 9075 = - 521 (P)
(98) 182 * 47 = 8554 - 165 * 57 = 9405 = - 851 ( = 23 * 37 )
(99) 182 * 49 = 8918 - 165 * 51 = 8415 = 503 (P)
(100)182 * 49 = 8918 - 165 * 53 = 8745 = 173 (P)
(101)182 * 49 = 8918 - 165 * 55 = 9075 = - 157(P)
(102)182 * 49 = 8918 - 165 * 57 = 9405 = - 487 (P)
(103)182 * 49 = 8918 - 165 * 59 = 9735 = - 817 ( = 19 * 43 )
( To be continued )
It should be better to list the real values of their difference
instead of the absolute values of their difference .
( X ) Summary :
If a , b and c are relatively prime , then
( 1 ) a ± b is not divisible by a or b .
( 2 incomplete groups )
( 2 ) ab ± 1 is not divisible by a or b .
( 1 complete group with 1 void group . )
( 3 ) ab ± a ± b is not divisible by a or b .
( 1 complete group with 2 incomplete groups . )
( 4 ) ab ± ac ± bc is not divisible by a , b or c .
( 3 incomplete groups . )
( 5 ) abc ± ab ± ac ± bc is not divisible by a , b or c .
( 1 complete group with 3 incomplete groups . )
Defects of this algorithm :
The limitation is usually not large enough compared with the n's so generated . If we want to generate bigger primes , the no. of basic primes needed should be more thus the limitation will become greater . But it is not easy to distribute the basic primes into , say 2 incomplete groups so that the n's generated will be smaller than the limitation .
In the following we shall demonstrate an example how to arrange the basic primes for 2 incomplete groups .
Let S = { 2 , 3 , 5 , 7 , 11 , 13 } be split into 2 groups A and B . Thus L1 = 17 ^2 = 289
and L2 = 17 * 19 = 323 .
Take A = { 2 , 7 , 13 } and B = { 3 , 5 , 11 } , thus π A = 182 and πB = 165
(1) n1= 182 - 165 = 17 (P)
(2) Add 2 to A i.e. A1 = { 2 , 2 , 7 , 13 } thus π A1 = 2 *182 = 364
and get n2 = π A1 - πB = 364 - 165 = 199 (P)
(3) Add 3 to B i.e. B1 = { 9 , 5 , 11 } thus πB1 = 3 *165 = 495
and get n3 = πB1 - π A = 495 - 182 = 313 (P)
(4) Get n4 = πB1 - π A1 = 495 - 364 = 131 (P)
(5) Add 4 to A and 5 to B , thus π A2 = 4 * 182 = 728 while πB2 = 5 * 165 = 825
and get n5 = 825 - 728 = 97 (P)
(6) Get n6 = 4 * 182 - 3 * 165 = 728 - 495 = 233 (P)
(7) Add 7 to A and 9 to B , thus 7 * π A = 7 * 182 = 1274 while 9 * 165 = 1485
and get n7 = 1485 - 1274 = 211 (P)
(8) Add 8 to A and 9 to B , thus 8 * 182 = 1456 , and get n8 = 1485 - 1456 = 29 (P)
(9) Add 13 to A and 15 to B , thus 13 * 182 = 2366 while 15 * 165 = 2475
and get n9 = 2475 - 2366 = 109 (P)
(10) Add 14 to A and 15 to B , thus 14 * 182 = 2548 , and get n10 = 2548 - 2475 = 73 (P)
(11) Add 14 to A and 17 to B , ( notice that L1 will become 19 ^2 = 361 and L2 will become
19 * 23 = 437 ) thus 17 * 165 = 2805 , and get n11 = 2805 - 2548 = 257 (P)
(12) Add 14 to A and 19 to B , thus 14 * 182 = 2548 while 19 * 165 = 3135 , and get
n12 = 3135 - 2548 = 587 . By actual division we find that 587 is not divisible by 17 .
Thus L1 becomes 23 ^2 = 529 and L2 becomes 23 * 29 = 667 . Therefore 529 is a prime .
(13) Add 16 to A and 17 to B , thus 16 *182 = 2912 and 17 * 165 = 2805 , and get
n13 = 2912 - 2805 = 107 (P)
(14) Add 16 to A and 19 to B , thus 16 * 182 = 2912 and 19 * 165 = 3135 , and get
n14 = 3135 - 2912 = 223 (P)
(15) Add 17 to A and 19 to B , ( then L1 becomes 23 ^2 = 529 and L2 becomes 23 * 29 = 667 )
thus 17 * 182 = 3094 while 19 * 165 = 3135 , and get n15 = 3135 - 3094 = 41 (P)
(16) Add 19 to A and 23 to B , thus 19 * 182 = 3458 and 23 * 165 = 3795
and get n16 = 3795 -3458 = 337 . Since 337 is not divisible by 17 , therefore 337 is a prime .
