A104513 The number of consecutive integers > 1 beginning with A104512(n), the sum of which equals n, or 0 if impossible.

0, 0, 2, 0, 2, 3, 2, 0, 3, 4, 2, 3, 2, 4, 5, 0, 2, 4, 2, 5, 6, 4, 2, 3, 5, 4, 6, 7, 2, 5, 2, 0, 6, 4, 7, 8, 2, 4, 6, 5, 2, 7, 2, 8, 9, 4, 2, 3, 7, 5, 6, 8, 2, 9, 10, 7, 6, 4, 2, 8, 2, 4, 9, 0, 10, 11, 2, 8, 6, 7, 2, 9, 2, 4, 10, 8, 11, 12, 2, 5, 9, 4, 2, 8, 10, 4, 6, 11, 2, 12, 13, 8, 6, 4, 10, 3, 2, 7, 11, 8

Offset: 1

Author

Alfred S. Posamentier (asp2(AT)juno.com) and Robert G. Wilson v, Feb 23 2005

Keywords

• nonn
• look

Examples

			a(18) = 4 because 3+4+5+6 = 18.
		

References

• Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 67.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16416

Crossrefs

Cf. A104512, A104514, A104515, A104516.

Programs

Mathematica

f[n_] := Block[{r = Ceiling[n/2]}, If[ IntegerQ[ Log[2, n]], 0, m = Range[r]; lst = Flatten[ Table[ m[[k]], {i, r}, {j, i + 1, r}, {k, i, j}], 1]; Length[ lst[[ Position[ Plus @@@ lst, n, 1, 1][[1, 1]]] ]]]]; Table[ f[n], {n, 100}]

PARI

A104513(n) = if(!bitand(n,n-1), 0, my(b,d,u=1+sqrtint(2*n)); for(k=0,n-2,b = binomial(k+1,2); forstep(j=min(n,k+u),k+2,-1, d = binomial(j+1,2) - b; if(d==n,return(j-k),if(dAntti Karttunen, Mar 30 2021

Formula

a(n)=0 iff n=2^k.