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.

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A139002 Weights of Connes-Moscovici Hopf subalgebra.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 4, 4, 3, 6, 1, 1, 6, 3, 4, 4, 1, 1, 3, 1, 5, 15, 5, 5, 10, 10, 10, 10, 15, 10, 1, 1, 10, 15, 10, 10, 10, 10, 5, 5, 15, 5, 1, 3, 1, 1, 4, 4, 3, 6, 1, 6, 36, 18, 24, 24, 6, 6, 18, 6, 15, 15, 45, 45, 15, 15, 15, 15, 10, 20, 60, 20, 10, 60, 20, 15, 45, 15, 1, 1
Offset: 1

Views

Author

Tom Copeland, May 31 2008

Keywords

Comments

Gives multiplicity for tree shapes in "naturally grown" forests of rooted trees (from file for CM(t) referred to in Broadhurst).
A refinement of the enumeration of the trees of the first few forests in terms of planar rather than nonplanar rooted trees is presented on p. 21 of the Munthe-Kaas and Lundervold paper. - Tom Copeland, Jul 16 2018 (The refinement is presented also in Lundervold and on p. 35 of the Lundervold and Munthe-Kaas paper. - Tom Copeland, Jul 21 2021)
Enumerates the elementary differentials of the Butcher group that Cayley showed are in bijection with these nonplanar rooted trees when considering multivariable vector functions. When considering a scalar function of one independent variable, the associated differentials are no longer in bijection with the planar trees and are enumerated by A139605. Two nonplanar trees are considered equivalent if the branches of one may be rotated about its nodes to match those of the other. - Tom Copeland, Jul 21 2021

Examples

			From _Tom Copeland_, Dec 06 2017: (Start)
The number of distinct rooted tree types, or shapes, with n nodes is given by A000081(n+1), so the multiplicities for the tree shapes of the forests of naturally grown trees given here may be grouped according to A000081. For example, A000081(5)=4 corresponds to the four tree types depicted in Fig. 6 of Mathemagical Forests with four nodes, or vertices, with the four multiplicities (1,1,3,1); A000081(4)=2 corresponds to the two tree types depicted in Fig. 3 with three nodes and the two multiplicities (1,1); A000081(3)=1, with one tree type with two nodes and multiplicity (1); and A000081(2)=1, with one tree type with one node and multiplicity (1). Then the sequence here begins (1)(1)(1,1)(1,1,3,1).
First few rows (with last row reordered according to Fig. 7 of Mathemagical Forests):
  1
  1
  1, 1
  1, 1, 3, 1
  1, 1, 3, 4, 4, 3, 6, 1, 1
This last row corresponds to the one listed in Broadhurst as
  1, 3, 1, 1, 4, 4, 3, 6, 1.
(End)
		

References

  • J. Butcher, Numerical Methods for Ordinary Differential Equations, 3rd Ed., Wiley, 2016, Table 310(II) on p. 165.
  • E. Hairer, C. Lubich, G. Wanner, Geometric Numerical Integration - Structure-Preserving Algorithms for Ordinary Differential Equations, 2nd Ed., Springer, 2006, pp. 52 and 53.

Crossrefs

A206496 is a permutation of this sequence.

Programs

  • Mathematica
    nmax = 7;
    SetAttributes[t, Orderless];
    size[tree_] := Count[tree, _, All];
    lst = {{t[]}};
    forests[0, 0] = {{}}; forests[?Positive, 0] = {}; forests[?Negative, _] = {};
    forests[n_, k_] := forests[n, k] = With[{tree = Flatten[lst][[k]]}, Join[Append[tree] /@ forests[n - size@tree, k], forests[n, k-1]]];
    Do[AppendTo[lst, t @@@ forests[n-1, Length[Flatten@lst]]], {n, 2, nmax}];
    assoc = Association[{# -> 0} & /@ Flatten@lst];
    assoc[t[]] = 1;
    Do[assoc[Insert[tree, t[], Append[Most@p, 1]]] += assoc[tree], {n, 2, nmax}, {tree, lst[[n-1]]}, {p, Position[tree, t]}];
    Last /@ Normal@assoc (* Andrey Zabolotskiy, Mar 15 2024 *)

