A177713 Sums of two or more positive consecutive odd numbers.

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

Author

Vladimir Joseph Stephan Orlovsky, May 11 2010

Keywords

• nonn
• easy

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.

Examples

			1+3=4, 3+5=8, 1+3+5=9, 5+7=12, 3+5+7=15, 7+9=16, ...
		

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.

Crossrefs

Cf. A008586, A016945.

Programs

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