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Triangular numbers are (n*(n+1))/2 with n or n+1 twin prime

a(n)+a(n+1) is square: 1,4,9,16,25,36,121,144,289,324...

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)

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

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 10 elements are not necessarily consecutive primes.

All the terms are positive multiples of 210 = 7# but are not multiples of 11 and also must not belong to A206045, where the first term is 1536160080; indeed, terms d' in A206045 correspond to the longest possible APs of primes that have exactly 11 elements with these common differences d'.

A342309(d) gives the first element of the smallest AP with 10 elements whose common difference is a(n) = d.

Similar sequence with odd integers d is A040976 \ {0}.

Terms are even numbers that are not divisible by 3 and that are not also in A206037.

These longest corresponding APs are of the form (q, q+d) with q odd primes (see examples).

This subsequence of A359408 corresponds to the second case '2 is one less than prime 3' (see A173919); the first case is linked to A040976.

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

See the assembly [a(n),a(n+1)] as a decimal number. Round this number to the closest integer. All rounded assemblies will produce a prime number.

"Rounding to the closest integer" is ambiguous for decimal numbers like (k.5) where k is an integer. Here we round such numbers to be rounded to k+1. The only occurrence of such a "rounding ambiguity" in the sequence happens with a(1) = 1 and a(2) = 5. Indeed, no more (k.5) "dilemmas" like that one will ever occur again as the integers 50, 500, 5000,... (that might produce together with the previous term k the decimal number k.50 or k.500 or k.5000...) cannot be part of the sequence; this is because 50, 500, 5000,... are not primes themselves (they end with 0) and neither are 51, 501, 5001,... (they are divisible by 3).