Formula

The table on p. 364 of Ginocchio contains the Connes-Moscovici weights correlated with associated derivatives D^k. From the relation of this entry to A139605 and A145271, the action of the weighted differentials on an exponential is associated with the operation exp(x g(u)D_u) e^(ut) = e^(t H^{(-1)}(H(u)+x)) with g(x) = 1/D(H(x)) and H^{(-1)} the compositional inverse of H. With H^{(-1)}(x) = -log(1-x), the inverse about x=0 is H(x) = 1-e^(-x), giving g(x) = e^x and the resulting action e^(-t log(1-x)) = (1-x)^(-t) for u=0, an e.g.f. for the unsigned Stirling numbers of the first kind A008275 and A048994. Consequently, summing the Connes-Moscovici weights over each associated derivative gives these Stirling numbers. E.g., the fifth row in the examples reduces to (1+3+1+1) D + (4+4+3) D^2 + 6 D^3 + D^4 = 6 D + 11 D^2 + 6 D^3 + D^4. - Tom Copeland, Jul 14 2021

A181612 Triangle T(n,m) of the coefficients JacobiDC(x,y) = sum_{n>=0} sum_{m=0..n} (-1)^m* T(n,m) *x^(2*n) *y^(2*m)/(2*n)!.

Original entry on oeis.org

1, 1, 1, 5, 6, 1, 61, 107, 47, 1, 1385, 3116, 2142, 412, 1, 50521, 138933, 130250, 45530, 3693, 1, 2702765, 8783986, 10430983, 5353260, 1036715, 33218, 1, 199360981, 747603679, 1074680289, 728130163, 226132303
Offset: 0

Views

Author

R. J. Mathar, Jan 30 2011

Keywords

Examples

			The triangle starts in row n=0 as
1;
1, 1;
5, 6, 1;
61, 107, 47, 1;
1385, 3116, 2142, 412, 1;
50521, 138933, 130250, 45530, 3693, 1;
		

References

  • M. Abramowitz, I. A. Stegun, Handbook of Mathematical Functions, Dover. Section 16.22.

Crossrefs

Cf. A060627, A060628, A181613, A000364 (apparently the column m=0).

Programs

  • Maple
    A181612 := proc(n,m) JacobiDC(z,k) ; coeftayl(%,z=0,2*n) ; (-1)^m*coeftayl(%,k=0,2*m)*(2*n)! ; end proc:
    seq( seq(A181612(n,m),m=0..n),n=0..10) ;
  • Mathematica
    nmax = 8; se = Series[JacobiDC[x, y], {x, 0, 2*nmax}]; t[n_, m_] := Coefficient[se, x, 2*n]*(2*n)! // Coefficient[#, y, m]& // Abs; Table[t[n, m], {n, 0, nmax}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jan 09 2014 *)

Formula

From Peter Bala, Aug 23 2011: (Start)
The elliptic function dc(x,k) (JacobiDC(x,k) in Maple notation) is defined as dn(x,k)/cn(x,k) where dn(x,k) and cn(x,k) are the Jacobian elliptic functions of modulus k. The Taylor expansions begin
dn(x,k) = 1-k^2*x^2/2!+k^2*(4+k^2)*x^4/4!-k^2*(16+44*k^2+k^4)*x^6/6!+...
cn(x,k) = 1-x^2/2!+(1+4*k^2)*x^4/4!-(1+44*k^2+16*k^4)*x^6/6!+... and hence
dc(x,k) = 1+(1-k^2)*x^2/2!+(5-6*k^2+k^4)*x^4/4!+(61-107*k^2+47*k^4-k^6)*x^6/6!+....
The coefficients for cn(x,k) are in A060627. The coefficients of dn(x,k) may be obtained by row reversal of A060627.
The expansion for dc(x,k) can also be obtained directly from that of dn(x,k) since by Jacobi's imaginary transformations we have dc(x,k) = dn(i*x,k'), where the complementary modulus k' is given by k' = sqrt(1-k^2).
By Jacobi's real transformation the reciprocal of dc(x,k) is given by 1/dc(x,k) = dc(x*k,1/k).
The row polynomials of this table can be calculated using nested derivatives as follows (see [Dominici, Theorem 4.1 and Example 4.5]):
Let f(x) = sqrt(1-(1-k^2)*sin^2(x)). Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0.
See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x).
Then the coefficient of x^(2*n)/(2*n)! in the expansion of dc(x,k) is given by (-1)^n*D^(2*n)[f](0).
(End)

A181613 Triangle T(n,m) of the coefficients JacobiNC(x,y) = sum_{n>0} sum_{m=0..n-1} (-1)^m* T(n,m) *x^(2*n) *y^(2*m)/(2*n)!.