(17) Add 23 to A and 25 to B , thus 23 * 182 = 4186 while 25 * 165 = 4125
and get n17 = 4186 - 4125 = 61 (P)
(18) Add 26 to A and 27 to B , thus 26 * 182 = 4732 and 27 * 165 = 4455 ,
and get n18 = 4732 - 4455 = 277 (P)
(19) Add 28 to A and 29 to B , thus 28 * 182 = 5096 while 29 * 165 = 4785 ,
and get n19 = 5096 - 4785 = 311 (P)
(20) Add 28 to A and 31 to B , thus 28 * 182 = 5096 while 31 * 165 = 5115 ,
and get n 20 = 5115 - 5096 = 19 (P)
(21) Add 29 to A and 31 to B , thus 29 * 182 = 5278 while 31 * 165 = 5115 ,
and get n21 = 5278 - 5115 = 163 (P)
(22) Add 31 to A and 33 to B , thus 31 * 182 = 5642 while 33 * 165 = 5445 ,
and get n22 = 5642 - 5445 = 197 (P)
Summary in Table
A={2 ,7 ,13} , π A = 182 ; B={3 ,5 ,11} ,πB = 165 ; difference ; L1 = 289 ; L2 = 323
(1) 182 * 1 = 182 165 * 1 = 165 17
(2) 182 * 1 = 182 165 * 3 = 495 313
(3) 182 * 2 = 364 165 * 1 = 165 199
(4) 182 * 2 = 364 165 * 3 = 495 131
(5) 182 * 4 = 728 165 * 5 = 825 97
(6) 182 * 4 = 728 165 * 3 = 495 233
(7) 182 * 7 = 1274 165 * 9 = 1485 211
(8) 182 * 8 = 1456 165 * 9 = 1485 29
(9) 182 * 13 = 2366 165 * 15 = 2475 109
(10) 182 * 14 = 2548 165 * 15 = 2475 73
(11) 182 * 14 = 2548 165 * 17 = 2805 257
(12) 182 * 14 = 2548 165 * 19 = 3135 587
(13) 182 * 16 = 2912 165 * 17 = 2805 107
(14) 182 * 16 = 2912 165 * 19 = 3135 223
(15) 182 * 17 = 3094 165 * 19 = 3135 41
(16) 182 * 19 = 3458 165 * 23 = 3795 337
(17) 182 * 23 = 4186 165 * 25 = 4125 61
(18) 182 * 26 = 4732 165 * 27 = 4455 277
(19) 182 * 28 = 5096 165 * 29 = 4785 311
(20) 182 * 28 = 5096 165 * 31 = 5115 19
(21) 182 * 29 = 5278 165 * 31 = 5115 163
(22) 182 * 31 = 564 165 * 33 = 5445 197
(23) 182 * 32 = 582 165 * 33 = 5445 379
(24) 182 * 34= 6188 165 * 33 = 5445 743
(25) 182 * 52 = 9464 165 * 55 = 9075 389
(26) 182 * 56 = 10192 165 * 57 = 9405 787
(27) 182 * 101 = 18382 165 * 109 = 17985 397
(28) 182 * 97 = 17654 165 * 103 = 16995 659
(iii) Set of primes without " 2 " and " 5 " for 3 incomplete groups
If the set S used to generate primes does not contain 2 and 5
as its elements while 3 incomplete groups are involved . Then the
combined sum or difference of them is not divisible by 2 but may
be divisible by 5 .
Example
Let S = { 3 , 7 , 11 , 13 } , then L1 = 17 ^2 = 289 ; L2 = 17 * 19 = 323 ;
and L3 = 23 ^2 = 529 .
