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.

A339100 a(n) = GCD({(2*n-k)*T(n,k)+(k+1)*T(n,k+1), k=0..n}), where T(n,k) stands for A214406 (the second-order Eulerian numbers of type B).

Original entry on oeis.org

1, 6, 1, 60, 1, 42, 1, 120, 1, 66, 1, 5460, 1, 6, 1, 4080, 1, 798, 1, 3300, 1, 138, 1, 10920, 1, 6, 1, 1740, 1, 14322, 1, 8160, 1, 6, 1, 3838380, 1, 6, 1, 270600, 1, 12642, 1, 1380, 1, 282, 1, 371280, 1, 66, 1, 3180, 1, 798, 1, 3480, 1, 354, 1, 567867300
Offset: 1

Views

Author

Luc Rousseau, Nov 23 2020

Keywords

Comments

Define recursively the rational fractions R_n by: R_0(x)=1; R_{n+1}(x) = (R_n(x)*x/(1-x^2))'. 2*a(n) is the maximal integer that can be factored out of the numerator of R'_n -- staying with polynomials with integer coefficients.
Empirical observations: the prime factorizations of the a(n) follow a pattern: the 2-adic valuation of a(n) is the 2-adic valuation of n; the 3-adic valuation of a(n) is (n mod 2); for p a prime >= 5, the p-adic valuation of a(n) is 0 (if p-1 does not divide n), 1 (if p-1 divides n but p does not) or 2 (if both p-1 and p divide n). So, a(n) = 1 when n is odd, and the prime factorizations of a(n) for the first few even n are:
\ p|
\ | 2 3 5 7 11 13 17 19 23 29 31 37 41 43
n \|
---+-------------------------------------------
2 | 1 1 . . . . . . . . . . . .
4 | 2 1 1 . . . . . . . . . . .
6 | 1 1 . 1 . . . . . . . . . .
8 | 3 1 1 . . . . . . . . . . .
10 | 1 1 . . 1 . . . . . . . . .
12 | 2 1 1 1 . 1 . . . . . . . .
14 | 1 1 . . . . . . . . . . . .
16 | 4 1 1 . . . 1 . . . . . . .
18 | 1 1 . 1 . . . 1 . . . . . .
20 | 2 1 2 . 1 . . . . . . . . .
22 | 1 1 . . . . . . 1 . . . . .
24 | 3 1 1 1 . 1 . . . . . . . .
26 | 1 1 . . . . . . . . . . . .
28 | 2 1 1 . . . . . . 1 . . . .
30 | 1 1 . 1 1 . . . . . 1 . . .
32 | 5 1 1 . . . 1 . . . . . . .
34 | 1 1 . . . . . . . . . . . .
36 | 2 1 1 1 . 1 . 1 . . . 1 . .
38 | 1 1 . . . . . . . . . . . .
40 | 3 1 2 . 1 . . . . . . . 1 .
42 | 1 1 . 2 . . . . . . . . . 1

Examples

			In A214406, row number 4 is:
(k=0) (k=1) (k=2) (k=3) (k=4)
  1    112   718   744   105
Now,
(2*4-0)*  1 + (0+1)*112 =  120
(2*4-1)*112 + (1+1)*718 = 2220
(2*4-2)*718 + (2+1)*744 = 6540
(2*4-3)*744 + (3+1)*105 = 4140
(2*4-4)*105 + (4+1)*  0 =  420
The GCD of {120, 2220, 6540, 4140, 420} is 60, so a(4)=60.
		

Crossrefs

Programs

  • Mathematica
    T[n_,k_]:=T[n,k]=If[n==0&&k==0,1,If[n==0||k<0||k>n,0,(4*n-2*k-1)*T[n-1,k-1]+(2*k+1)*T[n-1,k]]]
    A[n_]:=Table[(2*n-k)*T[n,k]+(k+1)*T[n,k+1],{k,0,n}]/.{List->GCD}
    Table[A[n],{n,1,100}]
  • PARI
    r(n)=if(n==0,1,(r(n-1)*x/(1-x^2))')
    a(n)=my(p=(r(n))'*(1-x^2)^(2*n+1)/2);p/factorback(factor(p))
    for(n=1,60,print1(a(n),", "))