Original entry on oeis.org

1, 5, 4, 61, 76, 16, 1385, 2424, 1104, 64, 50521, 113672, 79728, 16832, 256, 2702765, 7432604, 7052528, 2586112, 264448, 1024, 199360981, 647923188, 775638816, 408850432, 85975296, 4205568, 4096, 19391512145, 72718170544, 105138354912, 72490884224, 23551644928, 2939602944, 67162112, 16384
Offset: 1

Views

Author

R. J. Mathar, Jan 30 2011

Keywords

Comments

The column m=0 is apparently A000364.

Examples

			The triangle starts in row n=1 as:
1;
5, 4;
61, 76, 16;
1385, 2424, 1104, 64;
50521, 113672, 79728, 16832, 256;
		

References

  • M. Abramowitz, I. A. Stegun, Handbook of Mathematical Functions, Dover. Section 16.22.
  • H. S. Wall, Analytic Theory of Continued Fractions, Chelsea 1973, p. 374.

Crossrefs

Programs

  • Maple
    A181613 := proc(n,m) JacobiNC(z,k) ; coeftayl(%,z=0,2*n) ; (-1)^m*coeftayl(%,k=0,2*m)*(2*n)! ; end proc:
    seq( seq(A181613(n,m),m=0..n-1),n=1..10) ;
  • Mathematica
    nmax = 8; se = Series[JacobiNC[x, y], {x, 0, 2*nmax}]; t[n_, m_] := Coefficient[se, x, 2*n]*(2*n)! // Coefficient[#, y, m]& // Abs; Table[t[n, m], {n, 1, nmax}, {m, 0, n-1}] // Flatten (* Jean-François Alcover, Jan 10 2014 *)

Formula

From Peter Bala, Aug 23 2011: (Start)
The Taylor expansion of the Jacobian elliptic function cn(u,k) begins
cn(u,k) = 1-u^2/2!+(1+4*k^2)*u^4/4!-(1+44*k^2+16*k^4)*u^6/6!+... - see A060627.
The Taylor expansion of the reciprocal function 1/cn(u,k) can be obtained directly from this by using Jacobi's imaginary transformation
1/cn(u,k) = cn(i*u,sqrt(1-k^2)) [Abramowitz and Stegun, 16.20] to yield
1/cn(u,k) = 1+u^2/2!+(5-4*k^2)*u^4/4!+(61-76*k^2+16*k^4)*u^6/6!+....
The coefficient polynomials R(2*n,k) of this expansion can be calculated as follows (apply [Dominici, Theorem 4.1]):
Let f(x) = sqrt(k^2-cos^2(x)). Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. Then R(2*n,k) = D^(2*n)[f](0).
See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x).
(End)
G.f. 1/(1 - x/(1 - 2^2*(1 - k^2)*x/(1 - 3^2*x/(1 - 4^2*(1 - k^2)*x/(1 - 5^2*x/(1 - ...)))))) = 1 + x + (5 - 4*k^2)*x^2 + (61 - 76*k^2 + 16*k^4)*x^3 + ... (see Wall, 94.19, p. 374).

A190392 E.g.f. A(x) satisfies A'(x) = sin(A(x)) + cos(A(x)).

Original entry on oeis.org

1, 1, 0, -4, -12, 4, 240, 1184, -1008, -59504, -401280, 643136, 38584128, 323581504, -848090880, -51666451456, -509739310848, 2004840714496, 123888658698240, 1386061762251776, -7721141999864832, -483475390212586496, -5974101514137292800, 45231727252157947904
Offset: 1

Views

Author

Vladimir Kruchinin, May 09 2011

Keywords

Comments

Let f(x) be a smooth function. The autonomous differential equation A'(x) = f(A(x)), with initial condition A(0) = 0, is separable and the solution is given by A(x) = inverse function of Integral_{t = 0..x} 1/f(t) dt. The inversion of the integral Integral_{t = 0..x} 1/f(t) dt is most conveniently found by applying [Dominici, Theorem 4.1]. The result is A(x) = Sum_{n>=1} D^(n-1)[f](0)*x^n/n!, where the nested derivative D^n[f](x)is defined recursively as D^0[f](x) = 1 and D^(n+1)[f](x) = (d/dx)(f(x)*D^n[f](x)) for n >= 0. See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x). In the present case we take f(x) = sin(x)+cos(x). - Peter Bala, Aug 27 2011