Take A = {[ 3 , 7 ] , 11 , / } , thus π A = 231
Take B = {[ 3 , 7 ] , / , 13 } , thus π B = 273
Take C = { / , 11 , 13 } , thus π C = 143
n1 = 231 + 273 + 143 = 647 ( √ 647 = 25.4 ) then we have to check whether 647 is divisible by 17 , 19 or 23 . As 647 - 323 = 324 , thus 647 = 2 * 17 * 19 + 1
so 647 is not divisible by 17 nor 19 . Also 647 - 529 = 118 = 2 * 59
so 647 is not divisible by 23 , therefore 647 is a prime .
n2 = 231 + 273 - 143 = 361 = 323 + 38 = 323 + 2 * 19 , thus 361 is divisible
by 19 .
n3 = 231 - 273 + 143 = 101 ( p )
n4 = -231 + 273 + 143 = 185 while 185 / 5 = 37 ( p )
(ii) Set of primes without " 2 " for 3 incomplete groups
If the set S used to generate primes does not contained 2 as its element
with 3 ( or any odd nos. > 3 ) incomplete groups are involved . Then the
product of each group will all be odd , thus the combined sum or difference
of them will also be odd , i.e. , not divisible by 2 .
Example
Let S = { 3 , 5 , 7 , 11 } , thus L1 = 13^2 = 169 , L2 = 13*17 = 221 and L3 = 19 ^2 = 361
Take A = { [ 3 , 5] , 7 , / } , B = { [ 3 , 5] , / , 11 } and C = { / , 7 , 11 }
Thus π A = 105 , π B = 165 , π C = 77 ,
n1 = 105 + 165 + 77 = 347 ( p ) since 347 - 221 = 126 = 2 * 9 * 7
n2 = 105 + 165 - 77 = 193 ( p )
n3 = 105 - 165 + 77 = 17 ( p )
n4 = - 105 + 165 + 77 = 137 ( p )
If complete groups are joined to participate in addition or subtraction with the incomplete groups , more n's can be generated . We may add in an even no. of complete groups , or choose 1 complete group with an additional element of 2 .
For example , let T = { 2 , 3 , 5 , 7 , 11 } , thus π T = 2310
Take A2 = { 9 , 25 , 7 } , thus π A2 = 1575 while B and C remain unchanged .
n5 = 2310 - 1575 - 165 - 77 = 493 , while 493 - 221 = 272 = 4 * 68 = 4 * 4 * 17 , thus 493 contains 17 as a factor .
n6 = 2310 - 1575 - 165 + 77 = 647 ( √ 647 = 25.4 ) , thus we have to check whether 647 is divisible by 13 , 17 , 19 or 23 . As 647 - 13 * 17 = 426 = 6 * 71 , thus 647 is not divisible by 13 nor 17 .
Also 19 * 23 = 437 while 647 - 437 = 210 ( does not contain 19 nor 23 as factor ) , therefore 647 is a prime .
( IX ) Miscellaneous examples
( i ) Set of primes used without " 5 "
If the set S used to generate primes does not contained 5 as its element , then the n's generated may be divisible by 5 . If this happens it can be easily inspected for the n's so generated will end in 5 . We just exclude all factors of 5 by dividing n by 5 and get the quotient .
Example
Let S = { 2 , 3 , 7 , 11 } thus L1 = 13 ^2 = 169 ; L2 = 13 * 17 = 221
Take A1 = { 2 , 11 } and B1 = { 3 , 7 }
Therefore π A1 = 22 , πB1 = 21
n1 = 22 + 21 = 43 ( p ) ,
n2 = 22 - 21 = 1
Take A2 = { 4 , 11 } , thus π A2 = 44 ,
n3 = π A2 + πB1 = 44 + 21 = 65 while 65 / 5 = 13 ( p )
n4 = π A2 - πB1 = 44 - 21 = 23 ( p )
Take B2 = { 9 , 7 } , thus π B2 = 63 ,
n5 = π A1 + π B2 = 22 + 63 = 85 while 85 / 5 = 17 ( p )
n6 = π A1 - π B2 = 22 - 63 = - 41 ( p )
n7 = π A2 + πB2 = 44 + 63 = 107 ( p )
n8 = π A2 - πB2 = 44 - 63 = - 19 ( p )
( VIII ) Primes stored in compartments
In section ( VII ) it can be seen that there is no restriction for the no. of
complete groups . For incomplete groups it seems that the total no. should be
r . However , if originally the primes of S are stored in so-called compartments
inside S where certain compartments contain > 1 distinct primes ( with various exponents ) ,
then the total no. of incomplete groups can be reduced .
For example , let S = { 2 , 3 , 5 , 7 , 11 , 13 , 17 } ----- totally 7 distinct primes . Thus L1 = 19^2 = 361 ,L2 = 19 * 23 = 437 and L3 = 29^2 = 841 .