A039755 Triangle of B-analogs of Stirling numbers of the second kind.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 13, 9, 1, 1, 40, 58, 16, 1, 1, 121, 330, 170, 25, 1, 1, 364, 1771, 1520, 395, 36, 1, 1, 1093, 9219, 12411, 5075, 791, 49, 1, 1, 3280, 47188, 96096, 58086, 13776, 1428, 64, 1, 1, 9841, 239220, 719860, 618870, 209622, 32340, 2388, 81, 1, 1
Offset: 0

Views

Author

Ruedi Suter (suter(AT)math.ethz.ch)

Keywords

Comments

Let M be an infinite lower triangular bidiagonal matrix with (1,3,5,7,...) in the main diagonal and (1,1,1,...) in the subdiagonal. n-th row = M^n * [1,0,0,0,...]. - Gary W. Adamson, Apr 13 2009
From Peter Bala, Aug 08 2011: (Start)
A type B_n set partition is a partition P of the set {1, 2, ..., n, -1, -2, ..., -n} such that for any block B of P, -B is also a block of P, and there is at most one block, called a zero-block, satisfying B = -B. We call (B, -B) a block pair of P if B is not a zero-block. Then T(n,k) is the number of type B_n set partitions with k block pairs. See [Wang].
For example, T(2,1) = 4 since the B_2 set partitions with 1 block pair are {1,2}{-1,-2}, {1,-2}{-1,2}, {1,-1}{2}{-2} and {2,-2}{1}{-1} (the last two partitions contain a zero block).
(End)
Exponential Riordan array [exp(x), (1/2)*(exp(2*x) - 1)]. Triangle of connection constants for expressing the monomial polynomials x^n as a linear combination of the basis polynomials (x-1)*(x-3)*...*(x-(2*k-1)) of A039757. An example is given below. Inverse array is A039757. Equals matrix product A008277 * A122848. - Peter Bala, Jun 23 2014
T(n, k) also gives the (dimensionless) volume of the multichoose(k+1, n-k) = binomial(n, k) polytopes of dimension n-k with side lengths from the set {1, 3, ..., 1+2*k}. See the column g.f.s and the complete homogeneous symmetric function formula for T(n, k) below. - Wolfdieter Lang, May 26 2017
T(n, k) is the number of k-dimensional subspaces (i.e., sets of fixed points like rotation axes and symmetry planes) of the n-cube. See "Sets of fixed points..." in LINKS section. - Tilman Piesk, Oct 26 2019

Examples

			Triangle T(n,k) begins:
  n\k 0     1       2        3       4       5      6     7    8   9 10 ...
  0:  1
  1:  1     1
  2:  1     4       1
  3:  1    13       9        1
  4:  1    40      58       16       1
  5:  1   121     330      170      25       1
  6:  1   364    1771     1520     395      36      1
  7:  1  1093    9219    12411    5075     791     49     1
  8:  1  3280   47188    96096   58086   13776   1428    64    1
  9:  1  9841  239220   719860  618870  209622  32340  2388   81   1
 10:  1 29524 1205941  5278240 6289690 2924712 630042 68160 3765 100  1
 ... reformatted and extended by _Wolfdieter Lang_, May 26 2017
The sequence of row polynomials of A214406 begins [1, 1+x, 1+8*x+3*x^2, ...]. The o.g.f.'s for the diagonals of this triangle thus begin
1/(1-x) = 1 + x + x^2 + x^3 + ...
(1+x)/(1-x)^3 = 1 + 4*x + 9*x^2 + 16*x^3 + ...
(1+8*x+3*x^2)/(1-x)^5 = 1 + 13*x + 58*x^2 + 170*x^3 + ... . - _Peter Bala_, Jul 20 2012
Connection constants: x^3 = 1 + 13*(x-1) + 9*(x-1)*(x-3) + (x-1)*(x-3)*(x-5). Hence row 3 = [1,13,9,1]. - _Peter Bala_, Jun 23 2014
Complete homogeneous symmetric functions: T(3, 1) = h^{(2)}_2 = 1^2 + 3^2 + 1^1*3^1 = 13. The three 2D polytopes are two squares and a rectangle. T(3, 2) = h^{(3)}_1 = 1^1 + 3^1 + 5^1 = 9. The 1D polytopes are three lines. - _Wolfdieter Lang_, May 26 2017
T(4, 3) = 16 is the number of 3-dimensional subspaces (mirror hyperplanes) of the 4-cube. (These are 4 cubes and 12 cuboids.) See "Sets of fixed points..." in LINKS section. - _Tilman Piesk_, Oct 26 2019
		

