A276601 Values of k such that k^2 + 9 is a triangular number (A000217).

1, 6, 12, 37, 71, 216, 414, 1259, 2413, 7338, 14064, 42769, 81971, 249276, 477762, 1452887, 2784601, 8468046, 16229844, 49355389, 94594463, 287664288, 551336934, 1676630339, 3213427141, 9772117746, 18729225912, 56956076137, 109161928331, 331964339076

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+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)).

a(n) = (1/4)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))), where P(n) = A000129(n). - G. C. Greubel, Sep 15 2021

Example

6 is in the sequence because 6^2+9 = 45, which is a triangular number.

Mathematica

CoefficientList[Series[(1+x)*(1+5*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, 6, 12, 37}, 31] (* G. C. Greubel, Sep 15 2021 *)

Prog

(PARI) Vec(x*(1+x)*(1+5*x+x^2) / ((1+2*x-x^2)*(1-2*x-x^2)) + O(x^40))
(Magma) I:=[1, 6, 12, 37]; [n le 2 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)*(5*P(n+1) - P(n) + (-1)^n*(P(n-1) + 5*P(n-2))) 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), A276599 (k=5), A276600 (k=6), A276602 (k=10), where k is the value added to n^2.

Sequence in context: A298881 A256875 A117596 * A096377 A026083 A215179

Adjacent sequences: A276598 A276599 A276600 * A276602 A276603 A276604

Keyword

nonn,easy

Author

Colin Barker, Sep 07 2016

Status

approved