If the primes are stored in say 3 compartments like :
{ [ 2,3,5] , [ 7,11] , [ 13,17 ] } ;
then we may take G1 = { [ 2^2 ,3,5,] , [ 7,11] , / } = { 60 , 77 , 1 } ,
thus π G1 = 4620
G2 = { [ 2^2,3,5,] , / , [ 13,17 ]} = { 60 , 1 , 221} ,
thus π G2 = 13260
G3 = { / , [7,11] , [ 13,17 ]} = { 1 , 77 , 221 } ,
thus π G3 = 17017
We may get n1 = π G1 + π G1 - π G3
= 4620 + 13260 - 17017
= 863
Since √ 863 = 29.3 thus for 863 to be composite , it must contain prime factor ≦ 29 .
As 863 - 841 ( L 3 ) = 22 = 2 * 11 ⇒ 863 = 29 * 29 + 2 * 11 , thus 863 is not divisible by 29 ;
Also 863 - 437 ( L2 ) = 426 = 2 * 213 = 2 * 3 * 71 ⇒ 863 = 19 * 23 + 2*3*71 , thus 863 is not divisible by 19 nor 23 , since 863 is already not divisible by any prime ≦ 17 , therefore 863 is a prime .
( VII ) Incomplete groups and complete groups
In section ( VI ) we have generated primes with r groups each containing r-1 primes
taken from the set S . These groups will be referred as " incomplete groups " .
In fact other groups which contain all the r primes ( referred as " complete groups ") may also participate to generate primes with those incomplete groups . The result after combined sum or difference of them is still not divisible by every prime contained in S after the joining of complete groups .
Example
As in the table shown in section ( VI ) , if the set S being itself considered as a
complete group , with π S = 2*3*5 = 30 , join the operations of addition or
subtraction with the 3 incomplete groups G1 , G2 and G3 ,
then we may get n1 = 30 + 6 + 10 + 15 = 61 ( p ) since 61 < 77 ( L2 ) .
Also n2 = 30 + 6 + 10 - 15 = 31 ( p )
n3 = 30 + 6 + ( -10 ) + 15 = 41 ( p )
n4 = 30 +( -6 ) + 10 + 15 = 49 ( = L1 )
n5 = - 30 + 6 + 10 + 15 = 1
n6 = 30 + 6 - 10 - 15 = 11 (p)
n7 = 30 - 6 + 10 - 15 = 19 ( p )
n8 = 30 - 6 - 10 + 15 = 29 ( p )
( VI ) To generate primes with more than 2 groups
Instead of partition S , a set with r primes ( r ≧ 3 ) into 2 groups only ,
we may form r groups ( denoted by G ) of primes from S , such that each
group contains exactly r -1 primes taken from S , i.e. , each prime will not exist
in one and only one group .
Let π G denotes the product of primes ( each with exponents ≧ 1 )
contained in each G . If we take the combined sum or difference of every π G ,
the value n obtained will not be divisible by any prime contained in S .
If n < L , the limitation then n will be prime or = 1 .
Examples
( I ) Let S = { 2 , 3 , 5 } , thus L1 = 49 and L2 = 77 .
( 1 ) Take G1 = { 2 , 3 } , G2 = { 2 , 5 } and G3 = { 3 , 5 } ,
Thus G1 lacks the prime 5 , G2 lacks 3 and G3 lacks 2
as shown in the following table .
S | 2 , 3 , 5 |
G1| 2 , 3 , / | ----- π G1 = 6
G2| 2 , / , 5 | ----- π G2 = 10
G3| / , 3 , 5 | ----- π G3 = 15
Then n1 = 6 + 10 + 15 = 31 ( p )
n2 = 6 + 10 - 15 = 1
n3 = 6 - 10 + 15 = 11 ( p )
and n4 = -6 + 10 + 15 = 19 (p)
It seems more primes can be generated in this way .
( V ) The ultimate limitation L3
After the extension of limitation from L1 to L2 , the chance of the number generated
to be a prime becomes greater . Now we try to extend the limitation a bit further .
If a number n generated > L2 , i.e. [ P(n+1) * P(n+2) ] , but < [P ( n + 3 )] ^2 ( denoted by L3 ) , then n will be prime if n is not divisible by both P (n+1) and P (n+2) .
To find whether n is divisible by P (n+1) and P (n+2) or not , we can divide n by
P (n+1) and P( n+2) one by one , or once with the following method .
