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 1-8 of 8 results.

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

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)

A170911 Write exp(-x) = Product_{n>=1} (1 + g_n x^n); a(n) = denominator(g_n).

Original entry on oeis.org

1, 2, 3, 8, 5, 72, 7, 128, 81, 800, 11, 13824, 13, 6272, 30375, 32768, 17, 419904, 19, 20480000, 750141, 247808, 23, 1528823808, 15625, 1384448, 1594323, 5035261952, 29, 30233088000000, 31, 2147483648, 235782657, 37879808, 1313046875, 240734712102912, 37, 189267968
Offset: 1

Views

Author

N. J. A. Sloane, Jan 30 2010

Keywords

Examples

			-1, 1/2, 1/3, 3/8, 1/5, 13/72, 1/7, 27/128, 8/81, 91/800, 1/11, ...
		

Crossrefs

Cf. A170910 (numerators).

Programs

  • Maple
    L:=100; t1:=exp(-x); 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;

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

A220420 Express the Sum_{n>=0} p(n)*x^n, where p(n) is the partition function, as a product Product_{k>=1} (1 + a(k)*x^k).

Original entry on oeis.org

1, 2, 1, 4, 1, 0, 1, 14, 1, -4, 1, -8, 1, -16, 1, 196, 1, -54, 1, -92, 1, -184, 1, 144, 1, -628, 1, -1040, 1, -2160, 1, 41102, 1, -7708, 1, -12932, 1, -27592, 1, 54020, 1, -98496, 1, -173720, 1, -364720, 1, 853624, 1, -1341970, 1, -2383916, 1, -4918536, 1
Offset: 1

Views

Author

Michel Marcus, Dec 14 2012

Keywords

Comments

This is the PPE (power product expansion) of A000041.
When n is odd, a(n) = 1.
When n is even, a(n) = 2, 4, 0, 14, -4, -8, -16, 196, -54, -92, -184, 144, -628, -1040, -2160, 41102, ...
Alkauskas (2016, Problem 3, p. 3) conjectured that a(8*k+2), a(8*k+4), and a(8*k+6) are all negative, and a(8*k) is positive for k >= 1. [This statement is not wholly true for k = 0.] - Petros Hadjicostas, Oct 07 2019

Crossrefs

Programs

  • Mathematica
    terms = 55; sol[0] = {};
    sol[m_] := sol[m] = Join[sol[m - 1], If[OddQ[m], {a[m] -> 1}, First @ Solve[Thread[Table[PartitionsP[n], {n, 0, m}] == CoefficientList[ (Product[1 + a[n]*x^n, {n, 1, m}] /. sol[m - 1]) + O[x]^(m + 1), x]]]]];
    Array[a, terms] /. sol[terms] (* Jean-François Alcover, Dec 06 2018, corrected Oct 03 2019 *)
    (* Second program: *)
    A[m_, n_] := A[m, n] = Which[m == 1, PartitionsP[n], 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] (* Jean-François Alcover, Oct 03 2019, using the formula given by Petros Hadjicostas *)
  • PARI
    a(m) = {default(seriesprecision, m+1); ak = vector(m); pol = 1 / eta(x + x * O(x^m)); ak[1] = polcoeff(pol, 1); for (k=2, m, pol = taylor(pol / (1+ak[k-1]*x^(k-1)), x); ak[k] = polcoeff(pol, k, x);); for (k=1, m, print1(ak[k], ", "););}

Formula

From Petros Hadjicostas, Oct 04 2019: (Start)
Define (A(m,n): n,m >= 1) by A(m=1,n) = p(n) = A000041(n) 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).]
a(n) = Sum_{s|n} s/n + Sum_{s|n, s > 1} (-a(n/s))^s/s. [Eq. (1) in Alkauskas (2008, 2009).]
(End)

A170918 a(n) = numerator of the coefficient c(n) of x^n in (tan x)/Product_{0 < k < n} 1 + c(k)*x^k, n = 1, 2, 3, ...

Original entry on oeis.org

1, -1, 7, -14, 54, -1112, 6574, -48488, 1143731, -14813072, 16252211, -3500388967, 125127865048, -158589803803, 33133618166566, -30512906279732, 4378989933312913, -330336346477870319, 1981395373839282068, -251479418962683770473, 79893293800974935213, -31493610597939643431532
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. A170919 (denominators), 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
    t1:=tan(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;
    g1:=map(numer,g);
    g2:=map(denom,g);
  • PARI
    t=tan(x+O(x)^25); vector(#t,n,c=polcoef(t,n);t/=1+c*x^n;numerator(c)) \\ M. F. Hasler, May 07 2022

Extensions

Following a suggestion from Ilya Gutkovskiy, name corrected so that it matches the data by M. F. Hasler, May 07 2022

A354016 Product_{n>=1} (1 + x^n/n!)^a(n) = exp(x).

Original entry on oeis.org

1, 1, -2, 9, -24, 70, -720, 5985, -39200, 337176, -3628800, 40907790, -479001600, 6128488080, -87104969952, 1318070979225, -20922789888000, 354250929192160, -6402373705728000, 121882099274319384, -2432849766865689600, 51041047393559059200
Offset: 1

Views

Author

Ilya Gutkovskiy, May 14 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nn = 22; f[x_] := Product[(1 + x^n/n!)^a[n], {n, 1, nn}]; sol = SolveAlways[0 == Series[f[x] - Exp[x], {x, 0, nn}], x]; Table[a[n], {n, 1, nn}] /. sol // Flatten
    a[1] = 1; a[n_] := a[n] = (n - 1)! ((-1)^n + Sum[d (-d!)^(-n/d) a[d], {d, Divisors[n] ~ Complement ~ {1, n}}]); Table[a[n], {n, 1, 22}]

Formula

a(1) = 1; a(n) = (n-1)! * ((-1)^n + Sum_{d|n, 1 < d < n} d * (-d!)^(-n/d) * a(d)).
Showing 1-8 of 8 results.