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A131576
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Number of ways to represent n as a sum of an even number of consecutive integers.
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19
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0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 2, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 2, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 2, 1, 1, 1, 2, 0, 2, 1, 1, 1, 1, 1, 2, 0, 2, 1, 1, 1, 2, 1, 1, 0, 1, 1, 3, 1, 1, 2, 1, 0, 2, 1, 1, 1, 2, 1, 2, 0, 1, 2, 1, 1, 2, 1, 2, 0, 1, 1, 2, 1, 1, 2, 1, 0, 4
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OFFSET
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1,21
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COMMENTS
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Equals number of odd divisors of n greater than sqrt(2*n). [Hirschhorn and Hirschhorn]
Conjecture: a(n) is also the number of pairs of subparts in the symmetric representation of sigma(n) which are mirror images of each other in the main diagonal. (Cf. A279387). - Omar E. Pol, Feb 22 2017 [Conjecture clarified by N. J. A. Sloane, Dec 16 2020]
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REFERENCES
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M. D. Hirschhorn and P. M. Hirschhorn, Partitions into Consecutive Parts, Mathematics Magazine, 78:5 (2005), 396-398. [Please do not delete this reference. - N. J. A. Sloane, Dec 16 2020]
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LINKS
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FORMULA
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G.f.: Sum_{k>=1} x^(k*(2*k+1))/(1-x^(2*k)). [Corrected by N. J. A. Sloane, Dec 18 2020]
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EXAMPLE
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a(11)=1 because we have 11=5+6; a(21)=2 because we have 21=10+11=1+2+3+4+5+6; a(75)=3 because we have 75=37+38=10+11+12+13+14+15=3+4+5+6+7+8+9+10+11+12.
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MAPLE
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G:=sum(x^(k*(2*k+1))/(1-x^(2*k)), k=1..10): Gser:=series(G, x=0, 85): seq(coeff(Gser, x, n), n=1..80); # Emeric Deutsch, Sep 08 2007
A131576 := proc(n) local dvs, a, k, r; dvs := numtheory[divisors](n) ; a := 0 ; for k in dvs do r := n/k+1 ; if r mod 2 = 0 then if r/2-k >= 1 then a := a+1 ; fi ; fi ; od: RETURN(a) ; end: seq(A131576(n), n=1..120) ; # R. J. Mathar, Sep 13 2007
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MATHEMATICA
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With[{m = 105}, Rest@ CoefficientList[Series[Sum[x^(k (2 k + 1))/(1 - x^(2 k)), {k, m}], {x, 0, m}], x]] (* Michael De Vlieger, Mar 04 2018 *)
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PROG
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(PARI) a(n) = my(s=sqrt(2*n)); sumdiv(n, d, (d % 2) && (d > s)); \\ Michel Marcus, Jan 15 2020
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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