cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 20 results. Next

A290261 Write 1 - x/(1-x) as an inverse power product 1/(1 + a(1)*x) * 1/(1 + a(2)*x^2) * 1/(1 + a(3)*x^3) * 1/(1 + a(4)*x^4) * ...

Original entry on oeis.org

1, 2, 2, 6, 6, 10, 18, 54, 54, 114, 186, 334, 630, 1314, 2106, 5910, 7710, 15642, 27594, 57798, 97902, 207762, 364722, 712990, 1340622, 2778930, 4918482, 10437702, 18512790, 37500858, 69273666, 154021590, 258155910, 535004610, 981288906
Offset: 1

Views

Author

Gus Wiseman, Jul 24 2017

Keywords

Comments

The initial terms are given on page 1234 of Gingold, Knopfmacher, 1995.
From Petros Hadjicostas, Oct 04 2019: (Start)
In Section 3 of Gingold and Knopfmacher (1995), it is proved that, if f(z) = Product_{n >= 1} (1 + g(n))*z^n = 1/(Product_{n >= 1} (1 - h(n))*z^n), then g(2*n - 1) = h(2*n - 1) and Sum_{d|n} (1/d)*h(n/d)^d = -Sum_{d|n} (1/d)*(-g(n/d))^d. The same results were proved more than ten years later by Alkauskas (2008, 2009). [If we let a(n) = -g(n), then Alkauskas works with f(z) = Product_{n >= 1} (1 - a(n))*z^n; i.e., a(2*n - 1) = -h(2*n - 1) etc.]
The PPE of 1 - x/(1-x) is given in A220418, which is also studied in Gingold and Knopfmacher (1995) starting at p. 1223.
(End)

Crossrefs

Programs

  • Mathematica
    nn=20;Solve[Table[Expand[SeriesCoefficient[Product[1/(1+a[k]x^k),{k,n}],{x,0,n}]]==-1,{n,nn}],Table[a[n],{n,nn}]][[1,All,2]]
    (* Second program: *)
    A[m_, n_] := A[m, n] = Which[m == 1, 2^(n-1), m > n >= 1, 0, True, A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1] ];
    a[n_] := A[n, n];
    a /@ Range[1, 55] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

Formula

a(n) = Sum_t (-1)^v(t) where the sum is over all enriched p-trees of weight n (see A289501 for definition) and v(t) is the number of nodes (branchings and leaves) in t.
From Petros Hadjicostas, Oct 04 2019: (Start)
a(n) satisfies Sum_{d|n} (1/d)*(-a(n/d))^d = -(2^n - 1)/n. Thus, a(n) = Sum_{d|n, d>1} (1/d)*(-a(n/d))^d + (2^n - 1)/n.
a(2*n - 1) = A220418(2*n - 1) for n >= 1 because A220418 gives the PPE of 1 - x/(1-x).
Define (A(m,n): n,m >= 1) by A(m=1,n) = 2^(n-1) 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 a(n) = A(n,n). [Theorem 3 in Gingold et al. (1988).]
(End)

A220418 Express 1 - x - x^2 - x^3 - x^4 - ... as product (1 + g(1)*x) * (1 + g(2)*x^2) *(1 + g(3)*x^3) * ... and use a(n) = - g(n).

Original entry on oeis.org

1, 1, 2, 3, 6, 8, 18, 27, 54, 84, 186, 296, 630, 1008, 2106, 3711, 7710, 12924, 27594, 48528, 97902, 173352, 364722, 647504, 1340622, 2382660, 4918482, 9052392, 18512790, 33361776, 69273666, 127198287, 258155910, 475568220, 981288906, 1814542704, 3714566310
Offset: 1