Crossrefs

Programs

  • Maple
    A := x ; for i from 1 to 35 do sin(A)+cos(A) ; convert(taylor(%,x=0,25),polynom) ; A := int(%,x) ; print(A) ; end do:
    for i from 1 to 25 do printf("%d,", coeftayl(A,x=0,i)*i!) ; end do: # R. J. Mathar, Jun 03 2011
  • Mathematica
    terms = 25; A[] = 0; Do[A[x] = Integrate[Sin[A[x]] + Cos[A[x]], x] + O[x]^terms, terms]; CoefficientList[A[x], x]*Range[0, terms-1]! (* Jean-François Alcover, Feb 21 2013, updated Jan 15 2018 *)
  • Maxima
    g(n):=(-1)^floor(n/2)*1/n!;
    a(n):=T190015_Solve(n,g);

Formula

E.g.f.: A(x) = inverse of Integral_{t = 0..x} 1/(sin(t)+cos(t)) dt = series reversion (x - x^2/2! + 3*x^3/3! - 11*x^4/4! + 57*x^5/5! - ...) = x + x^2/2! - 4*x^4/4! - 12*x^5/5! + .... a(n) = D^(n-1)[sin(x)+cos(x)](0), where the nested derivative operator D^n is defined above. Compare with A012244. -Peter Bala, Aug 27 2011
E.g.f.: A(x) = 2*arctan((sqrt(2)-1)*exp(sqrt(2)*x))-Pi/4. Compare with A028296. - Peter Bala, Sep 02 2011
G.f.: 1/G(0) where G(k) = 1 - 2*x*(k+1)/(1 + 1/(1 - 2*x*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 10 2013.
G.f.: -(1/x)/Q(0), where Q(k)= 2*k+1 - 1/x + (k+1)*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Apr 15 2013
G.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 + (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013
E.g.f. if offset 0: 2^(1/2)/(2^(1/2)*cosh(x*2^(1/2))-sinh(x*2^(1/2))). - Sergei N. Gladkovskii, Nov 10 2013
a(n) ~ -(n-1)! * 2^(3*n/2+1) * sin(n*arctan(Pi/log(3 - 2*sqrt(2)))) / (Pi^2 + log(3 - 2*sqrt(2))^2)^(n/2). - Vaclav Kotesovec, Jan 07 2014

A124796 Coefficients in expansion of powers of the operator "multiplication by f(x) followed by differentiation", in the prime factorization order.

Original entry on oeis.org

1, 1, 1, 1, 0, 3, 0, 1, 1, 1, 0, 6, 0, 0, 0, 1, 0, 7, 0, 4, 0, 0, 0, 10, 0, 0, 1, 1, 0, 4, 0, 1, 0, 0, 0, 25, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 15, 0, 0, 0, 0, 0, 15, 0, 5, 0, 0, 0, 30, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 65, 0, 0, 0, 0, 0, 0, 0, 20, 1, 0, 0, 7, 0, 0, 0, 1, 0, 11, 0, 0, 0, 0, 0, 21, 0, 0, 0, 4, 0
Offset: 1

Views

Author

Max Alekseyev, Nov 29 2006

Keywords

Comments

Let d o f(x) be an operator of multiplication by f(x) followed by differentiation. (d o f)^m = Sum a([k0,k1,...])*((d^0 f)^k0*(d^1 f)^k1*...)*d^(m-k1-2*k2-...) where the sum is taken over all nonnegative integer vectors [k0,k1,...] such that k0+k1+...=m and k1+2*k2+...<=m.
For all k >= 0 it holds that a(2^k) = a(3^k) = 1 and also a(p) = 0 for all primes p > 3. - Alexander Adamchuk, Dec 03 2006 and Antti Karttunen, Feb 28 2023

Examples

			From _Antti Karttunen_, Feb 28 2023: (Start)
For n=6, a(6) = a(2^1 * 3^1) = a([1,1,0,0,0,...]) = a([0,1,0,0,...]) + (1+1)*a([1,0,0,0,...]) + 0 = a(3) + 2*a(1) = 3.
For n=10, a(10) = a(2^1 * 5^1) = a([1,0,1,0,0,0...]) = a([0,0,1,0,0,0,...]) + 2*0 + 1*a([0,1,0,0,0,...]) = a(5) + 0 + 1*a(3) = 1.
For n=20, a(20) = a(2^2 * 5^1) = a([2,0,1,0,0,0...]) = a([1,0,1,0,0,0,...]) + 3*0 + 1*a([1,1,0,0,0,...]) = a(10) + 0 + 1*a(6) = 1+3 = 4.
(End)
		

