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.

A372793 Sequence related to the asymptotic expansion of Sum_{k=1..n} tau(m*k).

Original entry on oeis.org

1, 2, 3, 16, 5, 864, 7, 4096, 729, 64000, 11, 6879707136, 13, 2809856, 61509375, 4294967296, 17, 812479653347328, 19, 26843545600000000, 26795786661, 2791309312, 23, 4019988717840603673710821376, 9765625, 73719087104, 7625597484987, 25962355635465062711296, 29
Offset: 1

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Author

Vaclav Kotesovec, May 13 2024

Keywords

Comments

For m>=1, Sum_{k=1..n} tau(m*k) = A018804(m) * n * log(n) + O(n).
If p is prime, then Sum_{k=1..n} tau(p*k) ~ (2*p - 1) * n * (log(n) - 1 + 2*gamma)/p + n*log(p)/p, where gamma is the Euler-Mascheroni constant A001620.

Examples

			Sum_{k=1..n} tau(4*k) ~ (8*n*(log(n) + 2*gamma - 1) + n*4*log(2)) / 4, a(4) = exp(4*log(2)) = 16.
Sum_{k=1..n} tau(6*k) ~ (15*n*(log(n) + 2*gamma - 1) + n*(5*log(2) + 3*log(3))) / 6, a(6) = exp(5*log(2) + 3*log(3)) = 864.
Sum_{k=1..n} tau(8*k) ~ (20*n*(log(n) + 2*gamma - 1) + n*12*log(2)) / 8, a(8) = exp(12*log(2)) = 4096.
Sum_{k=1..n} tau(9*k) ~ (21*n*(log(n) + 2*gamma - 1) + n*6*log(3)) / 9, a(9) = exp(6*log(3)) = 729.
Sum_{k=1..n} tau(10*k) ~ (27*n*(log(n) + 2*gamma - 1) + n*(9*log(2) + 3*log(5))) / 10, a(10) = exp(9*log(2) + 3*log(5)) = 64000.
Sum_{k=1..n} tau(12*k) ~ (40*n*(log(n) + 2*gamma - 1) + n*(20*log(2) + 8*log(3))) / 12, a(12) = exp(20*log(2) + 8*log(3)) = 6879707136.
		

Crossrefs

Cf. A000005 (m=1), A099777 (m=2), A372713 (m=3), A372784 (m=4), A372785 (m=5), A372786 (m=6), A372787 (m=7), A372788 (m=8), A372789 (m=9), A372790 (m=10), A372791 (m=11), A372792 (m=12).

Formula

Sum_{k=1..n} tau(m*k) ~ A018804(m) * n * (log(n) - 1 + 2*gamma)/m + n*log(a(m))/m.
a(m) = exp(limit_{n->oo} (m * (Sum_{k=1..n} tau(m*k)) - A018804(m)*n*(log(n) - 1 + 2*gamma))/n).
If p is prime, then a(p) = p.
If p is prime, then a(p^k) = p^(k*p^(k-1)).
If p and q are distinct primes, then a(p*q) = p^(2*q-1) * q^(2*p-1).