Views

Author

Michel Marcus, Dec 14 2012

Keywords

Comments

This is the PPE (power product expansion) of A153881 (with offset 0).
When p is prime, a(p) = (2^p-2)/p (A064535).
From Petros Hadjicostas, Oct 04 2019: (Start)
This sequence appears as an example in Gingold and Knopfmacher (1995) starting at p. 1223.
In Section 3 of Gingold and Knopfmacher (1995), it is proved that, if f(z) = Product_{n >= 1} (1 + g(n))*z^n = 1/(Product_{n >= 1} (1 - h(n))*z^n), then g(2*n - 1) = h(2*n - 1) and Sum_{d|n} (1/d)*h(n/d)^d = -Sum_{d|n} (1/d)*(-g(n/d))^d. The same results were proved more than ten years later by Alkauskas (2008, 2009). [If we let a(n) = -g(n), then Alkauskas works with f(z) = Product_{n >= 1} (1 - a(n))*z^n; i.e., a(2*n - 1) = -h(2*n - 1) etc.]
The PPE of 1/(1 - x - x^2 - x^3 - x^4 - ...) is given in A290261, which is also studied in Gingold and Knopfmacher (1995, p. 1234).
(End)
The number of terms in the Zassenhaus formula exponent of order n, as computed by the algorithm by Casas, Murua & Nadinic, is equal to a(n) at least for n = 2..24. - Andrey Zabolotskiy, Apr 09 2023

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i<1, 1,
          b(n, i-1)+a(i)*b(n-i, min(n-i, i)))
        end:
    a:= proc(n) option remember; 2^n-b(n, n-1) end:
    seq(a(n), n=1..40);  # Alois P. Heinz, Jun 22 2018
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0 || i < 1, 1, b[n, i - 1] + a[i]*b[n - i, Min[n - i, i]]];
    a[n_] := a[n] = 2^n - b[n, n - 1] ;
    Array[a, 40] (* Jean-François Alcover, Jul 09 2018, after Alois P. Heinz *)
  • PARI
    a(m) = {default(seriesprecision, m+1); gk = vector(m); pol = 1 + sum(n=1, m, -x^n); gk[1] = polcoeff( pol, 1); for (k=2, m, pol = taylor(pol/(1+gk[k-1]*x^(k-1)), x); gk[k] = polcoeff(pol, k, x);); for (k=1, m, print1(-gk[k], ", "););}

Formula

g(1) = -1 and for k > 1, g(k) satisfies Sum_{d|k} (1/d)*(-g(k/d))^d = (2^k - 1)/k, where a(k) = -g(k). - Gevorg Hmayakyan, Jun 05 2016 [Corrected by Petros Hadjicostas, Oct 04 2019. See p. 1224 in Gingold and Knopfmacher (1995).]
From Petros Hadjicostas, Oct 04 2019: (Start)
a(2*n - 1) = A290261(2*n - 1) for n >= 1 because A290261 gives the PPE of 1/(1 - x - x^2 - x^3 - ...) = (1 - x)/(1 - 2*x).
Define (A(m,n): n,m >= 1) by A(m=1,n) = -1 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 a(n) = A(n,n). [Theorem 3 in Gingold et al. (1988).]
(End)

Extensions

Name edited by Petros Hadjicostas, Oct 04 2019

A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)*x^k) = 1 - h*x/(1-x).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 10, 17, 20, 17, 10, 4, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 5, 16, 36, 57, 66, 57, 36, 16, 5, 1
Offset: 1

Views

Author

Gevorg Hmayakyan, Jun 01 2016

Keywords

Comments

The a(k,m) form a table where each row has k-1 elements starting from 2 and the a(1,1) = 1.

Examples

			a{1}(h) = h,
a{2}(h) = h,
a{3}(h) = h^2 + h,
a{4}(h) = h^3 + h^2 + h,
a{5}(h) = h^4 + 2*h^3 + 2*h^2 + h,
a{6}(h) = h^5 + 2*h^4 + 2*h^3 + 2*h^2 + h,
a{7}(h) = h^6 + 3*h^5 + 5*h^4 + 5*h^3 + 3*h^2 + h,
a{8}(h) = h^7 + 3*h^6 + 6*h^5 + 7*h^4 + 6*h^3 + 3*h^2 + h,
a{9}(h) = h^8 + 4*h^7 + 9*h^6 + 13*h^5 + 13*h^4 + 9*h^3 + 4*h^2 + h
...
and the corresponding a(k,m) table is:
  1,
  1,
  1,  1,
  1,  1,  1,
  1,  2,  2,  1,
  1,  2,  2,  2,  1,
  1,  3,  5,  5,  3,  1,
  1,  3,  6,  7,  6,  3,  1,
  1,  4,  9, 13, 13,  9,  4,  1,
  ...
a(7,3) = 5 because there are six strict trees contributing positive one {{5,1},1}, {{4,2},1}, {{4,1},2}, {{3,2},2}, {4,{2,1}}, {{3,1},3} and there is one strict tree contributing negative one {4,2,1}. - _Gus Wiseman_, Nov 14 2016
		

