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.

A129557 Numbers k > 0 such that k^2 is a centered pentagonal number (A005891).

Original entry on oeis.org

1, 4, 34, 151, 1291, 5734, 49024, 217741, 1861621, 8268424, 70692574, 313982371, 2684456191, 11923061674, 101938642684, 452762361241, 3870983965801, 17193046665484, 146995452057754, 652883010927151, 5581956194228851, 24792361368566254
Offset: 1

Views

Author

Alexander Adamchuk, Apr 20 2007

Keywords

Comments

Corresponding numbers m such that centered pentagonal number A005891(m) = (5*m^2 + 5*m + 2)/2 is a perfect square are listed in A129556 = {0, 2, 21, 95, 816, 3626, 31005, ...}.
Also positive integers x in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of y being A254332. - Colin Barker, Jan 28 2015

Links

Crossrefs

Cf. A005891 (centered pentagonal numbers).
Cf. A129556 (k such that A005891(k) is a perfect square).
Cf. A000290, A005667, A254332, A254333.

Programs

  • Mathematica
    Do[ f=(5n^2+5n+2)/2; If[ IntegerQ[ Sqrt[f] ], Print[ Sqrt[f] ] ], {n,1,40000} ]
CoefficientList[Series[(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)),{x,0,30}],x] (* Vincenzo Librandi, Apr 11 2012 *)
  • PARI
    A129557()={ for(n=1,1000000000, f=(5*n^2+5*n+2)/2 ; if(issquare(f), print1(sqrtint(f), ", ") ; ); ) ; } \\ R. J. Mathar, Oct 11 2007
  • PARI
    Vec(x*(1-x)*(1+5*x+x^2)/((1+6*x-x^2)*(1-6*x-x^2)) + O(x^100)) \\ Colin Barker, Jan 28 2015

Formula

a(n) = sqrt( (5*A129556(n)^2 + 5*A129556(n) + 2)/2 ).
For n >= 5, a(n) = 38*a(n-2) - a(n-4). - Max Alekseyev, May 08 2009
G.f.: x*(1-x)*(1 + 5*x + x^2)/((1 + 6*x - x^2)*(1 - 6*x - x^2)). - Colin Barker, Apr 11 2012
From Andrea Pinos, Oct 07 2022: (Start)
The ratios of successive terms converge to two different limits:
lower: D = lim_{n->oo} a(2n)/a(2n-1) = (7 + 2*sqrt(10))/3;
upper: E = lim_{n->oo} a(2n+1)/a(2n) = (13 + 4*sqrt(10))/3.
So lim_{n->oo} a(n+2)/a(n) = D*E = 19 + 6*sqrt(10).
a(n) = (A005667(n) - (-1)^n*A005667(n-1))/4. (End)

Extensions

More terms from R. J. Mathar, Oct 11 2007
More terms from Max Alekseyev, May 08 2009

A254332 Indices of centered pentagonal numbers (A005891) which are also squares (A000290).

Original entry on oeis.org

1, 3, 22, 96, 817, 3627, 31006, 137712, 1177393, 5229411, 44709910, 198579888, 1697799169, 7540806315, 64471658494, 286352060064, 2448225223585, 10873837476099, 92968086837718, 412919472031680, 3530339074609681, 15680066099727723, 134059916748330142
Offset: 1

Views

Author

Colin Barker, Jan 28 2015

Keywords

Comments

Also positive integers y in the solutions to 2*x^2 - 5*y^2 + 5*y - 2 = 0, the corresponding values of x being A129557.

Examples

			3 is in the sequence because the 3rd centered pentagonal number is 16, which is also the 4th square number.

Links

Crossrefs

Cf. A000290, A005891, A129557, A254333.

Programs

  • Mathematica
    LinearRecurrence[{1,38,-38,-1,1},{1,3,22,96,817},30] (* Harvey P. Dale, Mar 27 2017 *)
  • PARI
    Vec(x*(2*x^3+19*x^2-2*x-1) / ((x-1)*(x^2-6*x-1)*(x^2+6*x-1)) + O(x^100))

Formula

a(n) = a(n-1) + 38*a(n-2) - 38*a(n-3) - a(n-4) + a(n-5).
G.f.: x*(2*x^3 + 19*x^2 - 2*x - 1) / ((x-1)*(x^2 - 6*x - 1)*(x^2 + 6*x - 1)).
a(n) = (1/40)*(20 - b^n*(19 + 3*b) + (3 + b)*c^n - (b^n*(3 + b) + (1 - 3*b)*c^n)*(-1)^n) with b = sqrt(10) - 3 and c = sqrt(10) + 3. - Alan Michael Gómez Calderón, Jul 02 2024

A277792 Squares that are also pentagonal pyramidal numbers.

Original entry on oeis.org

0, 1, 196, 2601, 15376, 60025, 181476, 461041, 1032256, 2099601, 3960100, 7027801, 11861136, 19193161, 29964676, 45360225, 66846976, 96216481, 135629316, 187662601, 255360400, 342287001, 452583076, 591024721, 763085376, 975000625, 1233835876, 1547556921, 1925103376, 2376465001, 2912760900
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 31 2016

Keywords

Comments

Intersection of A000290 and A002411.

Examples

			a(2) = 196 because 196 = 14^2 is a perfect square and 196 = 7^2*(7 + 1)/2 is the 7th pentagonal pyramidal number.

Links

Crossrefs

Cf. A000290, A001653, A002411, A007588, A036353, A056220, A254333.

Programs

  • Magma
    [n^2*(2*n^2-1)^2: n in [0..30]]; // Vincenzo Librandi, Nov 01 2016
  • Mathematica
    Table[n^2 (2 n^2 - 1)^2, {n, 0, 30}]
LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,1,196,2601,15376,60025,181476},40] (* Harvey P. Dale, Nov 01 2024 *)

Formula

O.g.f.: x*(1 + 189*x + 1250*x^2 + 1250*x^3 + 189*x^4 + x^5)/(1 - x)^7.
E.g.f.: x*(1 + 97*x + 336*x^2 + 256*x^3 + 60*x^4 + 4*x^5)*exp(x).
a(n) = a(-n).
a(n) = n^2*(2*n^2 - 1)^2.
a(n) = A000290(A007588(n)).
a(n) = A000290(n)*A000290(A056220(n)).
Sum_{n>=1} 1/a(n) = (2*Pi^2+9*sqrt(2)*Pi*cot(Pi/sqrt(2))+3*Pi^2*csc(Pi/sqrt(2))^2-24)/12 = 1.0055779712856...
Showing 1-3 of 3 results.

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