Crossrefs

Programs

  • Magma
    [[(&+[(-1)^(k-j)*(2*j+1)^n*Binomial(k, j): j in [0..k]])/( 2^k*Factorial(k)): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 14 2019
    
  • Maple
    A039755 := proc(n,k) if k < 0 or k > n then 0 ; elif n <= 1 then 1; else procname(n-1,k-1)+(2*k+1)*procname(n-1,k) ; end if; end proc:
    seq(seq(A039755(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Oct 30 2009
  • Mathematica
    t[n_, k_] = Sum[(-1)^(k-j)*(2j+1)^n*Binomial[k, j], {j, 0, k}]/(2^k*k!); Flatten[Table[t[n, k], {n, 0, 10}, {k, 0, n}]][[1 ;; 56]]
    (* Jean-François Alcover, Jun 09 2011, after Peter Bala *)
  • PARI
    T(n,k)=if(k<0 || k>n,0,n!*polcoeff(polcoeff(exp(x+y/2*(exp(2*x+x*O(x^n))-1)),n),k))
    
  • Sage
    [[sum((-1)^(k-j)*(2*j+1)^n*binomial(k, j) for j in (0..k))/( 2^k*factorial(k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 14 2019

Formula

E.g.f. row polynomials: exp(x + y/2 * (exp(2*x) - 1)).
T(n,k) = T(n-1,k-1) + (2*k+1)*T(n-1,k) with T(0,k) = 1 if k=0 and 0 otherwise. Sum_{k=0..n} T(n,k) = A007405(n). - R. J. Mathar, Oct 30 2009; corrected by Joshua Swanson, Feb 14 2019
T(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^(k-j)*C(k,j)*(2*j+1)^n.
T(n,k) = (1/(2^k*k!)) * A145901(n,k). - Peter Bala
The row polynomials R(n,x) satisfy the Dobinski-type identity:
R(n,x) = exp(-x/2)*Sum_{k >= 0} (2*k+1)^n*(x/2)^k/k!, as well as the recurrence equation R(n+1,x) = (1+x)*R(n,x)+2*x*R'(n,x). The polynomial R(n,x) has all real zeros (apply [Liu et al., Theorem 1.1] with f(x) = R(n,x) and g(x) = R'(n,x)). The polynomials R(n,2*x) are the row polynomials of A154537. - Peter Bala, Oct 28 2011
Let f(x) = exp((1/2)*exp(2*x)+x). Then the row polynomials R(n,x) are given by R(n,exp(2*x)) = (1/f(x))*(d/dx)^n(f(x)). Similar formulas hold for A008277, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
From Peter Bala, Jul 20 2012: (Start)
The o.g.f. for the n-th diagonal (with interpolated zeros) is the rational function D^n(x), where D is the operator x/(1-x^2)*d/dx. For example, D^3(x) = x*(1+8*x^2+3*x^4)/(1-x^2)^5 = x + 13*x^3 + 58*x^5 + 170*x^7 + ... . See A214406 for further details.
An alternative formula for the o.g.f. of the n-th diagonal is exp(-x/2)*(Sum_{k >= 0} (2*k+1)^(k+n-1)*(x/2*exp(-x))^k/k!).
(End)
From Tom Copeland, Dec 31 2015: (Start)
T(n,m) = Sum_{i=0..n-m} 2^(n-m-i)*binomial(n,i)*St2(n-i,m), where St2(n,k) are the Stirling numbers of the second kind, A048993 (also A008277). See p. 755 of Dolgachev and Lunts.
The relation of this entry's e.g.f. above to that of the Bell polynomials, Bell_n(y), of A048993 establishes this formula from a binomial transform of the normalized Bell polynomials, NB_n(y) = 2^n Bell_n(y/2); that is, e^x exp[(y/2)(e^(2x)-1)] = e^x exp[x*2*Bell.(y/2)] = exp[x(1+NB.(y))] = exp(x*P.(y)), so the row polynomials of this entry are given by P_n(y) = [1+NB.(y)]^n = Sum_{k=0..n} C(n,k) NB_k(y) = Sum_{k=0..n} 2^k C(n,k) Bell_k(y/2).