Crossrefs

Programs

  • Maple
    with(ListTools), with(numtheory), with(combinat);
    L := product(1-a[k]*x^k, k = 1 .. 600);
    S := Flatten([seq(-h, i = 1 .. 100)]);
    Sabs := Flatten([seq(i, i = 1 .. 100)]);
    seq(assign(a[i] = solve(coeff(L, x^i) = `if`(is(i in Sabs), S[Search(i, Sabs)], 0), a[i])), i = 1 .. 20);
    map(coeffs, [seq(simplify(a[i]), i = 1 .. 20)]);
  • Mathematica
    strictrees[n_Integer?Positive]:=Prepend[Join@@Function[ptn,Tuples[strictrees/@ptn]]/@Select[IntegerPartitions[n],And[Length[#]>1,UnsameQ@@#]&],n];
    Table[Sum[(-1)^(Count[tree,,{0,Infinity}]-1),{tree,Select[strictrees[n],Length[Flatten[{#}]]===m&]}],{n,1,9},{m,1,n-1/.(0->1)}] (* _Gus Wiseman, Nov 14 2016 *)
    (* second program *)
    A[m_, n_] :=
      A[m, n] =
       Which[m == 1, -h, m > n >= 1, 0, True,
        A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
    a[n_] := Expand[-A[n, n]];
    a /@ Range[1, 25] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

Formula

a(k,m) = a(k, k-m).
For prime p: Sum_{m = 1..p-1} a(p, m) = (2^p - 2)/p.
a{k}(h) satisfies Sum_{d|k} (1/d)*(a{k/d}(h))^d = ((h+1)^k - 1)/k. [Corrected by Petros Hadjicostas, Oct 04 2019]
For prime p: a{p}(h) = ((h+1)^p - h^p - 1)/p.
See A273873 for the definition of strict tree. Then a(n,m) = Sum_t (-1)^{v(t)-1} where the sum is over all strict trees of weight n with m leaves, and v(t) is the number of nodes in t (including the leaves, which are positive integers). See example 2 and the first Mathematica program. - Gus Wiseman, Nov 14 2016

A170912 Write cos(x) = Product_{n >= 1} (1 + g_n*x^(2*n)); a(n) = numerator(g_n).

Original entry on oeis.org

-1, 1, 7, 131, 1843, 97261, 4683059, 1331727679, 568285777, 9521655609199, 175554688130609, 11334988388673161, 3457026400678609391, 6594042537777612027841, 249248595232521829462213, 268938575250382935485761673113, 3929672369519648081411955883, 4719016202742955262333630268611
Offset: 1

Views

Author

N. J. A. Sloane, Jan 30 2010

Keywords

Examples

			-1/2, 1/24, 7/360, 131/13440, 1843/453600, 97261/47900160, ...
		

Crossrefs

Cf. A170913.

Programs

  • Maple
    t1:=cos(x);
    L:=100;
    t0:=series(t1,x,L):
    g:=[]; M:=40; t2:=t0:
    for n from 1 to M do
    t3:=coeff(t2,x,n); t2:=series(t2/(1+t3*x^n),x,L); g:=[op(g),t3];
    od:
    g;
    h:=[seq(g[2*n],n=1..nops(g)/2)];
    h1:=map(numer,h);
    h2:=map(denom,h);
  • Mathematica
    A[m_, n_] :=
      A[m, n] =
       Which[m == 1, (-1)^n/(2*n)!, m > n >= 1, 0, True,
        A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
    a[n_] := Numerator[A[n, n]];
    a /@ Range[1, 55] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

A170913 Write cos(x) = Product_{n>=1} (1 + g_n*x^(2*n)); a(n) = denominator(g_n).

Original entry on oeis.org

2, 24, 360, 13440, 453600, 47900160, 5448643200, 2988969984000, 3126159036000, 101370917007360000, 4390627842881280000, 552984315270266880000, 393839317506450816000000, 1465809349094778175488000000, 129517997955171415349760000000, 263130836933693530167218012160000000
Offset: 1

Views

Author

N. J. A. Sloane, Jan 30 2010

Keywords

Examples

			-1/2, 1/24, 7/360, 131/13440, 1843/453600, 97261/47900160, ...
		

