cp's OEIS Frontend

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.

%I A111787 #9 May 28 2014 00:16:41
%S A111787 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,
%T A111787 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,
%U A111787 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
%N 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.
%C A111787 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).
%D A111787 Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC Problem C part 3, Jun 2005, pp. 181-182
%H A111787 Donald Alan Morrison, <a href="/A111787/b111787.txt">Table of n, a(n) for n=1..10000</a>
%H A111787 Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C: <a href="http://www.jaapspies.nl/mathfiles/problem2005-2C.pdf">solution</a>
%H A111787 J. Spies, <a href="http://www.jaapspies.nl/oeis/a111787.sage">Sage program for computing A111787</a>
%e A111787 a(15)=3 because 15=4+5+6 (k=3) and 15=2+3+4+5 (k=4)
%p A111787 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);
%Y A111787 Cf. A111774, A111775, A109814.
%K A111787 easy,nonn
%O A111787 1,6
%A A111787 _Jaap Spies_, Aug 16 2005