The umbral compositional inverses of the Bell polynomials are the falling factorials Fct_n(y) = y! / (y-n)!; i.e., Bell_n(Fct.(y)) = y^n = Fct_n(Bell.(y)). Since P_n(y) = [1+2Bell.(y/2)]^n, the umbral inverses are determined by [1 + 2 Bell.[ 2 Fct.[(y-1)/2] / 2 ] ]^n = [1 + 2 Bell.[ Fct.[(y-1)/2] ] ]^n = [1+y-1]^n = y^n. Therefore, the umbral inverse sequence of this entry's row polynomials is the sequence IP_n( y) = 2^n Fct_n[(y-1)/2] = (y-1)(y-3) .. (y-2n+1) with IP_0(y) = 1 and, from the binomial theorem, with e.g.f. exp[x IP.(y)]= exp[ x 2Fct.[(y-1)/2] ] = (1+2x)^[(y-1)/2] = exp[ [(y-1)/2] log(1+2x) ].
(End)
Let B(n,k) = T(n,k)*((2*k)!)/(2^k*k!) and P(n,x) = Sum_{k=0..n} B(n,k)*x^(2*k+1). Then (1) P(n+1,x) = (x+x^3)*P'(n,x) for n >= 0, and (2) Sum_{n>=0} B(n,k)/(n!)*t^n = binomial(2*k,k)*exp(t)*(exp(2*t)-1)^k/4^k for k >= 0, and (3) Sum_{n>=0} t^n* P(n,x)/(n!) = x*exp(t)/sqrt(1+x^2-x^2*exp(2*t)). - Werner Schulte, Dec 12 2016
From Wolfdieter Lang, May 26 2017: (Start)
G.f. column k: x^k/Product_{j=0..k} (1 - (1+2*j)*x), k >= 0.
T(n, k) = h^{(k+1)}_{n-k}, the complete homogeneous symmetric function of degree n-k of the k+1 symbols a_j = 1 + 2*j, j = 0, 1, ..., k. (End)
With p(n, x) = Sum_{k=0..n} A001147(k) * T(n, k) * x^k for n >= 0 holds:
(1) Sum_{i=0..n} p(i, x)*p(n-i, x) = 2^n*(Sum_{k=0..n} A028246(n+1, k+1)*x^k);
(2) p(n, -1/2) = (n!) * ([t^n] sqrt(2 / (1 + exp(-2*t)))). - Werner Schulte, Feb 16 2024

A085527 a(n) = (2n+1)^n.

Original entry on oeis.org

1, 3, 25, 343, 6561, 161051, 4826809, 170859375, 6975757441, 322687697779, 16679880978201, 952809757913927, 59604644775390625, 4052555153018976267, 297558232675799463481, 23465261991844685929951, 1977985201462558877934081, 177482997121587371826171875
Offset: 0

Views

Author

N. J. A. Sloane, Jul 05 2003

Keywords

Comments

a(n) is the determinant of the zigzag matrix Z(n) (see A088961). - Paul Boddington, Nov 03 2003
a(n) is also the number of rho-labeled graphs with n edges. A graph with n edges is a rho-labeled graph if there exists a one-to-one mapping from its vertex set to {0,1,...,2n} such that every edge receives as a label the absolute difference of its end-vertices and the edge labels are x1,x2,...,xn where xi=i or xi=2n+1-i. - Christian Barrientos and Sarah Minion, Feb 20 2015
a(n) is the number of nodes in the canonical automaton for the affine Weyl group of types B_n and C_n. - Tom Edgar, May 12 2016
a(n) is the number of rooted (at an edge) 2-trees with n+2 edges. See also A052750. - Nikos Apostolakis, Dec 05 2018

References

  • Anders Björner and Francesco Brenti, Combinatorics of Coxeter groups. Graduate Texts in Mathematics, 231. Springer, New York, 2005.

