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.

A272718 Partial sums of gcd-sum sequence A018804.

Original entry on oeis.org

1, 4, 9, 17, 26, 41, 54, 74, 95, 122, 143, 183, 208, 247, 292, 340, 373, 436, 473, 545, 610, 673, 718, 818, 883, 958, 1039, 1143, 1200, 1335, 1396, 1508, 1613, 1712, 1829, 1997, 2070, 2181, 2306, 2486, 2567, 2762, 2847, 3015, 3204, 3339, 3432, 3672, 3805, 4000, 4165
Offset: 1

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Author

Gareth McCaughan, May 05 2016

Keywords

Comments

a(n) is the sum of all pairs of greater common divisors for (i,j) where 1 <= i <= j <= n. - Jianing Song, Feb 07 2021

Examples

			The gcd-sum function takes values 1,3,5 for n = 1,2,3; therefore a(3) = 1+3+5 = 9.
		

Crossrefs

Partial sums of A018804.

Programs

  • Mathematica
    b[n_] := GCD[n, #]& /@ Range[n] // Total;
    Array[b, 100] // Accumulate (* Jean-François Alcover, Jun 05 2021 *)
    f[p_, e_] := (e*(p - 1)/p + 1)*p^e; s[n_] := Times @@ f @@@ FactorInteger[n]; Accumulate[Array[s, 100]] (* Amiram Eldar, Dec 10 2024 *)
  • PARI
    first(n)=my(v=vector(n),f); v[1]=1; for(i=2,n, f=factor(i); v[i] = v[i-1] + prod(j=1, #f~, (f[j, 2]*(f[j, 1]-1)/f[j, 1] + 1)*f[j, 1]^f[j, 2])); v \\ Charles R Greathouse IV, May 05 2016

Formula

According to Bordellès (2007), a(n) = (n^2 / (2*zeta(2)))*(log n + gamma - 1/2 + log(A^12/(2*Pi))) + O(n^(1+theta+epsilon)) where gamma = A001620, A ~= 1.282427129 is the Glaisher-Kinkelin constant A074962, theta is a certain constant defined in terms of the divisor function and known to lie between 1/4 and 131/416, and epsilon is any positive number.
G.f.: (1/(1 - x))*Sum_{k>=1} phi(k)*x^k/(1 - x^k)^2, where phi(k) is the Euler totient function. - Ilya Gutkovskiy, Jan 02 2017
a(n) = (1/2)*Sum_{k=1..n} phi(k) * floor(n/k) * floor(1+n/k), where phi(k) is the Euler totient function. - Daniel Suteu, May 28 2018
From Jianing Song, Feb 07 2021: (Start)
a(n) = Sum_{i=1..n, j=i..n} gcd(i,j).
a(n) = (A018806(n) + n*(n+1)/2) / 2 = (Sum_{k=1..n} phi(k)*(floor(n/k))^2 + n*(n+1)/2) / 2, phi = A000010.
a(n) = A178881(n) + n*(n+1)/2.
a(n) = A018806(n) - A178881(n). (End)