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Search: id:A125079
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| A125079 |
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Excess of number of divisors of 2n+1 of form 12k+1, 12k+5 over those of form 12k+7, 12k+11. |
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+0
4
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| 1, 1, 2, 0, 1, 0, 2, 2, 2, 0, 0, 0, 3, 1, 2, 0, 0, 0, 2, 2, 2, 0, 2, 0, 1, 2, 2, 0, 0, 0, 2, 0, 4, 0, 0, 0, 2, 3, 0, 0, 1, 0, 4, 2, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 2, 2, 0, 2, 0, 1, 2, 4, 0, 0, 0, 0, 2, 2, 0, 0, 0, 4, 1, 2, 0, 2, 0, 2, 2, 0, 0, 0, 0, 3, 0, 2, 0, 0, 0, 2, 2, 4, 0, 0, 0, 2, 4, 2, 0, 0, 0, 4, 0, 0
(list; graph; listen)
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OFFSET
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0,3
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REFERENCES
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N. J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 82, Eq. (32.56).
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FORMULA
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Expansion of q^(-1/2)*eta(q^3)^3*eta(q^4)*eta(q^12)/(eta(q)*eta(q^6)^2) in powers of q.
Expansion of phi(-q^3)*psi(-q^3)/(chi(-q)*chi(-q^2)) in powers of q where phi(),psi(),chi() are Ramanujan theta functions.
Euler transform of period 12 sequence [ 1, 1, -2, 0, 1, 0, 1, 0, -2, 1, 1, -2, ...].
a(n)=b(2n+1) where b(n) is multiplicative and b(2^e) = 0^e, b(3^e) = 1, b(p^e) = e+1 if p == 1 (mod 4), b(p^e) = (1+(-1)^e)/2 if p == 3 (mod 4).
a(6n+3)=a(6n+5)=0.
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PROGRAM
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(PARI) {a(n)=if(n<0, 0, n=2*n+1; sumdiv(n, d, kronecker(6, d)*(-1)^(d\12)))}
(PARI) {a(n)=if(n<0, 0, if(n%6==1, n\=3, 1); sumdiv(2*n+1, d, kronecker(-4, d)) )}
(PARI) {a(n)=local(A, p, e); if(n<0, 0, n=2*n+1; A=factor(n); prod( k=1, matsize(A)[1], if(p=A[k, 1], e=A[k, 2]; if(p==2, 0, if(p==3, 1, if(p%4==1, e+1, !(e%2)))))))}
(PARI) {a(n)=local(A); if(n<0, 0, A=x*O(x^n); polcoeff( eta(x^3+A)^3* eta(x^4+A)* eta(x^12+A)/ eta(x+A)/ eta(x^6+A)^2, n))}
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CROSSREFS
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A002175(n)=a(6n). A008441(n)=a(2n)=a(6n+1). A121444(n)=a(6n+2)/2.
Sequence in context: A036579 A139353 A029397 this_sequence A129447 A104597 A072662
Adjacent sequences: A125076 A125077 A125078 this_sequence A125080 A125081 A125082
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KEYWORD
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nonn
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AUTHOR
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Michael Somos, Nov 18 2006
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