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Length of n-th row of A124064.

The corresponding smallest term of the first such longest possible arithmetic progression of primes with common difference n is A342309(n). - Bernard Schott, Oct 12 2021

From Bernard Schott, Feb 24 2023: (Start)

For every positive integer n, there exists a smallest prime p that does not divide n = A053669(n); then, an AP of k primes with common difference n cannot contain more terms than this value of p so k <= p, moreover, the longest possible APs of primes have p-1 or p elements.

Proof: consider the AP of p elements (q, q+n, q+2*n, q+3*n, ..., q+(p-1)*n) with common difference n, q prime and p is the smallest prime that does not divide n; the modular arithmetic modulo p gives this set of remainders with p elements: {0, 1, 2, ..., p-1}, so there is always a multiple of p in each such AP with p terms, hence length k of longest possible AP of primes is >= p-1 and <= p.

Moreover, when the longest possible AP contains k = p elements, then this unique longest AP must start with p (corresponding to remainder = 0) and the common difference n is a multiple of A151799(p)# and not of p#, where # = primorial = A002110.

Now, always with a common difference n, when the longest possible AP contains k = p-1 elements, these longest APs with p-1 primes can start with p or with another prime q != p, and there are infinitely many such longest APs with p-1 terms (see Properties in Wikipedia link) in this case. When this AP starts with p, the set of remainders is {0, 1, ..., p-2} and when this AP starts with q, then the set of remainders becomes {1, 2, ..., p-1}.

Terms are ordered without repetition in A173919. (End)

Inspired by problem A1880 in Diophante (see link).

a(n) is the last term of the n-th row of A124064.

About the k-tuples conjecture in A123556: let p = A053669(n), if the arithmetic progression of p elements starting at p with difference n consists of primes, then A123556(n) = p, otherwise A123556(n) = p-1. The first 21 terms of this sequence are precisely the same as A053669, then a(22) = 7 (corresponding to the arithmetic progression (7,29)) while A053669(22) = 3.

Comment from Hugo van der Sanden Aug 14 2021: (Start)

Row d=12 starts 4 9 9 10 10 469 3937 7343 7343 44719 78937 78937 78937 78937 55952333 233761133 597191343199.

Row d=18 starts 4 4 15 15 15 695 695 1727 7711 13951 13951 46159 400847 400847 400847 65737811 13388955301 934046384293.

Row d=24 starts 4 9 9 10 10 793 4819 6415 7271 14069 14069 14069 31589 67344271 616851797 48299373047 48299373047 20302675273219.

Row d=30 starts 4 4 9 25 25 2779 2779 6347 6347 6347 10811 10811 87109 87109 87109 1513723 15009191 15009191 316612697 316612697 1275591688621.

Row d=36 starts 4 10 10 10 15 1333 3161 4997 6865 34885 142171 834863 1327447 35528747 720945097 63389173477 63389173477 16074207679897 41728758250241.

Row d=42 starts 4 4 9 35 35 2701 2987 2987 7729 26995 26995 185795 307553 708385 708385 708385 1090198367 1819546069 20263042201 5672249016001.

Later terms in these rows are always >10^14. (End)

If p is the least prime that does not divide d, then T(d,k) <= p^2 if k >= p^2 (i.e. any a.p. of length >= p^2 with difference d contains a term divisible by p^2, and the only semiprime divisible by p^2 is p^2). Thus every row is eventually 0. - Robert Israel, Aug 11 2024

This sequence is to A124570 as A123556 is to A124064.

a(n) is at most A053669(n)^2, with equality if and only if A053669(n)^2 is the first semiprime in the corresponding arithmetic progression. - Charlie Neder, Jan 10 2019

By subsampling a given arithmetic progression with k terms and distance d one may generate an arithmetic progression of a larger distance d*b, b=1,2,3...., with 1+(k-1)/b terms: a(b*n) >= 1+floor( (a(n)-1)/b ), b=1,2,3..... - R. J. Mathar, Aug 02 2021

18 <= a(6) <= 24. - Jinyuan Wang, Aug 06 2021

a(12) >= 17. a(18) >= 18. a(24) >= 18. a(30) >= 21. a(36) >= 19. a(42) >= 20. a(6006) >= 24 (starting 652744562555081). Hugo van der Sanden, Aug 14 2021