A211221 For any partition of n consider the product of the sigma of each element. Sequence gives the maximum of such values.

1, 3, 4, 9, 12, 27, 36, 81, 108, 243, 324, 729, 972, 2187, 2916, 6561, 8748, 19683, 26244, 59049, 78732, 177147, 236196, 531441, 708588, 1594323, 2125764, 4782969, 6377292, 14348907, 19131876, 43046721, 57395628, 129140163, 172186884, 387420489, 516560652

Offset: 1

Author

Paolo P. Lava, Apr 13 2012

Keywords

• nonn
• easy

Examples

			For n=21 the partition (2,2,2,2,2,2,2,2,2,3) gives sigma(2)^9*sigma(3)=3^9*4=78732 that is the maximum value that can be reached.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,3).

Crossrefs

Cf. A000203, A085884, A211217, A211218, A211219, A211220.

Programs

Maple

with(numtheory); with(combinat);
A211221:=proc(q)
local b,c,i,j,k,m,n,t;
for n from 1 to q do
  k:=partition(n); b:=numbpart(n); m:=0;
  for i from 1 to b do
    c:=nops(k[i]); t:=1;
    for j from 1 to c do t:=t*sigma(k[i][j]); od; if t>m then m:=t; fi; od;
  print(m);
od; end:
A211221(100)

Mathematica

LinearRecurrence[{0,3},{1,3,4},40] (* Harvey P. Dale, Jun 06 2015 *)

Formula

For n>1, a(n) = 3^n/2 for n even and a(n) = 4*3^(n-3)/2 for n odd.

For n>3, a(n) = 3*a(n-2). G.f.: x*(1+3*x+x^2)/(1-3*x^2). [Colin Barker, Apr 18 2012]

Closed form: a(1)=1, then a(n) = 1/6*(7-(-1)^(n-2))*3^(1/4*(-1)^(n-2))*3^(1/2*(n-2))*27^(1/4) = 3^((2*n+(-1)^n-5)/4)*(7-(-1)^n)/2. [Paolo P. Lava, Apr 20 2012]