A276599 Values of n such that n^2 + 5 is a triangular number (A000217).

1, 4, 10, 25, 59, 146, 344, 851, 2005, 4960, 11686, 28909, 68111, 168494, 396980, 982055, 2313769, 5723836, 13485634, 33360961, 78600035, 194441930, 458114576, 1133290619, 2670087421, 6605301784, 15562409950, 38498520085, 90704372279, 224385818726

Offset

1,2

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Formula

a(n) = 6*a(n-2) - a(n-4) for n>4.

G.f.: x*(1+x)*(1+3*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)).

a(n) = (1/4)*(P(n+2) + P(n+1) + (-1)^n*(3*P(n) - 7*P(n-1))), where P(n) = A000129(n). - G. C. Greubel, Sep 15 2021

Example

4 is in the sequence because 4^2 + 5 = 21, which is a triangular number.

Mathematica

CoefficientList[Series[(1+x)*(1+3*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 07 2016 *)
LinearRecurrence[{0, 6, 0, -1}, {1, 4, 10, 25}, 30] (* Harvey P. Dale, Feb 13 2017 *)

Prog

(PARI) Vec(x*(1+x)*(1+3*x+x^2)/((1+2*x-x^2)*(1-2*x-x^2)) + O(x^40))
(PARI) a(n)=([0, 1; -1, 6]^(n\2)*if(n%2, [1; 10], [-1; 4]))[1, 1] \\ Charles R Greathouse IV, Sep 07 2016
(Magma) I:=[1, 4, 10, 25]; [n le 4 select I[n] else 6*Self(n-2) - Self(n-4): n in [1..31]]; // G. C. Greubel, Sep 15 2021
(Sage)
def P(n): return lucas_number1(n, 2, -1)
[(1/4)*(P(n+2) + P(n+1) + (-1)^n*(3*P(n) - 7*P(n-1))) for n in (1..30)] # G. C. Greubel, Sep 15 2021

Crossrefs

Cf. A000129, A000217, A230044.

Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276600 (k=6), A276601 (k=9), A276602 (k=10), where k is the value added to n^2.

Sequence in context: A159297 A248731 A279101 * A281867 A298806 A362179

Adjacent sequences: A276596 A276597 A276598 * A276600 A276601 A276602

Keyword

nonn,easy

Author

Colin Barker, Sep 07 2016

Status

approved