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A279789 Number of ways to choose a constant partition of each part of a constant partition of n.

%I A279789 #19 May 23 2019 08:55:13
%S A279789 1,1,3,3,8,3,17,3,30,12,41,3,130,3,137,45,359,3,656,3,1306,141,2057,3,
%T A279789 5446,36,8201,544,18610,3,34969,3,72385,2061,131081,165,290362,3,
%U A279789 524297,8205,1109206,3,2130073,3,4371490,33594,8388617,3,17445321,132,33556496
%N A279789 Number of ways to choose a constant partition of each part of a constant partition of n.
%C A279789 Also number of ways to choose a divisor d|n and then a sequence of n/d divisors of d.
%H A279789 Alois P. Heinz, <a href="/A279789/b279789.txt">Table of n, a(n) for n = 0..6643</a>
%H A279789 Gus Wiseman, <a href="/A063834/a063834.txt">Sequences enumerating triangles of integer partitions</a>
%F A279789 a(n) = Sum_{d|n} tau(n/d)^d for n > 0. - _Andrew Howroyd_, Aug 26 2018
%F A279789 G.f.: 1 + Sum_{k>=1} tau(k)*x^k/(1 - tau(k)*x^k). - _Ilya Gutkovskiy_, May 23 2019
%F A279789 a(n) = 3 <=> n is prime <=> n in { A000040 }. - _Alois P. Heinz_, May 23 2019
%e A279789 The a(6)=17 twice-constant partitions are:
%e A279789 ((6)),
%e A279789 ((3)(3)), ((33)),
%e A279789 ((3)(111)), ((111)(3)),
%e A279789 ((2)(2)(2)), ((222)),
%e A279789 ((2)(2)(11)), ((2)(11)(2)), ((11)(2)(2)),
%e A279789 ((2)(11)(11)), ((11)(2)(11)), ((11)(11)(2)),
%e A279789 ((1)(1)(1)(1)(1)(1)), ((11)(11)(11)), ((111)(111)), ((111111)).
%p A279789 with(numtheory):
%p A279789 a:= proc(n) option remember; `if`(n=0, 1,
%p A279789       add(tau(n/d)^d, d=divisors(n)))
%p A279789     end:
%p A279789 seq(a(n), n=0..70);  # _Alois P. Heinz_, Dec 20 2016
%t A279789 nn=20;Table[DivisorSum[n,Power[DivisorSigma[0,#],n/#]&],{n,nn}]
%o A279789 (PARI) a(n)=if(n==0, 1, sumdiv(n, d, numdiv(n/d)^d)) \\ _Andrew Howroyd_, Aug 26 2018
%Y A279789 Cf: A000005 (tau), A000040, A018818, A063834, A260685, A279375.
%K A279789 nonn
%O A279789 0,3
%A A279789 _Gus Wiseman_, Dec 18 2016