

A283814


Irregular triangle read by rows in which nth row lists the numbers m such that 2*prime(m) can be represented as the sum of two primes in exactly n ways.


1



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, 30, 34, 38, 46, 20, 28, 42, 26, 31, 32, 36, 37, 40, 50, 41, 43, 58, 33, 35, 39, 45, 47, 52, 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
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OFFSET

1,2


COMMENTS

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.
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}.


LINKS

Table of n, a(n) for n=1..80.
Zak Seidov, First 100 rows of the triangle.


EXAMPLE

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:
n=3: 2*prime(5) = 2*11 = 22 = 3 + 19 = 5 + 17 = 11 + 11,
2*prime(6) = 2*13 = 26 = 3 + 23 = 7 + 19 = 13 + 13,
2*prime(11) = 2*31 = 62 = 3 + 59 = 7 + 19 = 19 + 43 = 31 + 31.


MATHEMATICA

A116619=Table[Count[PrimeQ[2*Prime[n]Prime[Range[n]]], True], {n, 1000}];
Flatten[Position[A116619, #]& /@ Range[100]]


CROSSREFS

Cf. A116619 (number of ways of representing 2*prime(n) as the sum of two primes).
Sequence in context: A175060 A138773 A132989 * A114881 A246273 A082319
Adjacent sequences: A283811 A283812 A283813 * A283815 A283816 A283817


KEYWORD

nonn,tabf


AUTHOR

Zak Seidov, Mar 17 2017


STATUS

approved