Crossrefs

Programs

  • PARI
    A124796(n) = if(1==n,1,my(u=primepi(vecmax(factor(n)[, 1]))); if(n%3,0,((1+valuation(n,2)) * A124796(n/3))) + if(n%2,0,(A124796(n/2) + sum(i=3,u,if(n%prime(i),0,(valuation(n,prime(i-1))+1)*A124796((n/2)*prime(i-1)/prime(i))))))); \\ Antti Karttunen, Feb 28 2023

Formula

For n=p0^k0*p1^k1*... where 2=p0
a([k0,k1,0,0,...]) = S(k0+k1+1,k0+1), Stirling number of the 2nd kind, see A008277.

A190015 Triangle T(n,k) for solving differential equation A'(x)=G(A(x)), G(0)!=0.

Original entry on oeis.org

1, 1, 2, 1, 6, 8, 1, 24, 42, 16, 22, 1, 120, 264, 180, 192, 136, 52, 1, 720, 1920, 1248, 540, 1824, 2304, 272, 732, 720, 114, 1, 5040, 15840, 10080, 8064, 18720, 22752, 9612, 7056, 10224, 17928, 3968, 2538, 3072, 240, 1, 40320, 146160, 92160, 70560, 32256, 207360, 249120, 193536, 73728, 61560, 144720, 246816, 101844, 142704, 7936, 51048, 110448, 34304, 8334, 11616, 494, 1
Offset: 0

Author

Vladimir Kruchinin, May 04 2011

Keywords

Comments

For solving the differential equation A'(x)=G(A(x)), where G(0)!=0,
a(n) = 1/n!*sum(pi(i) in P(2*n-1,n), T(n,i)*prod(j=1..n, g(k_j-1))),
where pi(i) is the partition of 2*n-1 into n parts in lexicographic order P(2*n-1,n).
G(x) = g(0)+g(1)*x+g(2)*x^2+...
Examples
A003422 A'(x)=A(x)+1/(1-x)
A000108 A'(x)=1/(1-2*A(x)),
A001147 A'(x)=1/(1-A(x))
A007489 A'(x)=A(x)+x/(1-x)^2+1.
A006351 B'(x)=(1+B(x))/(1-B(x))
A029768 A'(x)=log(1/(1-A(x)))+1.
A001662 B'(x)=1/(1+B(x))
A180254 A'(x)=(1-sqrt(1-4*A(x)))/2
Compare with A145271. There (j')^k = [(d/dx)^j g(x)]^k evaluated at x=0 gives formulas expressed in terms of the coefficients of the Taylor series g(x). If, instead, we express the formulas in terms of the coefficients of the power series of g(x), we obtain a row reversed array for A190015 since the partitions there are in reverse order to the ones here. Simply exchange (j!)^k * (j")^k for (j')^k, where (j")^k = [(d/dx)^j g(x) / j!]^k, to transform from one array to the other. E.g., R^3 g(x) = 1 (0')^1 (1')^3 + 4 (0')^2 (1')^1 (2')^1 + 1 (0')^3 (3')^1 = 1 (O")^1 (1")^3 + 4 (0")^2 (1")^1 2*(2")^1 + 1 (0")^1 3!*(3")^1 = 1 (O")^1 (1")^3 + 8 (0")^2 (1")^1 (2")^1 + 6 (0")^1 (3")^1, the fourth partition polynomial here. - Tom Copeland, Oct 17 2014

Examples

			Triangle begins:
1;
1;
2,1;
6,8,1;
24,42,16,22,1;
120,264,180,192,136,52,1;
720,1920,1248,540,1824,2304,272,732,720,114,1;
5040,15840,10080,8064,18720,22752,9612,7056,10224,17928,3968,2538,3072,240,1;
40320,146160,92160,70560,32256,207360,249120,193536,73728,61560,144720,246816, 101844,142704,7936,51048,110448,34304,8334,11616,494,1;
Example for n=5:
partitions of number 9 into  5 parts in lexicographic order:
[1,1,1,1,5]
[1,1,1,2,4]
[1,1,1,3,3]
[1,1,2,2,3]
[1,2,2,2,2]
a(5) = (24*g(0)^4*g(4) +42*g(0)^3*g(1)*g(3) +16*g(0)^3*g(2)^2 +22*g(0)^2*g(1)^2*g(2) +g(0)*g(1)^4)/5!.
		

