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 1-4 of 4 results.

A300410 Number of centered square numbers dividing n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 2, 1, 2, 1, 2, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3
Offset: 1

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Author

Ilya Gutkovskiy, Mar 05 2018

Keywords

Examples

			a(26) = 2 because 26 has 4 divisors {1, 2, 13, 26} among which 2 divisors {1, 13} are centered square numbers.
		

Crossrefs

Programs

  • Maple
    N:= 100: # for a(1)..a(N)
    V:= Vector(N,1):
    for k from 1 do
      m:= 2*k*(k+1)+1;
      if m > N then break fi;
      r:= [seq(i,i=m..N,m)];
      V[r]:= map(t->t+1, V[r]);
    od:
    convert(V,list); # Robert Israel, Mar 05 2018
  • Mathematica
    nmax = 100; Rest[CoefficientList[Series[Sum[x^(2 k (k + 1) + 1)/(1 - x^(2 k (k + 1) + 1)), {k, 0, nmax}], {x, 0, nmax}], x]]

Formula

G.f.: Sum_{k>=0} x^(2*k*(k+1)+1)/(1 - x^(2*k*(k+1)+1)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A228048 = 1.440659... . - Amiram Eldar, Jan 02 2024

A358543 a(n) is the smallest number with exactly n divisors that are square pyramidal numbers.

Original entry on oeis.org

1, 5, 30, 140, 420, 1540, 4620, 13860, 78540, 157080, 471240, 1141140, 3603600, 3423420, 13693680, 30630600, 58198140, 116396280, 214414200, 428828400, 581981400, 1163962800, 5354228880, 4073869800, 8147739600, 26771144400, 36082846800, 80313433200, 93699005400, 187398010800
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 21 2022

Keywords

Comments

Any terms for n > 25 exceed 10^10. - Lucas A. Brown, Dec 24 2022
a(25) <= 8147739600, a(26) <= 26771144400, a(27) <= 36082846800, a(28) <= 80313433200. - Jon E. Schoenfield, Dec 16 2022

Examples

			a(3) = 30 because 30 has 3 square pyramidal divisors {1, 5, 30} and this is the smallest such number.
		

Crossrefs

Programs

  • PARI
    issqpyr(n) = my(m = sqrtnint(3*n, 3)); n==m*(m+1)*(2*m+1)/6; \\ A253903
    a(n) = my(k=1); while (sumdiv(k, d, issqpyr(d)) != n, k++); k; \\ Michel Marcus, Nov 21 2022

Extensions

a(15) from Michel Marcus, Nov 21 2022
a(16)-a(20) from Jinyuan Wang, Nov 28 2022
a(21)-a(22) from Lucas A. Brown, Dec 14 2022
a(23)-a(24) from Lucas A. Brown, Dec 18 2022
a(25) from Lucas A. Brown, Dec 22 2022
a(26)-a(30) from Bert Dobbelaere, May 18 2025

A334926 G.f.: Sum_{k>=1} x^(k*(2*k^2 + 1)/3) / (1 - x^(k*(2*k^2 + 1)/3)).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1
Offset: 1

Views

Author

Ilya Gutkovskiy, May 16 2020

Keywords

Comments

Number of octahedral numbers (A005900) dividing n.

Crossrefs

Programs

  • Mathematica
    nmax = 100; CoefficientList[Series[Sum[x^(k (2 k^2 + 1)/3)/(1 - x^(k (2 k^2 + 1)/3)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest

Formula

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A175577 = 1.278185... . - Amiram Eldar, Jan 02 2024

A334924 G.f.: Sum_{k>=1} x^(k^2*(k + 1)/2) / (1 - x^(k^2*(k + 1)/2)).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1
Offset: 1

Views

Author

Ilya Gutkovskiy, May 16 2020

Keywords

Comments

Number of pentagonal pyramidal numbers (A002411) dividing n.

Crossrefs

Programs

  • Mathematica
    nmax = 100; CoefficientList[Series[Sum[x^(k^2 (k + 1)/2)/(1 - x^(k^2 (k + 1)/2)), {k, 1, nmax}], {x, 0, nmax}], x] // Rest

Formula

Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi^2/3 - 2 = A195055 - 2 = 1.289868... . - Amiram Eldar, Jan 02 2024
Showing 1-4 of 4 results.