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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. 2
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 (list; graph; refs; listen; history; text; internal format)
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: A282499 A216194 A279168 * A200524 A308223 A222328

Adjacent sequences:  A111784 A111785 A111786 * A111788 A111789 A111790

KEYWORD

easy,nonn

AUTHOR

Jaap Spies, Aug 16 2005

STATUS

approved

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Last modified July 17 18:41 EDT 2019. Contains 325109 sequences. (Running on oeis4.)