A113039 Number of ways the set {1,2,...,n} can be split into three subsets of which the three sums are consecutive.
0, 0, 1, 0, 3, 5, 0, 23, 52, 0, 254, 593, 0, 3611, 8859, 0, 55554, 142169, 0, 946871, 2466282, 0, 17095813, 45359632, 0, 323760077, 870624976, 0, 6367406592, 17307580710, 0, 129063054631, 353941332518, 0, 2682355470491, 7410591325928, 0, 56930627178287
Offset: 1
Keywords
Examples
For n=5 we have splittings 4/23/15, 4/5/123, 13/5/24, so a(5)=3.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..100
Crossrefs
Cf. A112972.
Programs
-
Maple
A113039:=proc(n) local i,j,p,t; t:= 0,0; for j from 3 to n do p:=1; for i to j do p:=p*(x^(-2*i)+x^(i)*(y^i+y^(-i))); od; t:=t,coeff(coeff(p,x,3),y,1); od; t; end; # second Maple program: b:= proc() option remember; local i, j, t; `if` (args[1]=0, `if` (nargs=2, 1, b(args[t] $t=2..nargs)), add (`if` (args[j] -args[nargs] <0, 0, b(sort ([seq (args[i] -`if` (i=j, args[nargs], 0), i=1..nargs-1)])[], args[nargs]-1)), j=1..nargs-1)) end: a:= proc(n) local m; m:= n*(n+1)/2; `if` (n>2 and irem (m,3)=0, b(m/3-1, m/3, m/3+1, n), 0) end: seq (a(n), n=1..42); # Alois P. Heinz, Sep 03 2009
-
Mathematica
a[n_] := If[n <= 2, 0, Product[x^(-2k)+x^k(y^k+y^(-k)), {k, 1, n}] // SeriesCoefficient[#, {x, 0, 3}, {y, 0, 1}]&]; Table[an = a[n]; Print[n, " ", an]; an, {n, 1, 26}] (* Jean-François Alcover, Nov 17 2022 *)
Formula
a(n) is the coefficient of x^3y in product(x^(-2k)+x^k(y^k+y^(-k)), k=1..n) for n>2.
Extensions
Extended beyond a(25) by Alois P. Heinz, Sep 03 2009
Comments