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

1,2

Links

Table of n, a(n) for n=1..37.

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

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]

Example

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.

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 *)

Crossrefs

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

Sequence in context: A034418 A034421 A353306 * A029448 A249179 A103014

Adjacent sequences: A211218 A211219 A211220 * A211222 A211223 A211224

Keyword

nonn,easy,changed

Author

Paolo P. Lava, Apr 13 2012

Status

approved