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%I #13 Mar 20 2017 03:39:49
%S 1,2,3,4,8,5,6,11,7,9,10,18,12,13,14,15,22,16,17,19,21,23,24,25,27,29,
%T 30,34,38,46,20,28,42,26,31,32,36,37,40,50,41,43,58,33,35,39,45,47,52,
%U 53,59,44,48,49,65,51,61,62,55,57,60,66,67,70,85,54,56,63,68,72,73,75,77,79,64,76,78,80,81,83
%N Irregular triangle read by rows in which n-th row lists the numbers m such that 2*prime(m) can be represented as the sum of two primes in exactly n ways.
%C From b116619.txt it seems that the sequence is correct at least for first 677 terms (first 100 rows of triangle). But as it is usual in number theory better consider this sequence as conjectured.
%C Lengths of first 100 rows of triangle (see a283814.txt): {2,3,3,4,5,5,8,3,7,3,8,4,3,7,8,1,10,7,6,9,3,7,6,3,4,7,13,4,6,7,7,9,7,8,8,3,8,8,5,5,5,11,5,10,3,6,8,10,5,8,5,9,6,9,6,7,10,6,6,6,8,5,7,12,11,6,8,6,9,4,12,6,8,5,5,5,11,10,13,7,7,10,9,7,4,9,7,5,4,8,7,6,10,7,6,10,6,10,6,6}.
%H Zak Seidov, <a href="/A283814/a283814.txt">First 100 rows of the triangle.</a>
%e 3rd row is {5,6,11} because only the 5th, 6th and 11th primes can be represented as the sum of 2 primes in exactly 3 ways:
%e n=3: 2*prime(5) = 2*11 = 22 = 3 + 19 = 5 + 17 = 11 + 11,
%e 2*prime(6) = 2*13 = 26 = 3 + 23 = 7 + 19 = 13 + 13,
%e 2*prime(11) = 2*31 = 62 = 3 + 59 = 7 + 19 = 19 + 43 = 31 + 31.
%t A116619=Table[Count[PrimeQ[2*Prime[n]-Prime[Range[n]]],True],{n,1000}];
%t Flatten[Position[A116619,#]& /@ Range[100]]
%Y Cf. A116619 (number of ways of representing 2*prime(n) as the sum of two primes).
%K nonn,tabf
%O 1,2
%A _Zak Seidov_, Mar 17 2017
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