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A279789
Number of ways to choose a constant partition of each part of a constant partition of n.
14
1, 1, 3, 3, 8, 3, 17, 3, 30, 12, 41, 3, 130, 3, 137, 45, 359, 3, 656, 3, 1306, 141, 2057, 3, 5446, 36, 8201, 544, 18610, 3, 34969, 3, 72385, 2061, 131081, 165, 290362, 3, 524297, 8205, 1109206, 3, 2130073, 3, 4371490, 33594, 8388617, 3, 17445321, 132, 33556496
OFFSET
0,3
COMMENTS
Also number of ways to choose a divisor d|n and then a sequence of n/d divisors of d.
FORMULA
a(n) = Sum_{d|n} tau(n/d)^d for n > 0. - Andrew Howroyd, Aug 26 2018
G.f.: 1 + Sum_{k>=1} tau(k)*x^k/(1 - tau(k)*x^k). - Ilya Gutkovskiy, May 23 2019
a(n) = 3 <=> n is prime <=> n in { A000040 }. - Alois P. Heinz, May 23 2019
EXAMPLE
The a(6)=17 twice-constant partitions are:
((6)),
((3)(3)), ((33)),
((3)(111)), ((111)(3)),
((2)(2)(2)), ((222)),
((2)(2)(11)), ((2)(11)(2)), ((11)(2)(2)),
((2)(11)(11)), ((11)(2)(11)), ((11)(11)(2)),
((1)(1)(1)(1)(1)(1)), ((11)(11)(11)), ((111)(111)), ((111111)).
MAPLE
with(numtheory):
a:= proc(n) option remember; `if`(n=0, 1,
add(tau(n/d)^d, d=divisors(n)))
end:
seq(a(n), n=0..70); # Alois P. Heinz, Dec 20 2016
MATHEMATICA
nn=20; Table[DivisorSum[n, Power[DivisorSigma[0, #], n/#]&], {n, nn}]
PROG
(PARI) a(n)=if(n==0, 1, sumdiv(n, d, numdiv(n/d)^d)) \\ Andrew Howroyd, Aug 26 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Gus Wiseman, Dec 18 2016
STATUS
approved