A254627 Indices of centered pentagonal numbers (A005891) that are also triangular numbers (A000217).

1, 2, 11, 28, 189, 494, 3383, 8856, 60697, 158906, 1089155, 2851444, 19544085, 51167078, 350704367, 918155952, 6293134513, 16475640050, 112925716859, 295643364940, 2026369768941, 5305104928862, 36361730124071, 95196245354568, 652484772464329

Offset

1,2

Comments

Also positive integers y in the solutions to x^2 - 5*y^2 + x + 5*y - 2 = 0, the corresponding values of x being A254626.

Also indices of centered pentagonal numbers (A005891) that are also hexagonal numbers (A000384). - Colin Barker, Feb 11 2015

Links

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

Hermann Stamm-Wilbrandt, 6 interlaced bisections

Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Formula

a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).

G.f.: x*(1+x-9*x^2-x^3)/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)).

a(n) = (10 - sqrt(5)*(2-sqrt(5))^n - 5*(-2+sqrt(5))^n - 2*sqrt(5)*(-2+sqrt(5))^n + sqrt(5)*(2+sqrt(5))^n + (-2-sqrt(5))^n*(-5+2*sqrt(5)))/20. - Colin Barker, Jun 06 2016

a(2*n+2) = A232970(2*n+1); a(2*n+1) = A110679(2*n). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019

a(n) = (2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4. - G. C. Greubel, Apr 19 2019

Example

2 is in the sequence because the 2nd centered pentagonal number is 6, which is also the 3rd triangular number.

Mathematica

CoefficientList[Series[x (x^3 + 9 x^2 - x - 1)/((x - 1) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, 25}], x] (* Michael De Vlieger, Jun 06 2016 *)
LinearRecurrence[{1, 18, -18, -1, 1}, {1, 2, 11, 28, 189}, 30] (* Harvey P. Dale, Apr 23 2017 *)

Prog

(PARI) Vec(x*(x^3+9*x^2-x-1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^30))
(PARI) {a(n) = (2 +(1+3*(-1)^n)*fibonacci(3*n) - 2*(-1)^n*fibonacci(3*n+1))/4}; \\ G. C. Greubel, Apr 19 2019
(Magma) [(2 +(1+2*(-1)^n)*Fibonacci(3*n) -(-1)^n*Lucas(3*n))/4 : n in [1..30]]; // G. C. Greubel, Apr 19 2019
(Sage) [(2 +(1+3*(-1)^n)*fibonacci(3*n) -2*(-1)^n*fibonacci(3*n+1))/4 for n in (1..30)] # G. C. Greubel, Apr 19 2019

Crossrefs

Cf. A000217, A005891, A254626, A254628.

Cf. A000384, A254962.

Sequence in context: A034534 A345035 A220833 * A206583 A045493 A116038

Adjacent sequences: A254624 A254625 A254626 * A254628 A254629 A254630

Keyword

nonn,easy

Author

Colin Barker, Feb 03 2015

Status

approved