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

Author

Jaap Spies, Aug 16 2005

Keywords

• easy
• nonn

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).

Examples

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

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

Crossrefs

Cf. A111774, A111775, A109814.

Programs

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);