Let k = n - P(n+1) * P(n+2) ⇒ n = P(n+1) * P(n+2) + k , then factorize k .
If k does not contain either P(n+1) or P(n+2) as factor , then n also ,
thus n is a prime . Otherwise n will be composite .
Examples
Let S = { 2 , 3 , 5 , 7 } and divided into S and { } .
Then L1 = 11 ^2 = 121 ; L2 = 11 * 13 = 143 and L3 = 17 ^2 = 289 ,
πS = 210 and π { } = 1 .
Then n1 = 210 +1 = 211 , 211 > 143 but < 289 thus we test whether 211 is divisible
by 11 or 13 .
Since 211 - 143 = 68 ⇒ 211 = 143 + 68
= 11 * 13 + 4 * 17
Thus 68 then 211 is not factorable by 11 or 13 ( also not by 17 ) ,
therefore 211 is a prime .
For n2 = 210 - 1 = 209 , also > 143 but < 289 .
Since 209 - 143 = 66 ⇒ 209 = 143 + 66
= 11 * 13 + 6 * 11 ,
therefore 209 contains 11 as a factor , thus is a composite .
L3 or P ( n + 3 ) ^2 , will be referred as the ultimate limitation .
( IV ) To generate twin primes
If the difference of 2 primes = 2 , then they are called " twin primes " .
We may use the algorithm under certain arrangement to generate twin
primes .
( i ) For a set S containing primes , if we divide it into A , being
S itself and B , being a void set , then π B will be 1 . Thus
n1 = π A + 1 and n2 = π A - 1 , i.e. n1 - n2 = 2 .
If both n1 and n2 are primes , then they are twin primes .
Examples
( 1 ) Let S = { 2 , 3 , 5 } , thus limitation = 7 ^ 2 = 49 and L2 = 77
Then π S = 30 and π { } = 1 ,
n1 = 30 + 1 = 31 and n2 = 30 - 1 = 29 ,
Since both 31 and 29 < 49 ,
Thus they are twin primes .
( 2 ) Let S = { 2^2 , 3 , 5 } ,
Then π S = 60 and π { } = 1 ,
n1 = 60 + 1 = 61 and n2 = 60 - 1 = 59 ,
Though both 61 and 59 > 49 , but since < 77 ,
Therefore they are also twin primes .
( ii ) For 2 sets S1 and S2 such that S1 contains 2 and
other primes as elements while S2 contains 2^2 = 4 and the
same rest primes as elements .
Example
Let S1 = { 2 , 3 , 5 , 7 } and divided into A = { 2 } and
B = { 3 , 5 , 7 }
while S2 = { 4 , 3 , 5 , 7 } and divided into A = { 4 } and
B = { 3 , 5 , 7 } .
Then Limitation = 11^2 = 121 for both cases .
For S1 , πA = 2 and πB = 105 , thus n1 = 105 + 2 = 107
and n2 = 105 - 2 = 103 .
While for S2 , πA = 4 and πB = 105 , thus n1 = 105 + 4 = 109
and n2 = 105 - 4 = 101 .
The difference of both n1 = 2 , and the difference of both n2
also = 2 . Since all the 4 n's < 121 , thus they are all primes .
So 2 pairs of twin primes have been generated .
( III ) Extension of limitation from [ P(n+1)] ^2 to P(n+1) * P(n+2)
When the no. of primes contained in S increases , the
values of π A and π B also increase . The values of n1 and n2 ,
their sum and difference , will frequently be greater then [ P(n+1)] ^2 ,
the limitation . ( Though the limitation will also increase . )
To make the algorithm more usable , we extend the limitation from
[ P(n+1)] ^2 to P(n+1) * P(n+2) , where P(n+2) is the prime no.
just greater than P(n+1) . Since the composite no . just greater than
[ P(n+1)] ^2 and not divisible by primes ≦ Pn is P(n+1) * P(n+2) .
If n1 or n2 so generated , though > P(n+1)] ^2 ,
but < P(n+1) * P(n+2) , then it is still a prime .
Example
Let S = { 2^2, 3^3 , 5^2 } , then limitation L1 = 7^2 = 49 ,
extended limitation L2 = 7 * 11 = 77 .
(1) Take A = { 2^2 , 5^2 } and B = ( 3^3 }
thus πA = 100 and π B = 27 .
Then n2 = 100 - 27 = 73 . Since 73 > 49 but < 77 , therefore
73 is a prime .
(2) Take A = { 2^2 , 3^3 } and B = { 5^2 } , thus
πA = 4 * 27 = 108 and π B = 25 .