Crossrefs

Cf. A170912.

Programs

  • Maple
    t1:=cos(x);
    L:=100;
    t0:=series(t1, x, L):
    g:=[]; M:=40; t2:=t0:
    for n from 1 to M do
    t3:=coeff(t2, x, n); t2:=series(t2/(1+t3*x^n), x, L); g:=[op(g), t3];
    od:
    g;
    h:=[seq(g[2*n], n=1..nops(g)/2)];
    h1:=map(numer, h);
    h2:=map(denom, h);
  • Mathematica
    A[m_, n_] :=
      A[m, n] =
       Which[m == 1, (-1)^n/(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]];
    a /@ Range[1, 55] (* Petros Hadjicostas, Oct 04 2019, courtesy of  Jean-François Alcover *)

A353583 Numerators of coefficients c(n) in product expansion of 1 + tan x = Product_{k>=1} 1 + c(k)*x^k.

Original entry on oeis.org

1, 0, 1, -1, 7, -7, 199, -71, 484, -368, 187909, -610103, 2068657, -63614, 1530164189, -1715846683, 7628902283, -125125345078, 9521826231889, -17921564328719, 291162274608871, -47147385565688, 552647133893696333, -36898601487519532, 4761064630028162378
Offset: 1

Views

Author

M. F. Hasler, May 07 2022

Keywords

Comments

See A353584 for the denominators, and A353586 for the analog for (tan x)/x.

Examples

			1 + tan x = (1 + x)(1 + 1/3*x^3)(1 - 1/3*x^4/3)(1 + 7/15*x^5)(1 - 7/15*x^5)(...),
and this sequence lists the numerators of (1, 0, 1/3, -1/3, 7/15, -7/15, ...).
		

Crossrefs

Cf. A353584 (denominators), A353586 / A353587 (similar for (tan x)/x).
Cf. A170918 / A170919 for a variant.

Programs

  • PARI
    t=1+tan(x+O(x)^29); vector(#t-1,n,c=polcoef(t,n);t/=1+c*x^n;numerator(c))

A170915 Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n); a(n) = denominator(g_n).

Original entry on oeis.org

1, 1, 6, 6, 120, 120, 5040, 280, 72576, 362880, 39916800, 11975040, 1245404160, 88957440, 1307674368000, 11675664000, 71137485619200, 1067062284288000, 121645100408832000, 101370917007360000, 10218188434341888000, 5109094217170944000, 25852016738884976640000
Offset: 1

Views

Author

N. J. A. Sloane, Jan 30 2010

Keywords

Comments

From Petros Hadjicostas, Oct 06 2019: (Start)
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)

Examples

			g_n = 1, 0, -1/6, 1/6, -19/120, 19/120, -659/5040, 37/280, -7675/72576, ...
		

Crossrefs

Numerators are in A170914.

Programs

  • Maple
    # 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:
    h1 := map(denom, PPE_sin(100, 40)); # Petros Hadjicostas, Oct 06 2019 by modifying N. J. A. Sloane's program from A170912 and A170913.
  • Mathematica
    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]];
    a /@ Range[1, 55] (* Petros Hadjicostas, Oct 06 2019, courtesy of Jean-François Alcover *)

Formula

From Petros Hadjicostas, Oct 07 2019: (Start)
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)

A170914 Write 1 + sin x = Product_{n>=1} (1 + g_n * x^n); a(n) = numerator(g_n).

Original entry on oeis.org

1, 0, -1, 1, -19, 19, -659, 37, -7675, 40043, -3578279, 1123009, -95259767, 7091713, -85215100151, 832857559, -4180679675171, 63804880881241, -6399968826052559, 5697831990097981, -478887035449041839, 252737248941887573, -1123931378903214542099, 35703551772944759
Offset: 1

Views

Author

N. J. A. Sloane, Jan 30 2010

Keywords

Comments

From Petros Hadjicostas, Oct 06 2019: (Start)
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)

Examples

			g_n = 1, 0, -1/6, 1/6, -19/120, 19/120, -659/5040, 37/280, -7675/72576, ...
		

Crossrefs

Cf. Denominators are in A170915.

