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 1-2 of 2 results.

A357041 a(n) = Sum_{d|n} 2^(d-1) * binomial(d+n/d-1,d).

Original entry on oeis.org

1, 4, 7, 18, 21, 66, 71, 196, 305, 648, 1035, 2526, 4109, 8774, 16875, 34288, 65553, 134860, 262163, 531506, 1051237, 2109594, 4194327, 8425348, 16779257, 33611984, 67123631, 134350206, 268435485, 537178750, 1073741855, 2148064768, 4295048345, 8591114580
Offset: 1

Views

Author

Seiichi Manyama, Feb 26 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := DivisorSum[n, 2^(#-1) * Binomial[# + n/# - 1, #] &]; Array[a, 50] (* Amiram Eldar, Jul 31 2023 *)
  • PARI
    a(n) = sumdiv(n, d, 2^(d-1)*binomial(d+n/d-1, d));
    
  • PARI
    my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, (1/(1-2*x^k)^k-1))/2)
    
  • PARI
    my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, (2*x)^k/(1-x^k)^(k+1))/2)
    
  • Python
    from math import comb
    from sympy import divisors
    def A357041(n): return sum(comb(d+n//d-1,d)<Chai Wah Wu, Feb 27 2023

Formula

G.f.: (1/2) * Sum_{k>0} (1/(1 - 2 * x^k)^k - 1).
G.f.: (1/2) * Sum_{k>0} (2 * x)^k/(1 - x^k)^(k+1).
If p is prime, a(p) = p + 2^(p-1).

A360798 Expansion of Sum_{k>0} x^k / (1 - (2 * x)^k)^(k+1).

Original entry on oeis.org

1, 5, 13, 45, 81, 321, 449, 1745, 2945, 9153, 11265, 60609, 53249, 230401, 410625, 1259777, 1114113, 7263233, 4980737, 31337473, 44630017, 115367937, 96468993, 937283585, 551550977, 2399256577, 4594597889, 14579646465, 7784628225, 89894944769, 33285996545
Offset: 1

Views

Author

Seiichi Manyama, Feb 21 2023

Keywords

Crossrefs

Cf. A360797.

Programs

  • Mathematica
    a[n_] := DivisorSum[n, 2^(n-#) * Binomial[# + n/# - 1, #] &]; Array[a, 30] (* Amiram Eldar, Aug 02 2023 *)
  • PARI
    my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, x^k/(1-(2*x)^k)^(k+1)))
    
  • PARI
    a(n) = sumdiv(n, d, 2^(n-d)*binomial(d+n/d-1, d));

Formula

a(n) = Sum_{d|n} 2^(n-d) * binomial(d+n/d-1,d).
If p is prime, a(p) = 1 + p * 2^(p-1).
Showing 1-2 of 2 results.