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

Author

Colin Barker, Feb 03 2015

Keywords

• nonn
• easy

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

Examples

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

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).

Crossrefs

Cf. A000217, A005891, A254626, A254628.

Cf. A000384, A254962.

Programs

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

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 *)

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

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

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