A370905 Partial sums of the alternating sum of divisors function (A206369).

1, 2, 4, 7, 11, 13, 19, 24, 31, 35, 45, 51, 63, 69, 77, 88, 104, 111, 129, 141, 153, 163, 185, 195, 216, 228, 248, 266, 294, 302, 332, 353, 373, 389, 413, 434, 470, 488, 512, 532, 572, 584, 626, 656, 684, 706, 752, 774, 817, 838, 870, 906, 958, 978, 1018, 1048

Offset: 1

Author

Amiram Eldar, Mar 05 2024

Keywords

• nonn
• easy

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• László Tóth, A survey of the alternating sum-of-divisors function, Acta Universitatis Sapientiae, Mathematica, Vol. 5, No. 1 (2013), pp. 93-107.
• László Tóth, Alternating Sums Concerning Multiplicative Arithmetic Functions, Journal of Integer Sequences, Vol. 20 (2017), Article 17.2.1.

Crossrefs

Cf. A206369, A370906.

Programs

Mathematica

f[p_, e_] := Sum[(-1)^(e-k)*p^k, {k, 0, e}]; beta[1] = 1; beta[n_] := Times @@ f @@@ FactorInteger[n]; Accumulate[Array[beta[#] &, 100]]

PARI

beta(n) = {my(f = factor(n)); prod(i=1, #f~, p = f[i, 1]; e = f[i, 2]; sum(k = 0, e, (-1)^(e-k)*p^k));}
lista(kmax) = {my(s = 0); for(k = 1, kmax, s += beta(k); print1(s, ", "))};

PARI

a(n) = sum(k=1, n, (-1)^bigomega(k) * (n\k) * (n\k+1))/2; \\ Daniel Suteu, May 11 2024

Python

from math import prod
from sympy import factorint
def A370905(n): return sum(prod((lambda x:x[0]+int((x[1]<<1)>=p+1))(divmod(p**(e+1),p+1)) for p, e in factorint(k).items()) for k in range(1,n+1)) # Chai Wah Wu, Mar 05 2024

Formula

a(n) = Sum_{k=1..n} A206369(k).

a(n) = (Pi^2/30) * n^2 + O(n * log(n)^(2/3) * log(log(n))^(4/3)) (Tóth, 2013).

a(n) = (1/2) * Sum_{k=1..n} A008836(k) * floor(n/k) * floor(n/k + 1). - Daniel Suteu, May 11 2024