|
Search: id:A061391
|
|
|
| A061391 |
|
t(n,3) = Sum_{d|n} tau(d^3), where tau(n) = number of divisors of n, cf. A000005. |
|
+0
4
|
|
| 1, 5, 5, 12, 5, 25, 5, 22, 12, 25, 5, 60, 5, 25, 25, 35, 5, 60, 5, 60, 25, 25, 5, 110, 12, 25, 22, 60, 5, 125, 5, 51, 25, 25, 25, 144, 5, 25, 25, 110, 5, 125, 5, 60, 60, 25, 5, 175, 12, 60, 25, 60, 5, 110, 25, 110, 25, 25, 5, 300, 5, 25, 60, 70, 25, 125, 5, 60, 25, 125, 5, 264
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
FORMULA
|
t(n, k) = Sum_{d|n} tau(d^k) is multiplicative: if the canonical factorization of n = Product p^e(p) over primes then t(n, k) = Product t(p^e(p), k), t(p^e(p), k) = (1/2) *(k*e(p)+2)*(e(p)+1).
a(n) = sum( d dividing n, tau(nd)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 30 2002
Dirichlet g.f.: zeta^6(x)
Also tau_6(n) (see A007426)
Multiplicative with a(p^e) = (e+6 choose e). Mitch Harris (Harris.Mitchell(AT)mgh.harvard.edu) Jun 27, 2005.
|
|
EXAMPLE
|
For k=2 we get an interesting identity: Sum_{d|n} tau(d^2)=(tau(n))^2, cf. A048691, A035116.
|
|
CROSSREFS
|
Cf. t(n, 0) = A000005(n), t(n, 1) = A007425(n), t(n, 2) = A035116(n).
Sequence in context: A141244 A121849 A098331 this_sequence A123133 A122213 A049735
Adjacent sequences: A061388 A061389 A061390 this_sequence A061392 A061393 A061394
|
|
KEYWORD
|
nonn,mult
|
|
AUTHOR
|
Vladeta Jovovic (vladeta(AT)Eunet.yu), Apr 29 2001
|
|
|
Search completed in 0.002 seconds
|