Crossrefs

Programs

Formula

E.g.f.: sqrt(2)/(2*(1+LambertW(-2*x))*sqrt(-x/LambertW(-2*x))). - Vladeta Jovovic, Oct 16 2004
For r = 0, 1, 2, ..., the e.g.f. for the sequence whose n-th term is (2*n+1)^(n+r) can be expressed in terms of the function U(z) = Sum_{n >= 0} (2*n+1)^(n-1)*z^(2*n+1)/(2^n*n!). See A214406 for details. In the present case, r = 0, and the resulting e.g.f. is 1/z*U(z)/(1 - U(z)^2) taken at z = sqrt(2*x). - Peter Bala, Aug 06 2012
a(n) = [x^n] 1/(1 - (2*n+1)*x). - Ilya Gutkovskiy, Oct 10 2017
a(n) = (-2)^n * D(2*n + 1), where D(n) is the determinant of the n X n matrix M with elements M(j, k) = cos(Pi*j*k/n). See the Zhi-Wei Sun, Petrov link. - Peter Luschny, Sep 19 2021
a(n) ~ exp(1/2) * 2^n * n^n. - Vaclav Kotesovec, Dec 05 2021
Series reversion of (1 - x)^2 * log(1/(1 - x)) begins x + 3*x^2/2! + 25*x^3/3! + 343*x^4/4! + 6561*x^5/5! + .... - Peter Bala, Sep 27 2023
a(n) = Product_{k=1..n} tan(k*Pi/(1+2*n))^(2*n). - Chai Wah Wu, Nov 10 2024

A034940 Number of rooted labeled triangular cacti with 2n+1 nodes (n triangles).

Original entry on oeis.org

1, 3, 75, 5145, 688905, 152193195, 50174679555, 23089081640625, 14140034726843025, 11119632520038117075, 10920803043967635894075, 13100477280449146440878025, 18849023772776126861572265625, 32038907667175368299033846026875, 63516199119599233704934379969701875
Offset: 0

Views

Author

Christian G. Bower, Oct 15 1998

Keywords

Examples

			E.g. a(3) = 5!! 7^3 = (1*3*5) * 343 = 5145.
From _Peter Bala_, Jul 31 2012: (Start)
Relation with rows of A214406: F(x) := A(exp(x)).
(d/dx)^1(F) = F/(1-F^2)
(d/dx)^2(F)) = F*(1 + F^2)/(1 - F^2)^3
(d/dx)^3(F)) = F*(1 + 8*F^2 + 3*F^4)/(1 - F^2)^5
(d/dx)^4(F)) = F*(1 + 33*F^2 + 71*F^4 + 15*F^6)/(1 - F^2)^7
(End)
		

References

  • F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Camb. 1998, p. 307. (4.2.44)

Crossrefs

Programs

Formula

a(n) = b(2*n+1) where e.g.f. of b satisfies B(x)=x*exp(B(x)^2/2).
The closed form a(n) = (2n-1)!! (2n+1)^n can be obtained from the generating function. - Noam D. Elkies, Dec 16 2002
From Peter Bala, Jul 31 2012: (Start)
E.g.f. A(x) = series reversion of x*exp(-1/2*x^2) = sum {n >= 0} a(n)*x^(2*n+1)/(2*n+1)! = x + 3*x^3/3! + 75*x^5/5! + .... The Lagrange inversion formula gives a(n) = (2*n+1)^n*(2*n)!/(2^n*n!).
A(x)^2 = T(x^2), where T denotes the tree function T(x) := sum {n >= 1} n^(n-1)*x^n/n!. A(x)^r = sum {n >= 0} r*(2*n+r)^(n-1)*x^(2*n+r)/(2^n*n!).
x = A(x)*exp(-1/2*A(x)^2). dA/dx = exp(1/2*A^2)/(1-A^2).
Let the function F(x) = A(exp(x)). Then dF/dx = F/(1-F^2). More generally, (d/dx)^(n+1)(F) is a rational function in F(x) given by (d/dx)^(n+1)(F) = F*R(n,F^2)/(1-F^2)^(2*n+1), where R(n,x) is the n-th row generating polynomial of A214406.
(End)

Extensions

a(10) corrected by Jean-François Alcover, May 13 2013
a(12)-a(14) from Alois P. Heinz, Jul 08 2015

A085528 a(n) = (2*n+1)^(n+1).

