A065608 Sum of divisors of n minus the number of divisors of n.
0, 1, 2, 4, 4, 8, 6, 11, 10, 14, 10, 22, 12, 20, 20, 26, 16, 33, 18, 36, 28, 32, 22, 52, 28, 38, 36, 50, 28, 64, 30, 57, 44, 50, 44, 82, 36, 56, 52, 82, 40, 88, 42, 78, 72, 68, 46, 114, 54, 87, 68, 92, 52, 112, 68, 112, 76, 86, 58, 156, 60, 92, 98, 120, 80, 136, 66, 120, 92
Offset: 1
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)
- M. Alekseyev, E. Deutsch, and J. H. Steelman, Solution to problem 11281, Amer. Math. Monthly, 116, No. 5, 2009, p. 465.
- George E. Andrews, Stacked lattice boxes, Ann. Comb. 3 (1999), 115-130. See L_2(n).
- Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
- Masato Kobayashi, New recurrences for divisor sum functions and triangular numbers, arXiv:2207.05831 [math.NT], 2022.
Crossrefs
Starting (1, 2, 4, 4, 8, 6, ...), = row sums of triangle A077478. - Gary W. Adamson, Nov 12 2007
Starting with "1" = row sums of triangle A176919. - Gary W. Adamson, Apr 29 2010
Column k=2 of A125182.
Programs
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GAP
List([1..100],n->Sigma(n)-Tau(n)); # Muniru A Asiru, Mar 19 2018
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Maple
with(numtheory): seq(sigma(n)-tau(n),n=1..70); # Emeric Deutsch, Dec 22 2006
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Mathematica
Table[DivisorSigma[1,n]-DivisorSigma[0,n], {n,100}] (* Wesley Ivan Hurt, Dec 26 2013 *)
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PARI
a(n) = sigma(n) - numdiv(n); \\ Harry J. Smith, Oct 23 2009
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Python
from math import prod from sympy import factorint def A065608(n): f = factorint(n).items() return prod((p**(e+1)-1)//(p-1) for p, e in f)-prod(e+1 for p,e in f) # Chai Wah Wu, Jul 16 2022
Formula
a(n) = Sum_{d|n} (d-1). - Wesley Ivan Hurt, Dec 26 2013
G.f.: Sum_{k>=1} x^(2*k)/(1-x^k)^2. - Benoit Cloitre, Apr 21 2003
G.f.: Sum_{n>=1} (n-1)*x^n/(1-x^n). - Joerg Arndt, Jan 30 2011
L.g.f.: -log(Product_{k>=1} (1 - x^k)^(1-1/k)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, Mar 18 2018
G.f.: Sum_{n >= 1} q^(n^2)*( (n - 1) + q^n - (n - 1)*q^(2*n) )/(1 - q^n)^2 - differentiate equation 1 in Arndt with respect to t, then set x = q and t = q. - Peter Bala, Jan 22 2021
a(n) = n * A010054(n) - Sum_{k>=1} a(n - k*(k+1)/2), assuming a(n) = 0 for n <= 0 (Kobayashi, 2022). - Amiram Eldar, Jun 23 2023
Comments