A305292 Numbers k such that k-1 is a square and k+1 is a triangular number.

2, 5, 65, 170, 2210, 5777, 75077, 196250, 2550410, 6666725, 86638865, 226472402, 2943171002, 7693394945, 99981175205, 261348955730, 3396416785970, 8878171099877, 115378189547777, 301596468440090, 3919462027838450, 10245401755863185, 133146330756959525

Offset

1,1

Comments

It is easy to prove that there are no numbers k such that k-1 is a triangular number and k+1 is a square.

Links

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

Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Formula

G.f.: x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).

a(n) = a(-n-1) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).

a(n) = 35*a(n-2) - 35*a(n-4) + a(n-6) = 34*a(n-2) - a(n-4) + 2.

a(n) = (-2 + (13*sqrt(2) + 7*(-1)^n)*(1 + sqrt(2))^(2*n+1) - (13*sqrt(2) - 7* (-1)^n)*(1 - sqrt(2))^(2*n+1))/32.

a(n) = A214838(n) - 1.

a(n) = A077241(n-1)^2 + 1.

Mathematica

LinearRecurrence[{1, 34, -34, -1, 1}, {2, 5, 65, 170, 2210}, 25]

Prog

(PARI) Vec(x*(2 + 3*x - 8*x^2 + 3*x^3 + 2*x^4)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)) + O(x^30)) \\ Colin Barker, Jun 14 2018

Crossrefs

Cf. A000045, A000290, A077241, A214838.

Sequence in context: A270389 A350956 A086560 * A268211 A246280 A333165

Adjacent sequences: A305289 A305290 A305291 * A305293 A305294 A305295

Keyword

nonn,easy

Author

Bruno Berselli, Jun 11 2018

Status

approved