A156919 -id:A156919 - cp's OEIS frontend

A156920 Triangle of the normalized A142963 and A156919 sequences.

1, 1, 1, 1, 5, 1, 1, 15, 18, 1, 1, 37, 129, 58, 1, 1, 83, 646, 877, 179, 1, 1, 177, 2685, 8030, 5280, 543, 1, 1, 367, 10002, 56285, 82610, 29658, 1636, 1, 1, 749, 34777, 335162, 919615, 756218, 159742, 4916, 1

Offset: 0

Johannes W. Meijer, Feb 20 2009

The originator sequences are A142963 and A156919.
The Flower Triangle seems to be an appropriate name for the triangular array of this sequence. The zero patterns of the Flower Polynomials of the first, see A156921, the second, see A156925, the third, see A156927, and the fourth kind, see A156933, look like flowers.
The first Maple program generates the Flower Triangle sequence.
The second program generates the Right Hand Columns sequences and the third one generates the Left Hand Column sequences. For an explanation of these two algorithms see A142963.

			The first few rows of the triangle are:
  [1]
  [1, 1]
  [1, 5, 1]
  [1, 15, 18, 1]
  [1, 37, 129, 58, 1]
  [1, 83, 646, 877, 179, 1]

Shi-Mei Ma, Some combinatorial sequences associated with context-free grammars, arXiv:1208.3104v2 [math.CO]. - From N. J. A. Sloane, Aug 21 2012

Originator sequences A142963, A156919.
Related sequences A156921, A156925, A156927, A156933.
Left hand column sequences A050488, A142965, A142966, A142968.
Right hand column sequences A000340, A156922, A156923, A156924.
Row sums A014307(n+1).

A156920 := proc(n,m): if n=m then 1; elif m=0 then 1 ; elif m<0 or m>n then 0; else (m+1)*procname(n-1, m)+(2*n-2*m+1)*procname(n-1, m-1) ; end if; end proc: seq(seq(A156920(n, m), m=0..n), n=0..8);
RHCnr:=5; RHCmax:=10; RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1),p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx,x,k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr);
LHCnr:=5; LHCmax:=10: LHCend:=LHCnr+LHCmax: for k from LHCnr to LHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!)*x^p/(1-4*x)^(p+1),p=0..k)/ (((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); for n from 0 to nmax do d[n]:= coeff(fx,x,n)/2^n end do: LHC[n]:=d[LHCnr-1] end do: a:=n-> LHC[n]: seq(a(n), n=LHCnr..LHCend-1);

T[, 0] = 1; T[n, n_] = 1; T[n_, m_] := T[n, m] = (m + 1)*T[n - 1, m] + (2*n - 2*m + 1)*T[n - 1, m - 1];
Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017 *)

T(n,m) = (m+1)*T(n-1,m) + (2*n-2*m+1)*T(n-1,m-1) with T(n,m=0) = 1 and T(n,n) = 1, n>=0 and 0 <= m <= n.
From Peter Bala, Jul 22 2012: (Start)
T(n,k) = 1/(2^(n-k))*A156919(n,k).
E.g.f.: 1 + t*x + (t+t^2)*x^2/2! + (t+5*t^2+t^3)*x^3/3! + ... = sqrt(E(x,2*t)), where E(x,t) = (1-t)*exp(x*t)/(exp(x*t)-t*exp(x)) = 1 + t*x + (t+t^2)*x^2/2! + (t+4*t^2+t^3)*x^3/3! + ... is the e.g.f. for the Eulerian numbers A008292.
The row polynomials R(n,x) satisfy 1/sqrt(1-2*x)*(x*d/dx)^n(1/sqrt(1-2*x)) = R(n,x)/(1-2*x)^(n+1). (End)

Minor edits by Johannes W. Meijer, Sep 28 2011

A185412 Triangle T(n,m) read by rows: the matrix product A130595 * A156919.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 33, 33, 1, 1, 112, 378, 112, 1, 1, 353, 2938, 2938, 353, 1, 1, 1080, 18987, 44912, 18987, 1080, 1, 1, 3265, 111051, 520523, 520523, 111051, 3265, 1, 1, 9824, 612820, 5131040, 9998182, 5131040, 612820, 9824, 1

Offset: 0

Paul Barry, Jan 26 2011

Row sums are A165968(n+1).

