A047968 a(n) = Sum_{d|n} p(d), where p(d) = A000041 = number of partitions of d.
1, 3, 4, 8, 8, 17, 16, 30, 34, 52, 57, 99, 102, 153, 187, 261, 298, 432, 491, 684, 811, 1061, 1256, 1696, 1966, 2540, 3044, 3876, 4566, 5846, 6843, 8610, 10203, 12610, 14906, 18491, 21638, 26508, 31290, 38044, 44584, 54133, 63262, 76241
Offset: 1
Keywords
Examples
For n = 10 the divisors of 10 are 1, 2, 5, 10, hence the partition numbers of the divisors of 10 are 1, 2, 7, 42, so a(10) = 1 + 2 + 7 + 42 = 52. - _Omar E. Pol_, Feb 26 2014 From _Gus Wiseman_, Sep 16 2018: (Start) The a(6) = 17 constant multiset partitions: (111111) (111)(111) (11)(11)(11) (1)(1)(1)(1)(1)(1) (111222) (12)(12)(12) (111122) (112)(112) (112233) (123)(123) (111112) (111123) (111223) (111234) (112234) (112345) (123456) (End)
Links
- T. D. Noe, Table of n, a(n) for n=1..1000
- N. J. A. Sloane, Transforms
Crossrefs
Programs
-
Maple
with(combinat): with(numtheory): a := proc(n) c := 0: l := sort(convert(divisors(n), list)): for i from 1 to nops(l) do c := c+numbpart(l[i]) od: RETURN(c): end: for j from 1 to 60 do printf(`%d, `, a(j)) od: # Zerinvary Lajos, Apr 14 2007
-
Mathematica
a[n_] := Sum[ PartitionsP[d], {d, Divisors[n]}]; Table[a[n], {n, 1, 44}] (* Jean-François Alcover, Oct 03 2013 *)
Formula
G.f.: Sum_{k>0} (-1+1/Product_{i>0} (1-z^(k*i))). - Vladeta Jovovic, Jun 22 2003
G.f.: sum(n>0,A000041(n)*x^n/(1-x^n)). - Mircea Merca, Feb 24 2014.
a(n) = Sum_{y is a partition of n} A000005(GCD(y)). - Gus Wiseman, Sep 16 2018
Comments