A387469 Numbers s > 0 such that 8*s^2 + 64*s + 57 is a square.

4, 6, 41, 53, 258, 328, 1523, 1931, 8896, 11274, 51869, 65729, 302334, 383116, 1762151, 2232983, 10270588, 13014798, 59861393, 75855821, 348897786, 442120144, 2033525339, 2576865059, 11852254264, 15019070226, 69080000261, 87537556313, 402627747318, 510206267668

Offset

1,1

Comments

These are the solutions s (together with some k) of the Diophantine equation 1 + ... + k = (k+7) + ... + (k+7 + s) for k >= 1 and s >= 0, since that becomes k = ((2*s+1) + sqrt(8*s^2 + 64*s + 57)) / 2 which is an integer iff 8*s^2 + 64*s + 57 is a square (an odd square).

Links

Table of n, a(n) for n=1..30.

Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Formula

From Elmo R. Oliveira, May 12 2026: (Start)

G.f.: x*(4 + 2*x + 11*x^2 - x^4) / ((1-x) * (1-2*x-x^2) * (1+2*x-x^2)).

a(n) = a(n-1) + 6*a(n-2) - 6*a(n-3) - a(n-4) + a(n-5) for n > 5. (End)

Example

For s = 4: 8*4^2 + 64*4 + 57 = 21^2, thus 4 is a term.

Mathematica

LinearRecurrence[{1, 6, -6, -1, 1}, {4, 6, 41, 53, 258}, 30] (* Hugo Pfoertner, Aug 31 2025 *)

Prog

(PARI) is_a387469(s) = issquare(8*s^2+64*s+57) \\ Hugo Pfoertner, Aug 30 2025

Crossrefs

Cf. A000217.

Sequence in context: A319672 A355233 A377682 * A145387 A034923 A013022

Adjacent sequences: A387466 A387467 A387468 * A387470 A387471 A387472

Keyword

nonn,easy

Author

Ctibor O. Zizka, Aug 30 2025

Extensions

More terms from Hugo Pfoertner, Aug 31 2025

Status

approved