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A177713
Sums of two or more positive consecutive odd numbers.
6
4, 8, 9, 12, 15, 16, 20, 21, 24, 25, 27, 28, 32, 33, 35, 36, 39, 40, 44, 45, 48, 49, 51, 52, 55, 56, 57, 60, 63, 64, 65, 68, 69, 72, 75, 76, 77, 80, 81, 84, 85, 87, 88, 91, 92, 93, 95, 96, 99, 100, 104, 105, 108, 111, 112, 115, 116, 117, 119, 120, 121, 123, 124, 125, 128
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
The sums of two positive consecutive odd numbers are A008586 (without 0), the sums of three positive consecutive odd numbers are A016945 (without 3), etc.
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
Sequence in context: A044844 A145190 A327907 * A235992 A362006 A359783
KEYWORD
nonn,easy
AUTHOR
STATUS
approved