A383134 Array read by ascending antidiagonals: A(n,k) is the length of the arithmetic progression of only primes having difference n and first term prime(k).

2, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 5, 1, 1, 1, 2, 1, 2, 3, 1, 3, 1, 2, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset

1,1

References

Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 139.

Links

Table of n, a(n) for n=1..91.

Formula

A(A006512(n),k) = 1 for n > 1.

A(A040976(n),k) = A054977(k+1).

Example

The array begins as:
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 3, 2, 1, 2, 1, 2, 1, 1, 2, ...
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 3, 1, 2, 1, 2, 1, 2, 1, 1, ...
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 1, 5, 3, 4, 2, 3, 1, 2, 1, ...
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 3, 2, 1, 2, 1, 1, 1, 2, 2, ...
  2, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
  1, 3, 1, 2, 1, 2, 1, 2, 1, 1, ...
  ...
A(2,2) = 3 since 3 primes are in arithmetic progression with a difference of 2 and the first term equal to the 2nd prime: 3, 5, and 7.
A(6,3) = 5 since 5 primes are in arithmetic progression with a difference of 6 and the first term equal to the 3rd prime: 5, 11, 17, 23, and 29.

Mathematica

A[n_, k_]:=Module[{count=1, sum=Prime[k]}, While[PrimeQ[sum+=n], count++]; count]; Table[A[n-k+1, k], {n, 13}, {k, n}]//Flatten

Crossrefs

Cf. A000012, A006512, A040976, A054977, A175191 (n=2).

Cf. A088430, A206045, A237453.

Sequence in context: A368010 A237453 A395081 * A265754 A089309 A126387

Adjacent sequences: A383131 A383132 A383133 * A383135 A383136 A383137

Keyword

nonn,tabl

Author

Stefano Spezia, Apr 17 2025

Status

approved