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A104513
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The number of consecutive integers > 1 beginning with A104512(n), the sum of which equals n, or 0 if impossible.
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5
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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
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OFFSET
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1,3
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REFERENCES
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Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 67.
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LINKS
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FORMULA
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a(n)=0 iff n=2^k.
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EXAMPLE
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a(18) = 4 because 3+4+5+6 = 18.
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MATHEMATICA
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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}]
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PROG
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(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(d<n, break)))); (0)); \\ Antti Karttunen, Mar 30 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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