A183956 Number of strings of numbers x(i=1..5) in 0..n with sum i^2*x(i) equal to n*25.
2, 5, 10, 27, 53, 94, 161, 259, 399, 578, 811, 1120, 1505, 1985, 2562, 3267, 4104, 5092, 6249, 7595, 9146, 10923, 12948, 15245, 17831, 20735, 23980, 27592, 31597, 36020, 40894, 46252, 52114, 58520, 65494, 73076, 81300, 90195, 99807, 110163, 121306
Offset: 1
Keywords
Examples
All solutions for n=3 ..0....1....1....0....1....2....3....3....0....0 ..0....2....2....0....2....0....1....1....0....0 ..2....2....0....0....1....0....3....2....3....1 ..2....3....1....0....2....3....1....0....3....1 ..1....0....2....3....1....1....1....2....0....2
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
Programs
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Mathematica
r[n_, k_, s_] := r[n, k, s] = Which[s < 0, 0, n == 0, If[s == 0, 1, 0], True, Sum[r[n - 1, k, s - c*n^2], {c, 0, k}]]; T[n_, k_] := r[n, k, k*n^2]; a[n_] := T[5, n]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Jul 22 2022, after R. J. Mathar in A183953 *)
Formula
Empirical: a(n)=a(n-1)+a(n-4)-a(n-5)+a(n-9)-a(n-10)-a(n-13)+a(n-14)+a(n-16)-a(n-17)-a(n-20)+a(n-21)-a(n-34)+a(n-35)+a(n-38)-a(n-39)-a(n-41)+a(n-42)+a(n-45)-a(n-46)+a(n-50)-a(n-51)-a(n-54)+a(n-55)
Comments