A387469 - OEIS

%I #45 May 12 2026 12:14:52

%S 4,6,41,53,258,328,1523,1931,8896,11274,51869,65729,302334,383116,

%T 1762151,2232983,10270588,13014798,59861393,75855821,348897786,

%U 442120144,2033525339,2576865059,11852254264,15019070226,69080000261,87537556313,402627747318,510206267668

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

%C 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).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,6,-6,-1,1).

%F From _Elmo R. Oliveira_, May 12 2026: (Start)

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

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

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

%t LinearRecurrence[{1, 6, -6, -1, 1}, {4, 6, 41, 53, 258}, 30] (* _Hugo Pfoertner_, Aug 31 2025 *)

%o (PARI) is_a387469(s) = issquare(8*s^2+64*s+57) \\ _Hugo Pfoertner_, Aug 30 2025

%Y Cf. A000217.

%K nonn,easy

%O 1,1

%A _Ctibor O. Zizka_, Aug 30 2025

%E More terms from _Hugo Pfoertner_, Aug 31 2025