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Search: id:A003406
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| A003406 |
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Expansion of Ramanujan's function R(x) = 1 + Sum_{n >= 1} { x^(n*(n+1)/2) / ((1+x)(1+x^2)(1+x^3)...(1+x^n)) }. (Formerly M0206)
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+0 8
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| 1, 1, -1, 2, -2, 1, 0, 1, -2, 0, 2, 0, -1, -2, 2, 1, 0, -2, 2, -2, 0, 0, 3, 0, -2, -2, 1, 0, 2, 0, 0, 0, -2, 0, 0, 1, 0, 0, 0, 2, -1, 0, -2, -2, 0, 4, 0, 2, -2, 0, -2, -1, 2, 0, -2, 2, 0, 1, 0, 0, 0, 0, -2, 0, 0, 0, 0, -2, 4, 2, -1, 0, 0, -2, -2, -2, 2, 1, 2, 0, 0, 0, 0, -2, 2, 0, 0, -2, 2, -2, -2, 0, 3, 0, 0, 2, 0, 0, 0, -2, 1, -2, 0, -2, 0
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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a(n) = A117192(n) - A117193(n) for n>0, see also A000025. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Mar 03 2006
Ramanujan showed R(x) = 2*sum(n=0,oo, S(x)-P(n,x)) - 2*S(x)*D(x); where P(n,x) = prod(k=1,n, 1+x^k), S(x) = g.f. A000009 = P(oo,x) and D(x) = -1/2+sum(n=1,oo,x^n/(1-x^n)) = -1/2 + g.f. A000005 - Michael Somos.
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REFERENCES
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G. E. Andrews, Ramanujan's "lost" notebook V: Euler's partition identity, Adv. in Math. 61 (1986), no. 2, 156-164; Math. Rev. 87i:11137. [ The expansion in (2.8) is incorrect. ]
G. E. Andrews, Questions and conjectures in partition theory, Amer. Math. Monthly, 93 (1986), 708-711.
G. E. Andrews, Some debts I owe, Seminaire Lotharingien Combinatoire, Paper B42a, Issue 42, 2000.
F. J. Dyson, A walk through Ramanujan's garden, pp. 7-28 of G. E. Andrews et al., editors, Ramanujan Revisited. Academic Press, NY, 1988.
F. J. Dyson, Selected Papers, Am. Math. Soc., 1996, p. 200.
B. Gordon and D. Sinor, Multiplicative properties of eta-products, Number theory, Madras 1987, pp. 173-200, Lecture Notes in Math., 1395, Springer, Berlin, 1989. see page 182. MR1019331 (90k:11050)
S.-Y. Kang, Generalizations of Ramanujan's reciprocity theorem..., J. London Math. Soc., 75 (2007), 18-34.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..2000
G. E. Andrews, Some debts I owe
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FORMULA
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Also R(x)=1+Sum_{n >= 1} (-1)^(n-1)x^n(1-x)(1-x^2)...(1-x^(n-1)).
G.f.=1+sum(x^(n(n+1)/2)/product(1+x^j, j=1..n), n=1..infinity). - Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 30 2006
Define c(24k+1)=A003406(k), c(24k-1)=-2*A003475(k), c(n)=0 otherwise. Then c(n) is multiplicative with c(2^e)=c(3^e)=0^e, c(p^e) = (-1)^(e/2)*(1+(-1)^e)/2 if p == 7, 17 (mod 24), c(p^e) = (1+(-1)^e)/2 if p == 5, 11, 13, 19 (mod 24), c(p^e) = (e+1)*(-1)^(y*e) where p == 1, 23 (mod 24) and p = x^2-72*y^2 . - Michael Somos Aug 17 2006 */
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MAPLE
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g:=1+sum(x^(n*(n+1)/2)/product(1+x^j, j=1..n), n=1..20): gser:=series(g, x=0, 110): seq(coeff(gser, x, n), n=0..104); - Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 30 2006
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PROGRAM
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(PARI) {a(n)=local(t); if(n<0, 0, t=1+O(x^n); polcoeff( sum(k=1, n, t*= if(k>1, x^k-x, x)+ O(x^(n-k+2)), 1), n))} /* Michael Somos Mar 07 2006 */
(PARI) {a(n)=local(t); if(n<0, 0, t=1+O(x^n); polcoeff( sum(k=1, (sqrtint(8*n+1)-1)\2, t*= x^k/(1+x^k) +x*O(x^(n-(k^2-k)/2)), 1), n))} /* Michael Somos Aug 17 2006 */
(PARI) {a(n)=local(A, p, e, x, y); if(n<0, 0, n=24*n+1; A=factor(n); prod(k=1, matsize(A)[1], if(p=A[k, 1], e=A[k, 2]; if(p<5, 0, if(p%24>1&p%24<23, if(e%2, 0, if(p%24==7|p%24==17, (-1)^(e/2), 1)), x=y=0; if(p%24==1, forstep(i=1, sqrtint(p), 2, if(issquare((i^2+p)/2, &y), x=i; break)), for(i=1, sqrtint(p\2), if(issquare(2*i^2+p, &x), y=i; break))); (e+1)*(-1)^( (x+if((x-y)%6, y, -y))/6*e))))))} /* Michael Somos Aug 17 2006 */
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CROSSREFS
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Cf. A005895, A005896.
Sequence in context: A055800 A060572 A163543 this_sequence A107063 A108423 A092339
Adjacent sequences: A003403 A003404 A003405 this_sequence A003407 A003408 A003409
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KEYWORD
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sign,easy,nice
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com).
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