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Original name: Values of the difference d for 11 primes in arithmetic progression with the minimal start sequence {11 + j*d}, j = 0 to 10.

The computations were done without any assumptions on the form of d. 21st term is greater than 10^12.

All terms are multiples of 210=2*3*5*7. - Zak Seidov, May 16 2015

Equivalently, integers d such that the longest possible arithmetic progression (AP) of primes with common difference d has exactly 11 elements (see example). These 11 elements are not necessarily consecutive primes. In fact, here, for each term d, there exists only one such AP of primes, and this one always starts with A342309(d) = 11, so this unique AP is (11, 11+d, 11+2d, 11+3d, 11+4d, 11+5d, 11+6d, 11+7d, 11+8d, 11+9d, 11+10d). - Bernard Schott, Mar 08 2023

Conjectured (Polignac 1849) to be union of even numbers and the odd primes minus 2.

For n > 2: A092953(a(n)) > 0. - Reinhard Zumkeller, Nov 10 2012

The computations were done without any assumptions on the form of d.

All terms are multiples of 6. - Zak Seidov, Jan 07 2014

Equivalently, integers d such that the largest possible arithmetic progression (AP) of primes with common difference d has exactly 5 elements (see example). These 5 elements are not necessarily consecutive primes. In fact, for each term d, there exists only one such AP of primes, and this one always starts with A342309(d) = 5, so this unique AP is (5, 5+d, 5+2d, 5+3d, 5+4d). - Bernard Schott, Jan 25 2023

The computations were done without any assumptions on the form of d.

All terms are multiples of 30. - Zak Seidov, Jan 07 2014.

Equivalently, integers d such that the longest possible arithmetic progression (AP) of primes with common difference d has exactly 7 elements (see example). These 7 elements are not necessarily consecutive primes. In fact, for each term d, there exists only one such AP of primes, and this one always starts with A342309(d) = 7, so this unique AP is (7, 7+d, 7+2d, 7+3d, 7+4d, 7+5d, 7+6d). - Bernard Schott, Feb 12 2023

The row polynomials P(n,x) := Sum_{k=0..p(n)-1} a(n,k)*x^k, n >= 1, appear in the numerator of the g.f. G(p(n),x) for the numbers N(p(n),m) of inequivalent m-bead necklaces of two colors with p(n) beads of one color and m-p(n) beads of the other color. Here p(n)=A000040(n) (prime numbers). Equivalently, N(p(n),m) counts inequivalent necklaces with p(n) beads which are labeled with nonnegative numbers, such that the sum of the labels is m. For a proof of this equivalent formulation see a comment in A032191. Inequivalence is meant with respect to the cyclic group C_p(n).

This necklace g.f. is G(p(n),x) = P(n,x)/((1-x^p(n))*(1-x)^(p(n)-1)), n >= 1. The row polynomials P(n,x) are defined above. This g.f. is Z(C_p(n),x), the two variable (x[1] and x[p(n)]) cycle index polynomial for the cyclic group of prime order p(n), with substitution x[1]->1/(1-x^1)and x[p(n)]->1/(1-x^p(n)). This follows by Polya enumeration if the above mentioned labeled necklace problem is solved.

The row length sequence for this array a(n,k) is A000040(n) (n-th prime number), [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...].

The rows of this signed array are symmetric: a(n,k) = a(n,p(n)-1-k), n >= 2, k = 0..(p(n)-1)/2. See the explicit formula below.

The formulas for a(n,k), given below, produces in fact integers.

G.f. for column k, k>=0 (without leading zeros): sum(A103718(k,m)*p(n)^m,m=0..k)/k! produces for all n> pi(n) integers, where pi(n):=A000720(n), primes not exceeeding n.

As '2 is prime' and also '2 is one less than prime 3' (see A173919), there exist two subsequences with k = 2 elements in these APs of primes (see examples).

1. If d is an odd term, then d is in A040976 \ {0} with d = prime(m) - 2, for some m >= 2, and, for each such d, there exists only one longest possible AP of primes, and this AP is always: (2, prime(m)) = (2, d+2), so starts with 2. This subsequence corresponds to the first case: '2 is prime'.

2. If d is an even term, then d is in A360735 and the longest corresponding APs of primes are of the form (q, q+d) with q odd primes. This subsequence corresponds to the second case '2 is one less than prime 3'.

A342309(d) gives the first element of the smallest AP with 2 elements whose common difference is a(n) = d.

The two elements of these APs are not necessarily consecutive primes.

The computations were done without any assumptions on the form of d.

The computations were done without any assumptions on the form of d. 181st term is greater than 10^12.

Arises in Hardy-Littlewood k-tuple conjecture. Also a(n) is the exact number of d=2 and also d=4 differences in dRRS[modulus=n-th primorial]; see A049296 (dRRS[m]=set of first differences of reduced residue system modulo m).

For n>1 this is the determinant of the (n-1) X (n-1) matrix whose diagonal is A006093(n) = {1, 2, 4, 6, 10, 12, 16, 18..} = the first primes minus 1 and all other elements are 1's. The determinant begins: / (2-1) 1 1 1 1 1 1 ... / 1 (3-1) 1 1 1 1 1 ... / 1 1 (5-1) 1 1 1 1 ... / 1 1 1 (7-1) 1 1 1 ... / 1 1 1 1 (11-1) 1 1 ... / 1 1 1 1 1 (13-1) 1 ... - Alexander Adamchuk, May 21 2006

From Gary W. Adamson, Apr 21 2009: (Start)

Equals (-1)^n * (1, 1, 1, 3, 15, ...) dot (1, -2, 4, -6, 10, ...).

a(6) = 135 = (1, 1, 1, 3, 15) dot (1, -2, 4, -6, 10) = (1, -2, 4, -18, 150). (End)

Indices of the triangular numbers in A147846.

Integers k such that k or k+1 is prime. - Giovanni Teofilatto, Mar 05 2010

For a given common difference d, there always exists a longest possible arithmetic progression (AP) of primes, and the number of elements k in this AP of primes is necessarily a term of this sequence. See A123556 for explanations. - Bernard Schott, Mar 18 2023