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A170915
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Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n); a(n) = denominator(g_n).
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11
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1, 1, 6, 6, 120, 120, 5040, 280, 72576, 362880, 39916800, 11975040, 1245404160, 88957440, 1307674368000, 11675664000, 71137485619200, 1067062284288000, 121645100408832000, 101370917007360000, 10218188434341888000, 5109094217170944000, 25852016738884976640000
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OFFSET
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1,3
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COMMENTS
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The recurrence about (A(m,n): m,n >= 1) in the Formula section follows from Theorem 3 in Gingold et al. (1988); see also Gingold and Knopfmacher (1995, p. 1222). A(m=1,n) equals the n-th coefficient of the Taylor expansion of 1 + sin(x).
If 1 + sin(x) = 1/Product_{n>=1} (1 + f_n * x^n) (inverse power product expansion), then Gingold and Knopfmacher (1995) and Alkauskas (2008, 2009) proved that f_n = -g_n for n odd, and Sum_{s|n} (-g_{n/s})^s/s = -Sum_{s|n} (-f_{n/s})^s/s. [We caution that different authors may use -g_n for g_n, or -f_n for f_n, or both.] We have A328191(n) = numerator(f_n) and A328186(n) = denominator(f_n).
Wolfdieter Lang (see the link below) examined inverse power product expansions both for ordinary g.f.'s and for exponential g.f.'s.
In all cases, we assume the g.f.'s are unital, i.e., the g.f.'s start with a constant 1.
(End)
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LINKS
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FORMULA
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a(2*n+1) = A328186(2*n+1) for n >= 0.
Define (A(m,n): n,m >= 1) by A(m=1,2*n+1) = (-1)^n/(2*n+1)! for n >= 0, A(m=1,2*n) = 0 for n >= 1, A(m,n) = 0 for m > n >= 1 (upper triangular), and A(m,n) = A(m-1,n) - A(m-1,m-1) * A(m,n-m+1) for n >= m >= 2. Then g_n = A(n,n). (End)
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EXAMPLE
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g_n = 1, 0, -1/6, 1/6, -19/120, 19/120, -659/5040, 37/280, -7675/72576, ...
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MAPLE
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# Calculates the fractions g_n (choose L much larger than M):
PPE_sin := proc(L, M)
local t1, t0, g, t2, n, t3;
if L < 2.5*M then print("Choose larger value for L");
else
t1 := 1 + sin(x);
t0 := series(t1, x, L);
g := []; t2 := t0;
for n to M do
t3 := coeff(t2, x, n);
t2 := series(t2/(1 + t3*x^n), x, L);
g := [op(g), t3];
end do;
end if;
[seq(g[n], n = 1 .. nops(g))];
end proc;
# Calculates the denominators of g_n:
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MATHEMATICA
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A[m_, n_] :=
A[m, n] =
Which[m == 1, (1-(-1)^n)*(-1)^Floor[(n-1)/2]/(2*n!), m > n >= 1, 0, True,
A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
a[n_] := Denominator[A[n, n]];
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CROSSREFS
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KEYWORD
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nonn,frac
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AUTHOR
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STATUS
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approved
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