A123396 a(n) = number of earlier terms each of which, when added to n, give a triangular number.

0, 1, 1, 1, 0, 3, 2, 1, 1, 5, 3, 0, 2, 2, 5, 3, 2, 0, 3, 4, 5, 4, 0, 3, 2, 5, 5, 5, 5, 0, 0, 7, 2, 5, 6, 5, 7, 0, 2, 1, 9, 2, 5, 8, 6, 8, 1, 2, 2, 2, 10, 2, 5, 12, 8, 8, 1, 1, 4, 2, 2, 11, 3, 6, 14, 9, 9, 1, 1, 3, 4, 2, 3, 11, 4, 8, 15, 12, 8, 2, 2, 1, 3, 6, 2, 4, 11, 6, 9, 18, 13, 9, 1, 2, 3, 1, 5, 6, 2, 6

Offset

0,6

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Example

a(0)+10 = 10, a(1)+10 = 11, a(2)+10 = 11, a(3)+10 = 11, a(4)+10 = 10, a(5)+10 = 13, a(6)+10 = 12, a(7)+10 = 11, a(8)+10 = 11 and a(9)+10 = 15. Of these, three terms (a(0), a(4), a(9)) are such that, when each is added to 10, the result is a triangular number (i.e. of the form m(m+1)/2). Hence a(10) = 3.

Mathematica

f[l_List] := Append[l, Length[Select[l + Length[l], IntegerQ[Sqrt[8# + 1]] &]]]; Nest[f, {0}, 100] (* Ray Chandler, Oct 16 2006 *)

Prog

(PARI) {m=100; v=vector(m+1); print1(v[1]=0, ", "); for(n=1, m, c=0; for(k=1, n, a=n+v[k]; b=sqrtint(2*a); if(b*(b+1)/2==a, c++)); print1(v[n+1]=c, ", "))} \\ Klaus Brockhaus, Oct 16 2006

Crossrefs

Cf. A000217.

Sequence in context: A351073 A046225 A269233 * A225800 A176669 A327615

Adjacent sequences: A123393 A123394 A123395 * A123397 A123398 A123399

Keyword

easy,nonn

Author

Leroy Quet, Oct 14 2006

Extensions

More terms from Ray Chandler and Klaus Brockhaus, Oct 16 2006

Status

approved