cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Colin Barker, Feb 03 2015

Keywords

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

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

A254626 Indices of triangular numbers (A000217) that are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 3, 23, 61, 421, 1103, 7563, 19801, 135721, 355323, 2435423, 6376021, 43701901, 114413063, 784198803, 2053059121, 14071876561, 36840651123, 252509579303, 661078661101, 4531100550901, 11862575248703, 81307300336923, 212865275815561, 1459000305513721
Offset: 1

Views

Author

Colin Barker, Feb 03 2015

Keywords

Comments

Also positive integers x in the solutions to x^2 - 5*y^2 + x + 5*y - 2 = 0, the corresponding values of y being A254627.

Examples

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

Links

Crossrefs

Cf. A000217, A005891, A254627, A254628.

Programs

  • Mathematica
    LinearRecurrence[{1,18,-18,-1,1},{1,3,23,61,421},30] (* Harvey P. Dale, Jun 15 2024 *)
  • PARI
    Vec(-x*(x+1)^2*(x^2+1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^100))

Formula

a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5).
G.f.: -x*(x+1)^2*(x^2+1) / ((x-1)*(x^2-4*x-1)*(x^2+4*x-1)).
a(n) = (-2 + (2-r)^n - (-2-r)^n*(-2+r) + 2*(-2+r)^n + r*(-2+r)^n + (2+r)^n)/4 where r = sqrt(5). - Colin Barker, Nov 25 2016

A254962 Indices of hexagonal numbers (A000384) that are also centered pentagonal numbers (A005891).

Original entry on oeis.org

1, 2, 12, 31, 211, 552, 3782, 9901, 67861, 177662, 1217712, 3188011, 21850951, 57206532, 392099402, 1026529561, 7035938281, 18420325562, 126254789652, 330539330551, 2265550275451, 5931287624352, 40653650168462, 106432637907781, 729500152756861
Offset: 1

Views

Author

Colin Barker, Feb 11 2015

Keywords

Comments

Also positive integers x in the solutions to 4*x^2 - 5*y^2 - 2*x + 5*y - 2 = 0, the corresponding values of y being A254627.

Examples

			12 is in the sequence because the 12th hexagonal number is 276, which is also the 11th centered pentagonal number.

Links

Crossrefs

Cf. A000032 (Lucas numbers), A000384, A005891, A254627, A254628.

Programs

  • PARI
    Vec(-x*(x^4+x^3-8*x^2+x+1)/((x-1)*(x^2-4*x-1)*(x^2+4*x-1)) + O(x^100))

Formula

a(n) = a(n-1)+18*a(n-2)-18*a(n-3)-a(n-4)+a(n-5).
G.f.: -x*(x^4+x^3-8*x^2+x+1) / ((x-1)*(x^2-4*x-1)*(x^2+4*x-1)).
a(n) = (2 + (2-r)^n - (-2-r)^n*(-2+r) + 2*(-2+r)^n + r*(-2+r)^n + (2+r)^n)/8 where r = sqrt(5). - Colin Barker, Nov 25 2016
a(n+2) - a(n) = A000032(3*n + 2) if n is odd, A000032(3*n + 1) if n is even. - Diego Rattaggi, May 11 2020
Showing 1-3 of 3 results.

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