The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A351244

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A351244

a(n) = n^4 * Sum_{p|n, p prime} 1/p^4.
(history; published version)

#27 by Wesley Ivan Hurt at Thu Mar 09 20:51:23 EST 2023

STATUS

editing

approved

#26 by Wesley Ivan Hurt at Thu Mar 09 20:51:00 EST 2023

CROSSREFS

Cf. A000040, A085965.

STATUS

approved

editing

#25 by Vaclav Kotesovec at Sat Mar 04 02:19:36 EST 2023

STATUS

proposed

approved

#24 by Jon E. Schoenfield at Fri Mar 03 18:56:44 EST 2023

STATUS

editing

proposed

#23 by Jon E. Schoenfield at Fri Mar 03 18:56:42 EST 2023

FORMULA

Dirichlet g.f.: zeta(s-4)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^4/n^s) Sum_{p|n} 1/p^4. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^4*(p*j)^(s-4)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-4) = zeta(s-4)*primezeta(s). The result generalizes to higher powers of p. - Michael Shamos, Mar 02 2023

STATUS

proposed

editing

#22 by Vaclav Kotesovec at Fri Mar 03 10:31:18 EST 2023

STATUS

editing

proposed

#21 by Vaclav Kotesovec at Fri Mar 03 10:31:09 EST 2023

FORMULA

Sum_{k=1..n} a(k) ~ A085965 * n^5/5. - Vaclav Kotesovec, Mar 03 2023

STATUS

proposed

editing

#20 by Michel Marcus at Fri Mar 03 09:30:11 EST 2023

STATUS

editing

proposed

Discussion

Fri Mar 03

09:32

Michael Shamos: Yes.

#19 by Michel Marcus at Fri Mar 03 09:30:03 EST 2023

FORMULA

Dirichlet g.f. = : zeta(s-4)*primezeta(s). This follows because Sum_{n>=1} a(n)/n^s = Sum_{n>=1} (n^4/n^s) Sum_{p|n} 1/p^4. Since n = p*j, rewrite the sum as Sum_{p} Sum_{j>=1} 1/(p^4*(p*j)^(s-4)) = Sum_{p} 1/p^s Sum_{j>=1} 1/j^(s-4) = zeta(s-4)*primezeta(s). The result generalizes to higher powers of p in a(n). - Michael Shamos, Mar 02 2023

STATUS

proposed

editing

Discussion

Fri Mar 03

09:30

Michel Marcus: ok ?

#18 by Michel Marcus at Fri Mar 03 09:24:18 EST 2023

STATUS

editing

proposed