Programs

  • Maxima
    /* array of triangle */
    M:[1,1,2,1,6,8,1,24,42,16,22,1,120,264,180,192,136,52,1,720,1920,1248,540,1824,2304,272,732,720,114,1,5040,15840,10080,8064,18720,22752,9612,7056,10224,17928,3968,2538,3072,240,1,40320,146160,92160,70560,32256,207360,249120,193536,73728,61560,144720,246816,101844,142704,7936,51048,110448,34304,8334,11616,494,1];
    /* function of triangle */
    T(n,k):=M[sum(num_partitions(i),i,0,n-1)+k+1];
    /* count number of partitions of n into m parts */
    b(n,m):=if n
    				
  • Maxima
    /* Find triangle */
    Co(n,k):=if k=1  then a(n) else sum(a(i+1)*Co(n-i-1,k-1),i,0,n-k);
    a(n):=if n=1 then 1 else 1/n*sum(Co(n-1,k)*x(k),k,1,n-1);
    makelist(ratsimp(n!*a(n)),n,1,5);
    /* Vladimir Kruchinin, Jun 15 2012 */
    
  • PARI
    serlaplace( serreverse( intformal( 1 / sum(n=0, 9, eval(Str("g"n)) * x^n, x * O(x^9))))) /* Michael Somos, Oct 22 2014 */

A190904 a(n) = Sum_{k=0..n-1} cos(Pi*k/2)*binomial(n-1,k)*a(n-1-k)*a(k) for n > 0, a(0) = 1.

Original entry on oeis.org

1, 1, 1, 0, -3, -12, -27, 0, 441, 3024, 11529, 0, -442827, -4390848, -23444883, 0, 1636819569, 21224560896, 145703137041, 0, -16106380394643, -257991277243392, -2164638920874507, 0, 347592265948756521
Offset: 0

Author

Peter Luschny, Jul 26 2011

Keywords

Programs

  • Maple
    A190904 := proc(n) option remember; `if`(n=0,1,add(((1-irem(k,2))*(-1)^ iquo(k,2))*binomial(n-1,k)*A190904(n-1-k)*A190904(k),k=0..n-1)) end:
  • Mathematica
    a[0] = 1;
    a[n_] := a[n] =
      Sum[Mod[(k+1)^3, 4, -1] Binomial[n-1, k] a[n-k-1] a[k], {k, 0, n-1}];
    Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Jun 24 2019 *)

Formula

Let F(n,x) = Sum_{k=0..n-1} cos(Pi*k*x)*binomial(n-1,k)*F(n-1-k,x)* F(k,x), then
F(n, 0) = n! = A000142(n),
F(n, 1/2) = a(n),
F(n, 1) = 2^n*Euler_{n}(1) = A_{n}(-1) = A155585(n).
a(2*n) = A159601(n)*(-1)^floor((n-1)/2).
a(2*n+1) = A104203(2*n+1).
From Peter Bala, Aug 25 2011: (Start)
The sequence entries may be calculated as follows: Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0. The coefficients in the expansion of D^n[f](x) in powers of f(x) can be found in A145271. Then we have
a(2*n) = D^(2*n)[sqrt(1+sin^2(x))](0)
a(2*n+1) = D^(2*n)[sqrt(1-x^4)](0).
The generating function involves the Jacobian elliptic functions. Define E(u,k) := cn(i*u,k)-i*sn(i*u,k) = 1+u+u^2/2!+(1+k^2)*u^3/3!+(1+4*k^2)*u^4/4!+..., where cn(u,k) and sn(u,k) are Jacobian elliptic functions of modulus k (see A060627 and A060628). Then the e.g.f. A(u) for this sequence is
A(u) := E(u,i) = 1+u+u^2/2!-3*u^4/4!-12*u^5/5!-27*u^6/6!+....
Proof: Using well-known properties of the Jacobian elliptic functions (see for example Abramowitz and Stegun, Chapter 16) we find the generating function A(u) satisfies the differential equation
(d/du)A(u) = dn(i*u,i)*A(u) = 1/2*(A(i*u)+A(-i*u))*A(u), which leads to a recurrence for the coefficients of A(u):
a(n+1) = sum{k=0..floor(n/2)} (-1)^k*binomial(n,2*k)*a(2*k)*a(n-2*k) with a(0) = 1. This recurrence is equivalent to the defining recurrence for this sequence given above.
End proof.
The generating function A(u) satisfies 1/A(u) = A(-u).
Compare entries of this sequence with those of A104203, A159600, A193541 and A193544.
(End)

A253722 Triangle read by rows: coefficients of the partition polynomials for the reciprocal of the derivative of a power series, g(x)= 1/h'(x).