Original entry on oeis.org

1, 9, 125, 2401, 59049, 1771561, 62748517, 2562890625, 118587876497, 6131066257801, 350277500542221, 21914624432020321, 1490116119384765625, 109418989131512359209, 8629188747598184440949, 727423121747185263828481, 65273511648264442971824673
Offset: 0

Views

Author

N. J. A. Sloane, Jul 05 2003

Keywords

Comments

a(n) is the number of polynomials of degree at most n with integer coefficients all having absolute value <= n.
a(n-1) is the number of nodes in the canonical automaton for the affine Weyl group of type D_n. - Tom Edgar, May 12 2016

References

  • Anders Björner and Francesco Brenti, Combinatorics of Coxeter groups. Graduate Texts in Mathematics, 231. Springer, New York, 2005.

Crossrefs

Programs

Formula

From Peter Bala, Aug 06 2012: (Start)
E.g.f.: d/dx{(2*x/T(2*x))^(1/2)*1/(1 - T(2*x))} = 1 + 9*x + 125*x^2/2! + ..., where T(x) is the tree function sum {n >= 1} n^(n-1)*x^n/n! of A000169.
For r = 0, 1, 2, ... the e.g.f. for the sequence (2*n+1)^(n+r) can be expressed in terms of the function U(z) = sum {n >= 0} (2*n+1)^(n-1)*z^(2*n+1)/(2^n*n!). See A214406 for details. In the present case, r = 1, and the resulting e.g.f. is 1/z*U(z)*(1 + U(z)^2 )/(1 - U(z)^2)^3 taken at z = sqrt(2*x).
(End)
Sum_{n>=0} (-1)^n/a(n) = A253299. - Amiram Eldar, Jun 25 2021

A099753 a(n) = (2*n+1)^(n+2).

Original entry on oeis.org

1, 27, 625, 16807, 531441, 19487171, 815730721, 38443359375, 2015993900449, 116490258898219, 7355827511386641, 504036361936467383, 37252902984619140625, 2954312706550833698643, 250246473680347348787521, 22550116774162743178682911, 2154025884392726618070214209
Offset: 0

Views

Author

Kari Lajunen (Kari.Lajunen(AT)Welho.com), Nov 11 2004

Keywords

Crossrefs

Programs

  • GAP
    List([0..30], n-> (2*n+1)^(n+2)); # G. C. Greubel, Sep 03 2019
  • Magma
    [(2*n+1)^(n+2): n in [0..30]]; // G. C. Greubel, Sep 03 2019
    
  • Maple
    seq((2*n+1)^(n+2), n=0..30); # G. C. Greubel, Sep 03 2019
  • Mathematica
    Table[(2*n+1)^(n+2), {n,0,30}] (* G. C. Greubel, Sep 03 2019 *)
  • PARI
    vector(30, n, (2*n-1)^(n+1)) \\ G. C. Greubel, Sep 03 2019
    
  • Sage
    [(2*n+1)^(n+2) for n in (0..30)] # G. C. Greubel, Sep 03 2019
    

Formula

From Peter Bala, Aug 06 2012: (Start)
E.g.f.: d^2/dx^2{(2*x/T(2*x))^(3/2)*1/(1 - T(2*x))} = 1 + 27*x + 625*x^2/2! + ..., where T(x) is the tree function sum {n >=1} n^(n-1)*x^n/n! of A000169.
For r = 0, 1, 2, ..., the e.g.f. for the sequence (2*n+1)^(n+r) can be expressed in terms of the function U(z) = sum {n >= 0} (2*n+1)^(n-1)*z^(2*n+1)/(2^n*n!). See A214406 for details. In the present case, r = 2, and the resulting e.g.f. is 1/z*U(z)*(1 + 8*U(z)^2 + 3*U(z)^4)/(1 - U(z)^2)^5 taken at z = sqrt(2*x).
(End)

Extensions

Terms a(13) onward added by G. C. Greubel, Sep 03 2019

A288875 Triangle read by rows. The rows give the coefficients of the numerator polynomials for the o.g.f.s of the diagonal sequences of triangle A028338.

