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A177713
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Sums of two or more positive consecutive odd numbers.
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6
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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
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OFFSET
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1,1
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COMMENTS
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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.
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REFERENCES
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Paul Halmos, "Problems for Mathematicians, Young and Old", Dolciani Mathematical Expositions, 1991, Solution to problem 3G, p. 179.
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LINKS
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Erzsébet Orosz, On odd-summing numbers, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae 31 (2004) 125-129.
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EXAMPLE
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1+3=4, 3+5=8, 1+3+5=9, 5+7=12, 3+5+7=15, 7+9=16, ...
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MAPLE
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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
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MATHEMATICA
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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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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