A111787 a(n) is the least k >= 3 such that n can be written as sum of k consecutive integers. a(n)=0 if such a k does not exist.

0, 0, 0, 0, 0, 3, 0, 0, 3, 4, 0, 3, 0, 4, 3, 0, 0, 3, 0, 5, 3, 4, 0, 3, 5, 4, 3, 7, 0, 3, 0, 0, 3, 4, 5, 3, 0, 4, 3, 5, 0, 3, 0, 8, 3, 4, 0, 3, 7, 5, 3, 8, 0, 3, 5, 7, 3, 4, 0, 3, 0, 4, 3, 0, 5, 3, 0, 8, 3, 5, 0, 3, 0, 4, 3, 8, 7, 3, 0, 5, 3, 4, 0, 3, 5, 4, 3, 11, 0, 3, 7, 8, 3, 4, 5, 3, 0, 7, 3, 5, 0, 3, 0, 13

Offset

1,6

Comments

a(n)=0 if n is an odd prime or a power of 2. For numbers of the third kind we proceed as follows: suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Let p be the smallest odd prime divisor of n then a(n) = min(p,2n/p).

References

Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC Problem C part 3, Jun 2005, pp. 181-182

Links

Donald Alan Morrison, Table of n, a(n) for n=1..10000

Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C: solution

J. Spies, Sage program for computing A111787

Example

a(15)=3 because 15=4+5+6 (k=3) and 15=2+3+4+5 (k=4)

Maple

ispoweroftwo := proc(n) local a, t; t := 1; while (n > t) do t := 2*t end do; if (n = t) then a := true else a:= false end if; return a; end proc; A111787:= proc(n) local d, k; k:=0; if isprime(n) or ispoweroftwo(n) then return(0); fi; for d from 3 by 2 to n do if n mod d = 0 then k:=min(d, 2*n/d); break; fi; od; return(k); end proc; seq(A111787(i), i=1..150);

Crossrefs

Cf. A111774, A111775, A109814.

Sequence in context: A216194 A365462 A279168 * A200524 A308223 A222328

Adjacent sequences: A111784 A111785 A111786 * A111788 A111789 A111790

Keyword

easy,nonn

Author

Jaap Spies, Aug 16 2005

Status

approved