A191832 Number of solutions to the Diophantine equation x1*x2 + x2*x3 + x3*x4 + x4*x5 + x5*x6 = n, with all xi >= 1.
0, 0, 0, 0, 1, 2, 7, 10, 22, 29, 51, 61, 99, 115, 163, 192, 262, 287, 385, 428, 528, 600, 730, 780, 963, 1054, 1202, 1337, 1545, 1646, 1908, 2059, 2269, 2516, 2770, 2933, 3298, 3568, 3792, 4142, 4493, 4786, 5183, 5562, 5831, 6423, 6745, 7140, 7639, 8231, 8479, 9216, 9603, 10260, 10663, 11488, 11752, 12838, 13100, 13887
Offset: 1
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 1..1000
- George E. Andrews, Stacked lattice boxes, Ann. Comb. 3 (1999), 115-130. See L_5(n).
Programs
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Maple
with(numtheory); D00:=n->add(tau(j)*tau(n-j),j=1..n-1); D01:=n->add(tau(j)*sigma(n-j),j=1..n-1); D000:=proc(n) local t1,i,j; t1:=0; for i from 1 to n-1 do for j from 1 to n-1 do if (i+j < n) then t1 := t1+numtheory:-tau(i)*numtheory:-tau(j)*numtheory:-tau(n-i-j); fi; od; od; t1; end; L5:=n->D000(n)/6+D00(n)+D01(n)/2+(2*n-1/6)*tau(n)-11*sigma[2](n)/6; [seq(L5(n),n=1..60)]; # Alternate: g:= proc(n,k,j) option remember; if n < k-1 then 0 elif k = 2 then if n mod j = 0 then 1 else 0 fi else add(procname(n-j*x,k-1,x), x=1 .. floor((n-k+2)/j)) fi end proc: f:= n -> add(g(n,6,j),j=1..n-4); seq(f(n),n=1..100); # Robert Israel, Dec 02 2015
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Mathematica
g[n_, k_, j_] := g[n, k, j] = If[n < k - 1, 0, If[k == 2, If[ Mod[n, j] == 0, 1, 0], Sum[g[n - j x, k - 1, x], {x, 1, Floor[(n - k + 2)/j]}]]]; f[n_] := Sum[g[n, 6, j], {j, 1, n - 4}]; Array[f, 100] (* Jean-François Alcover, Sep 25 2020, after Robert Israel *)
Comments