A328186 Write 1/(1 + sin x) = Product_{n>=1} (1 + f_n x^n); a(n) = denominator(f_n).
1, 1, 6, 6, 120, 360, 5040, 2520, 72576, 1814400, 39916800, 59875200, 1245404160, 21794572800, 1307674368000, 81729648000, 71137485619200, 3201186852864000, 121645100408832000, 12164510040883200, 10218188434341888000, 281000181944401920000, 25852016738884976640000
Offset: 1
Examples
f_n = -1, 1, 1/6, 5/6, 19/120, -47/360, 659/5040, 1837/2520, 7675/72576, -154729/1814400, 3578279/39916800, 3984853/59875200, 95259767/1245404160, ...
Links
- Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.
- Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.
- H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.
- H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239.
- W. Lang, Recurrences for the general problem, 2009.
Programs
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Maple
# Calculates the fractions f_n (choose L much larger than M): PPE := 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/(1 + sin(x)); t0 := series(t1, x, L); f := []; t2 := t0; for n to M do t3 := coeff(t2, x, n); t2 := series(t2/(1 + t3*x^n), x, L); f := [op(f), t3]; end do; end if; [seq(f[n], n = 1 .. nops(f))]; end proc; # Calculates the denominators of f_n: h := map(denom, PPE(100, 40)); # Petros Hadjicostas, Oct 06 2019 by modifying N. J. A. Sloane's program from A170912 and A170913.
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Mathematica
A[m_, n_] := A[m, n] = Which[m == 1, 2*(-1)^n*I^(n + 2)*PolyLog[-(n + 1), -I]/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]]; a /@ Range[1, 55] (* Petros Hadjicostas, Oct 06 2019 using a program by Jean-François Alcover and a formula from A099612 and A099617 *)
Formula
a(2*n + 1) = A170915(2*n + 1) for n >= 0.
Define (A(m,n): n,m >= 1) by A(m=1, n) = 2 * (-1)^n * i^(n + 2) * PolyLog(-(n + 1), -i)/n! for n >= 1 (with i := sqrt(-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 f_n = A(n,n) and thus a(n) = denominator(A(n,n)).
If we write 1 + sin x = Product_{n>=1} (1 + g_n * x^n) and we know (g_n: n >= 1), then f_n = -g_n + Sum_{s|n, s > 1} (1/s) * ((-f_{n/s})^s + (-g_{n/s})^s). This proves of course that f_n = -g_n for n odd.
Comments