Original entry on oeis.org

1, -2, 4, -3, -8, 12, -4, 16, -36, 9, 16, -5, -32, 96, -54, -48, 24, 20, -6, 64, -240, 216, 128, -27, -144, -60, 16, 30, 24, -7, -128, 576, -720, -320, 216, 576, 160, -108, -96, -180, -72, 40, 36, 28, -8
Offset: 0

Author

Tom Copeland, May 02 2015

Keywords

Comments

This entry contains the integer coefficients of the partition polynomials P(n;h_1,h_2,...,h_(n+1)) for the reciprocal g(x) of the derivative of a power series in terms of the coefficients of the power series; i.e., g(x) = 1/[dh(x)/dx] = 1/[h_1 + 2*h_2 * x + 3*h_3 * x^2 + ...] = sum[n>=0, (h_1)^(-(n+1)) * P(n;h_1,...,h_(n+1)) * x^n].
This is a signed refinement of reversed A181289. See A145271, A133437, and A133314 for relations to compositional and multiplicative inversions.

Examples

			Let h(x) = h_0 + h_1 * x + h_2 * x^2 + ... . Then g(x) = 1/h'(x) = 1/[h_1 + 2*h_2 * x + 3*h_3 * x^2 + ...] = (h_1)^(-1) P(0;h_1) + (h_1)^(-2) * P(1;h_1,h_2) * x + (h_1)^(-3) * P(2;h_1,h_2,h_3) * x^2 + ... , and, with h_n = (n'), the first few partition polynomials are
P(0;..)=  1
P(1;..)= -2 (2')
P(2;..)=  4 (2')^2 - 3 (3')(1')
P(3;..)= -8 (2')^3 + 12 (3')(2')(1') - 4 (4')(1')^2
P(4;..)= 16 (2')^4 - 36 (2')^2(3')(1') + [9 (3')^2 + 16 (4')(2')](1')^2 - 5 (5')(1')^3
P(5;..)= -32 (2')^5 + 96 (2')^3(3')(1') + [-54 (3')^2(2') - 48 (4')(2')^2](1')^2 + [24 (3')(4') + 20 (5')(2')](1')^3 - 6 (6')(1')^4
P(6;..)= 64 (2')^6 - 240 (2')^4(3')(1') + [216 (3')^2(2') + 128 (4')(2')^3](1')^2 - [27 (3')^3 + 144 (4')(3')(2') + 60 (5')(2')^2](1')^3 + [16 (4')^2 + 30 (5')(3') + 24 (6')(2')](1')^4 - 7 (7')(1')^5
		

Crossrefs

Programs

  • Mathematica
    rows[n_] := {{1}}~Join~With[{s = 1/(1 + Sum[(k+1) u[k] x^k, {k, n}] + O[x]^(n+1))}, Table[Coefficient[s, x^k Product[u[t], {t, p}]], {k, n}, {p, Reverse@Sort[Sort /@ IntegerPartitions[k]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Feb 19 2024 *)
  • PARI
    C(v)={my(S=Set(v)); (-1)^(#v)*(#v)!*prod(i=1, #S, my(x=S[i], e=#select(y-> y==x, v)); (x+1)^e/e! )}
    row(n)=[C(Vec(p)) | p<-Vecrev(partitions(n))]
    { for(n=0, 7, print(row(n))) } \\ Andrew Howroyd, Feb 19 2024

Formula

For the partition (1')^e(1)*(2')^e(2)*...*(n')^e(n) in P(m;...), the unsigned integer coefficient is [e(2)+e(3)+...+e(n)]! * [2^e(2)*3^e(3)*...*n^e(n)]/[e(2)!*e(3)!*...*e(n)!] with the sign determined by (-1)^[e(1) + m].
The partitions of P(m;..) are formed by adding one to each index of the partitions of m of Abramowitz and Stegun's partition table (p. 831; in the reversed order) and appending (1')^e(1) as a factor to obtain a partition of 2m.
Row sums are 1,-2,1,0,0,0,... . Row sums of the unsigned coefficients are A003480.

