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
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).
Programs
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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)
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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]