Then n2 = 108 - 25 = 83 > 77 . We are not sure whether
83 is a prime or not .
By extending the limitation we have more chance to generate
more primes .
( II ) Primes with exponents > 1
In fact the primes contained in S are not necessarily distinct .
Each prime may repeat any times , but when divided into 2 groups
every same prime should be put into the same group . Thus the products
of elements of the 2 groups are still relatively prime .
If there are n no. of a certain prime Pr in a group , they may be
denoted concisely as ( Pr ) ^ n .
Example
Let S = { 2 , 3 ,3 ,3 , 5 , 5 } , limitation is still 49 .
(1) Take A = { 2 , 3 ,3 , 3} and B = { 5 , 5} ,
then A may also be denoted by { 2 , 3 ^ 3 } , while B by { 5 ^ 2 }
thus πA = 2 * 27 = 54 and π B = 25 .
Then n1 = 54 + 25 = 79 . Since 79 > 49 , we are not sure
whether it is a prime or not .
n2 = 54 - 25 = 29 ( p )
(2) Take A = { 2 , 5^2 } and B = ( 3 ^3 }
thus πA = 50 and π B = 27
Then n1 = 50 + 27 = 77 ( > 49 )
n2 = 50 - 27 = 23 ( p )
When the exponents of various primes may take values > 1 ,
we can generate more primes with the same set of distinct
primes .
Hi bobbym ,
No , it is not the case .
Hi bobbym ,
I need help to improve this algorithm latter !
( I ) Principle
Certain years ago I had also developed an algorithm to generate
primes from existing ones . The principle is not complicated .
Let S denotes a set of distinct consecutive primes starting
from 2 , 3 and onwards up to a certain prime Pn . If we
divide S into 2 parts A and B in any way , and let π A
denotes the product of primes contained in A ,
while π B denotes the product of primes contained in
B . Thus π A and π B are relatively prime . ( If one of
the 2 subsets of S , say A is a void set , then π A will be
defined to be 1 . )
Let n1 denotes π A + π B and n2 denotes | π A - π B | .
It is clear that both n1 and n2 are not divisable by any
of the primes ≦ Pn . n1 or n2 may be prime , composit ( with all
prime factors > Pn ) , or = 1 . ( Only for n2 .)
Let P (n+1) denotes the prime no. just > Pn , if n1 or n2 < P (n+1) ^ 2 ,
then n1 or n2 must be prime since the smallest composit no. with prime
factors all > Pn will be [ P (n+1)] ^ 2 . The value of P (n+1) ^ 2 will be
referred as the limitation .
Examples :
( I ) Let S = { 2 , 3 , 5 } , thus limitation = 7 ^ 2 = 49 .
(1) Take A = { 2 , 3 } and B = { 5 } , thus πA = 6 and π B = 5 .
Then n1 = 6 + 5 = 11 ( prime since 11 < 49 )
n2 = 6 - 5 = 1
(2) Take A = { 2 , 5 } and B = { 3 } , thus π A = 10 .
Then n1 = 10 + 3 = 13 ( prime )
n2 = 10 - 3 = 7 ( p )
(3) Take A = { 2 } and B = { 3 , 5 } , thus π B = 15 .
Then n1 = 2 + 15 = 17 ( p )
n2 = 15 - 2 = 13 ( p )
The prime no. so generated will not necessarily be distinct , as we have
13 = 10 + 3 = 15 - 2 .
It may be ambiguous to define what is half - sized for a
matrix ( no matter a square matrix or a triangular semi-matrix )
with n dots at each side . ( Assuming that the distance between
the dots are fixed and the dimension of the dots may be neglected . )
For n to be even , readily half - sized means the one with n / 2 dots .
But for n to be odd , quite probably we shall mean a sub - matrix
with [ n/2 ] + 1 = ( n + 1 ) / 2 dots to be half - sized so that its length
between the end points will be 1/2 of that of the big one .
For example for a matrix with 3 dots at each side , a half - sized
sub - matrix will be referred to one with 2 dots at each side .
Thanks thickhead ,
Now it seems if the no. of points of the big triangle with the 2 half - sized
small triangles trend to ∞ , then the probability will trend to 1/10 .
Hi thickhead ,
Have you omitted a " 25 " at the 5th row , a " 25" at the
6th row and a " 16 " at the 7th row in the table at # 41 ?
Also for the probability should the denominator be 55*21*21 ?