Extensions

Row 7 added by Andrey Zabolotskiy, Feb 19 2024

A339513 Define R_{1}(x)=1, R_{n+1}(x)=(R_n(x)*2*x/(1+x^2))'; then a(n)=R_{n}(1).

Original entry on oeis.org

1, 0, -1, 3, -2, -45, 347, -756, -13031, 184245, -810034, -11404503, 264733177, -1931955480, -21453955777, 796153961091, -8688345850874, -69492467459925, 4300450718587619, -65896562313762012, -307002797419794407, 37668399518087366325
Offset: 1

Author

Luc Rousseau, Dec 07 2020

Keywords

Comments

Let (R_n) be the sequence of rational functions satisfying: R_1(x) = 1; R_{n+1}(x) = (R_n(x) * 2*x/(1+x^2))'. By definition, a(n) = R_n(1).
Applying [Dominici, Theorem 4.1] proves that the e.g.f. of this sequence is the series reversion of log(1+x)/2 + x^2/4 + x/2.

Examples

			R_1(x) = 1,
  so a(1) = R_1(1) = 1.
R_2(x) = (R_1(x)*2*x/(1+x^2))' = (1 * 2*x/(1+x^2))' = 2*(1-x^2)/(1+x^2)^2,
  so a(2) = R_2(1) = 0.
R_3(x) = (R_2(x)*2*x/(1+x^2))' = (2*(1-x^2)/(1+x^2)^2 * 2*x/(1+x^2))' = 4*(1-8*x^2+3*x^4)/(1+x^2)^4, so a(3) = R_3(1) = -1.
		

Crossrefs

Programs

  • PARI
    list_a(nmax)=my(n,r);n=1;r=1;print1(subst(r,x,1),", ");while(n
    				
  • PARI
    my(x='x+O('x^33)); Vec(serlaplace(serreverse(log(1+x)/2 + x^2/4 + x/2))) \\ Joerg Arndt, Dec 22 2020

Formula

a(n) = (Sum_{k=0..n-1} (-1)^k*A214406(n-1,k))/2^(n-1).
a(n) = Sum_{P partition of n-1} A145271(P) * Product_{p part of P} A090932(p)*A075553(p+3).
E.g.f.: series reversion of log(1+x)/2 + x^2/4 + x/2.

A277394 Lagrange inversion, or reversion, for divided power series with odd powers only.

Original entry on oeis.org

1, -1, 10, -1, -280, 56, -1, 15400, -4620, 126, 120, -1, -1401400, 560560, -36036, -17160, 792, 220, -1, 190590400, -95295200, 10090080, 3203200, -126126, -360360, -50050, 1716, 2002, 364, -1
Offset: 1

Author

Tom Copeland, Oct 12 2016

Keywords

Comments

Coefficients for partition polynomials for compositional inversion order-by-order of odd functions, e.g.f.s, or formal Taylor series f(x) = a1 x + a3 x^3/3! + a5 x^5/5! + ... .
The compositional inverse of f(x) is g(x)
= a1^(-1) [1] x
+ a1^(-4) [-1 a3] x^3/3!
+ a1^(-7) [10 a3^2 - 1 a1 a5] x^5/5!
+ a1^(-10)[-280 a3^3 + 56 a1 a3 a5 - a1^2 a7] x^7/7!
+ a1^(-13)[15400 a3^4 - 4620 a1 a3^2 a5 + a1^2 (126 a5^2 + 120 a3 a7) - a1^3 a9] * x^9/9! ... .

Crossrefs

Cf. A133437, A134264, A134685, A133932, A145271, A176740 for other inversion formulas.

Programs

  • Mathematica
    rows[nn_] := With[{s = InverseSeries[x + Sum[a[k] x^(2k+1)/(2k+1)!, {k, nn}] + O[x]^(2nn+2)]}, Table[(2n-1)! Coefficient[s, x^(2n-1) Product[a[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n-1]]]}]];
    rows[5] // Flatten (* Andrey Zabolotskiy, Mar 07 2024 *)

Extensions

Corrected and extended by Andrey Zabolotskiy, Mar 07 2024
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