Original entry on oeis.org

1, 1, 1, 3, 8, 1, 15, 71, 33, 1, 105, 744, 718, 112, 1, 945, 9129, 14542, 5270, 353, 1, 10395, 129072, 300291, 191384, 33057, 1080, 1, 135135, 2071215, 6524739, 6338915, 2033885, 190125, 3265, 1, 2027025, 37237680, 150895836, 204889344, 103829590, 18990320, 1038780, 9824, 1
Offset: 0

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Author

Wolfdieter Lang, Jul 21 2017

Keywords

Comments

The Sheffer triangle A028338 of the type (1/sqrt(1-2*x), -(1/2)*log(1 - 2*x)) is called here |S1hat[2,1]|. The o.g.f. of the sequence of diagonal d, d >= 0 is D(d, t) = Sum_{m=0..d} A028338(d+m, m)*t^m. The e.g.f. of these o.g.f.s is taken as ED(y,t) := Sum_{d >= 0} D(d, t)*y^(d+1)/(d+1)!.
This e.g.f. is found to be ED(y,t) = 1 - sqrt(1 - 2*x(t;y)), where x = x(t;y) is the compositional inverse of y = y(t;x) = x*(1 - t*(-log(1-2*x)/(2*x))) = x + t*log(1-2*x)/2. The o.g.f.s are then D(d, t) = P(d, t)/(1 - t)^(2*d+1), with the row polynomials P(d, t) = Sum_{m=0..d} T(d, m)*t^m, d >= 0.
This computation was inspired by an article of P. Bala (see a link under, e.g., A112007) for Sheffer triangles of the Jabotinsky type (1, F(x)). There Sheffer is called exponential Riordan, and the diagonals are labeled by n = d+1, n >= 1.

Examples

			The triangle T(n, m) begins:
n\m      0        1         2         3         4        5       6    7  8 ...
0:       1
1:       1        1
2:       3        8         1
3:      15       71        33         1
4:     105      744       718       112         1
5:     945     9129     14542      5270       353        1
6:   10395   129072    300291    191384     33057     1080       1
7:  135135  2071215   6524739   6338915   2033885   190125    3265    1
8: 2027025 37237680 150895836 204889344 103829590 18990320 1038780 9824  1
...
		

Crossrefs

Programs

  • Mathematica
    De[d_, t_] := Sum[A028338[d+m, m] t^m, {m, 0, d}]; A028338[n_, k_] := SeriesCoefficient[Times @@ Table[x+i, {i, 1, 2n-1, 2}], {x, 0, k}]; P[n_, x_] := De[n, x] (1-x)^(2n+1); T[n_, m_] := Coefficient[P[n, x], x, m]; Table[T[n, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 24 2017 *)
    T[n_,m_]:=Sum[(-1)^(i-n+m)*Binomial[2*n+1,n-m-i]*(1/(2^i*i!)*Sum[(-1)^(i-j)*Binomial[i,j]*(2*j+1)^(n+i),{j,0,i}]),{i,0,n-m}];Flatten[Table[T[n,m],{n,0,8},{m,0,n}]] (* Detlef Meya, Dec 18 2023, after Peter Bala from A214406 *)

Formula

T(n, m) = [x^m] P(n, x), with the numerator polynomial of the o.g.f. of the diagonal n (main diagonal n=0) D(n, x) = P(n, x)/(1-x)^(2*n+1). See a comment above.
T(n, m) = Sum_{i=0..n-m} ( (-1)^(i-n+m)*binomial(2*n+1,n-m-i)*(1/(2^i*i!))*Sum_{j=0..i} (-1)^(i-j)*binomial(i,j)*(2*j+1)^(n+i) ). - Detlef Meya, Dec 18 2023, after Peter Bala from A214406.

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

Views

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.
Showing 1-8 of 8 results.