OFFSET
1,1
COMMENTS
Sums of two or more positive consecutive odd numbers are Sum_{k=0..m} (2*j+i+2*k) = (m+1)*(m+2*j+i) with m >= 1 and 2*j+i >=1. Testing a number n against being a member can be done by scanning all divisors d, building m=d-1, if this is >= 1 building n/d-m, and testing this for being an odd number >= 1. - R. J. Mathar, Jan 25 2011
REFERENCES
Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, Solution to problem 3G, p. 179.
LINKS
Erzsébet Orosz, On odd-summing numbers, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae 31 (2004) 125-129.
EXAMPLE
1+3=4, 3+5=8, 1+3+5=9, 5+7=12, 3+5+7=15, 7+9=16, ...
MAPLE
isA177713 := proc(n) local d, l; for d in numtheory[divisors](n) do l := d-1 ; if l >=1 then l := n/d -l; if type(l, 'odd') and l>=1 then return true; end if; end if; end do: return false; end proc:
for n from 2 to 130 do if isA177713(n) then printf("%d, ", n) ; end if; end do; # R. J. Mathar, Jan 25 2011
MATHEMATICA
z=200; lst={}; Do[c=a; Do[c+=b; If[c<=2*z, AppendTo[lst, c]], {b, a-2, 1, -2}], {a, 1, z, 2}]; Union@lst
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, May 11 2010
STATUS
approved