A180361 -id:A180361 - cp's OEIS frontend

A189020 a(n) = Sum_{k=1..10^n} tau_4(k), where tau_4 is the number of ordered factorizations into 4 factors (A007426).

1, 89, 3575, 93237, 1951526, 35270969, 578262093, 8840109380, 128217432396, 1784942188189, 24045237260214, 315312623543840, 4042957241191810, 50862246063060180, 629513636928477232, 7681900592647818929, 92587253467765253144, 1103781870246459696784, 13031388731053572679450, 152516435040764735691556, 1771079109308495896176156

Offset: 0

Andrew Lelechenko, Apr 15 2011

nonn

Using that tau_4 = tau_2 ** tau_2, where ** means Dirichlet convolution and tau_2 is (A000005), one can calculate a(n) faster than in O(10^n) operations - namely in O(10^(3n/4)) or even in O(10^(2n/3)). See links for details.

A. V. Lelechenko, The summation of the multiplicative functions (in Russian).

Cf. A057494 - partial sums up to 10^n of the divisors function tau_2 (A000005), A180361 - of the unitary divisors function tau_2* (A034444), A180365 - of the 3-divisors function tau_3 (A007425).

Also see A072692 for such sums of the sum of divisors function (A000203), A084237 for sums of Moebius function (A008683), A064018 for sums of Euler totient function (A000010).

a(n) = A061202(10^n) = Sum_{k=1..10^n} A007426(n).

a(16)-a(20) from Henri Lifchitz, Feb 05 2025

cp's OEIS Frontend

A189020 a(n) = Sum_{k=1..10^n} tau_4(k), where tau_4 is the number of ordered factorizations into 4 factors (A007426).

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