A221672 Length of shortest non-constant arithmetic progression (AP) containing n squares.

1, 2, 3, 5, 8, 13, 16, 23, 27, 36, 41, 52

Offset: 1

Author

Jonathan Sondow, Jan 28 2013

Keywords

• nonn
• hard
• more

Comments

Same as where records occur in A221671 (maximum number of squares in a non-constant AP of length n).

González-Jiménez and Xarles (2013) conjecture that for n >= 5 the sequence a(n)-1 equals the tail 7, 12, 15, 22, 26, 35, 40, 51, ... of A001318 (generalized pentagonal numbers k*(3*k-1)/2 for k = 0, +-1, +-2, ...). They prove it up to a(12)-1 = 51 = 6*(3*6-1)/2.

See A221671 for additional comments.

Also 8, 13, 16, 23, 27, 36, 41, 52 are where records occur for 8 <= n <= 52 in A193832 (number of squares in the arithmetic progression {24k + 1: 0 <= k <= n-1} [Granville]). - Jonathan Sondow, Dec 15 2017

Examples

			The AP 1, 25, 49 = 1^2, 5^2, 7^2 shows that a(n) = n for n = 1, 2, 3 (see A216869).
By Fermat and Euler, no four squares are in AP, so the AP 49, 169, 289, 409, 529 = 7^2, 13^2, 17^2, 409, 23^2 shows that a(4) = 5 (see Dickson and A216870).
As k*(3*k-1)/2 = 0, 1, 2, 5, 7 for k = 0, +-1, +-2, and 24*k*(3*k-1)/2 + 1 = (6*k-1)^2 is a square, the AP 24*n+1 for the 8 numbers n = 0, 1, ..., 7 contains 5 squares, so a(5) <= 8. González-Jiménez and Xarles (2013) prove a(5) > 7, so a(5) = 8.
		

References

• L. E. Dickson, History of the Theory of Numbers, Vol. II, Chelsea, New York, 1952, pp. 435-440.

Links

• Enrique González-Jiménez and Xavier Xarles, On a conjecture of Rudin on squares in Arithmetic Progressions, arXiv 2013.

Crossrefs

Cf. A001318, A080995, A193832, A216869, A216870, A221671.

Formula

A221671(a(n)) = n.

a(n) <= A001318(n)+1. (Proof. As 24*k*(3*k-1)/2 + 1 = (6*k-1)^2, a term in the AP 24*m+1 is a square when m is in A001318. Thus the AP 24*m+1 for m = 0, 1, ..., A001318(n) contains n squares and has length A001318(n)+1.)