Programs

  • Maple
    # 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 numerators of g_n:
    h1 := map(numer, PPE_sin(100, 40)); # Petros Hadjicostas, Oct 06 2019 by modifying N. J. A. Sloane's program from A170912 and A170913.
  • Mathematica
    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_] := Numerator[A[n, n]];
    a /@ Range[1, 55] (* Petros Hadjicostas, Oct 06 2019, courtesy of Jean-François Alcover *)

Formula

From Petros Hadjicostas, Oct 07 2019: (Start)
a(2*n+1) = -A328191(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)

A157164 1/product(1 - a(n)*(x^n)/n!, n=1..infinity) = 1 + sum((x^k)/k, k=1..infinity) = 1 - log(1-x).

Original entry on oeis.org

1, -1, -1, -8, -16, -74, -526, -6768, -30024, -291072, -2805408, -29134896, -374664720, -5276228736, -67692115440, -1404936248064, -16610113920768, -301439595923712, -5110200130727808, -103584959322338304
Offset: 1

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Author

Wolfdieter Lang Aug 10 2009

Keywords

Examples

			Recurrence I: a(4) = 6 - 4*a(1)*a(3) - 6*a(2)^2 -12*a(1)^2*a(2) - 24*a(1)^4 = 6-4*(-1)-6-12*(-1) = -8.
Recurrence II: a(4) = -6*(1+1+ 2*(-1/2)^2) + 0!*6 - 1*(8+3) + 2*6 = -8.
Recurrence II (rewritten): a(4) = -6*(1+2*(-1/2)^2) +1 = -8.
		

Crossrefs

A137852 (with different signs) for exp(x) = 1/product(1 - a(n)*(x^n)/n!, n=1..infinity).

Formula

Recurrence I: With P(n,m) the set of partitions of n with m parts:
a(n)= (n-1)! - sum(sum(M1(n;vec(e))*product(a(j)^e(j),j=1..m), p=(1^e(1),...,n^e(n)) from P(n,m)), m=2..n) for n>=2, with sum(j*e(j),j=1..n)=n, sum(e(j),j=1..n)=m for the partition p of n with m parts. Input a(1)=0!=1. See the array A008284(n,m) for the cardinalities of the sets P(n,m). The M1 numbers are given, for any partition p=(1^e(1),...,n^e(n)) from P(n,m) by M1(n;vec(e)):=n/product(j!^e(j), j=1..n). See the array A036038.
Recurrence II: a(n)= -(n-1)!*((1+(-1)^n) + sum(d*(a(d)/d!)^(n/d),d|n with 1=2; a(1)=0!=1. The set P(n,m) has been defined in recurrence I above. M2(n;vec(e):=n!/product(e(j)!, j=1..n).
Recurrence II (rewritten, due to email from V. Jovovic, Mar 10 2009. See the link for the general derivation):
a(n) = -(n-1)!*sum(d*(a(d)/d!)^(n/d),d|n with 1<=dA089064(n)=[1,0,1,1,8,26,194,1142,...].

A170919 a(n) = denominator of the coefficient c(n) of x^n in (tan x)/Product_{k=1..n-1} 1 + c(k)*x^k, n = 1, 2, 3, ...

Original entry on oeis.org

1, 1, 3, 3, 5, 45, 105, 315, 2835, 14175, 5775, 467775, 6081075, 2837835, 212837625, 70945875, 3618239625, 97692469875, 206239658625, 9280784638125, 1031198293125, 142924083427125, 322279795963125, 101111706320625, 136968913284328125, 161872352063296875
Offset: 1

Views

Author

N. J. A. Sloane, Jan 30 2010

Keywords

Examples

			1, -1, 7/3, -14/3, 54/5, -1112/45, 6574/105, -48488/315, 1143731/2835, ...
		

Crossrefs

Cf. A170918 (numerators), A170910-A170917.
Cf. A353583 / A353584 for power product expansion of 1 + tan x.
Cf. A353586 / A353587 for power product expansion of (tan x)/x.

Programs

  • Maple
    L := 28: g := NULL:
    t := series(tan(x), x, L):
    for n from 1 to L-2 do
       c := coeff(t, x, n);
       t := series(t/(1 + c*x^n), x, L);
       g := g, c;
    od: map(denom, [g]); # Based on Maple in A170918. - Peter Luschny, Oct 05 2019

Extensions

Following a suggestion from Ilya Gutkovskiy, name corrected so that it matches the data by Peter Luschny, May 12 2022
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