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Search: id:A111787
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| A111787 |
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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. |
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+0
1
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| 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; listen)
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OFFSET
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1,6
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COMMENT
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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).
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REFERENCES
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Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC Problem C part 3, Jun 2005, pp. 181-182
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LINKS
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Nieuw Archief voor Wiskunde 5/6 nr. 2 Problems/UWC, Problem C: solution
J. Spies, SAGE program for computing A111787
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EXAMPLE
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a(15)=3 because 15=4+5+6 (k=3) and 15=2+3+4+5 (k=4)
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MAPLE
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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);
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CROSSREFS
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Cf. A111774, A111775, A109814.
Sequence in context: A033685 A063691 A075874 this_sequence A091921 A037285 A005082
Adjacent sequences: A111784 A111785 A111786 this_sequence A111788 A111789 A111790
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KEYWORD
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easy,nonn
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AUTHOR
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Jaap Spies (j.spies(AT)hccnet.nl), Aug 16 2005
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