Because A130595 is the inverse of the Pascal triangle A007318, we have A007318 *(this lower triangular matrix) = A156919.

			Triangle begins in row n=0 with columns 0 <= m <= n:
  1;
  1,    1;
  1,    8,      1;
  1,   33,     33,       1;
  1,  112,    378,     112,       1;
  1,  353,   2938,    2938,     353,       1;
  1, 1080,  18987,   44912,   18987,    1080,      1;
  1, 3265, 111051,  520523,  520523,  111051,   3265,    1;
  1, 9824, 612820, 5131040, 9998182, 5131040, 612820, 9824, 1;

A156919 := proc(n,m) if n=m then 1; elif m=0 then 2^n ; elif m<0 or m>n then 0; else 2*(m+1)*procname(n-1,m)+(2*n-2*m+1)*procname(n-1,m-1) ; end if; end proc:
A130595 := proc(n,m) (-1)^(n+m)*binomial(n,m) ; end proc:
A185412 := proc(n,m) local a,j; a := 0 ; for j from m to n do a := a+A130595(n,j)*A156919(j,m) ; end do: a ; end proc: # R. J. Mathar, Feb 03 2011

G.f.: 1/(1+x-xy-2x/(1-3xy/(1+x-4x/(1-5xy/(1+x-6x/(1-7xy/(1+x-8x/(1-9xy/(1+ ... (continued fraction).

A000182 Tangent (or "Zag") numbers: e.g.f. tan(x), also (up to signs) e.g.f. tanh(x).

Original entry on oeis.org

1, 2, 16, 272, 7936, 353792, 22368256, 1903757312, 209865342976, 29088885112832, 4951498053124096, 1015423886506852352, 246921480190207983616, 70251601603943959887872, 23119184187809597841473536, 8713962757125169296170811392, 3729407703720529571097509625856

Offset: 1

N. J. A. Sloane

Number of Joyce trees with 2n-1 nodes. Number of tremolo permutations of {0,1,...,2n}. - Ralf Stephan, Mar 28 2003
The Hankel transform of this sequence is A000178(n) for n odd = 1, 12, 34560, ...; example: det([1, 2, 16; 2, 16, 272, 16, 272, 7936]) = 34560. - Philippe Deléham, Mar 07 2004
a(n) is the number of increasing labeled full binary trees with 2n-1 vertices. Full binary means every non-leaf vertex has two children, distinguished as left and right; labeled means the vertices are labeled 1,2,...,2n-1; increasing means every child has a label greater than its parent. - David Callan, Nov 29 2007
From Micha Hofri (hofri(AT)wpi.edu), May 27 2009: (Start)
a(n) was found to be the number of permutations of [2n] which when inserted in order, to form a binary search tree, yield the maximally full possible tree (with only one single-child node).
The e.g.f. is sec^2(x)=1+tan^2(x), and the same coefficients can be manufactured from the tan(x) itself, which is the e.g.f. for the number of trees as above for odd number of nodes. (End)
a(n) is the number of increasing strict binary trees with 2n-1 nodes. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
For relations to alternating permutations, Euler and Bernoulli polynomials, zigzag numbers, trigonometric functions, Fourier transform of a square wave, quantum algebras, and integrals over and in n-dimensional hypercubes and over Green functions, see Hodges and Sukumar. For further discussion on the quantum algebra, see the later Hodges and Sukumar reference and the paper by Hetyei presenting connections to the general combinatorial theory of Viennot on orthogonal polynomials, inverse polynomials, tridiagonal matrices, and lattice paths (thereby related to continued fractions and cumulants). - Tom Copeland, Nov 30 2014
The Zigzag Hankel transform is A000178. That is, A000178(2*n - k) = det( [a(i+j - k)]{i,j = 1..n} ) for n>0 and k=0,1. - _Michael Somos, Mar 12 2015
a(n) is the number of standard Young tableaux of skew shape (n,n,n-1,n-2,...,3,2)/(n-1,n-2,n-3,...,2,1). - Ran Pan, Apr 10 2015
For relations to the Sheffer Appell operator calculus and a Riccati differential equation for generating the Meixner-Pollaczek and Krawtchouk orthogonal polynomials, see page 45 of the Feinsilver link and Rzadkowski. - Tom Copeland, Sep 28 2015
For relations to an elliptic curve, a Weierstrass elliptic function, the Lorentz formal group law, a Lie infinitesimal generator, and the Eulerian numbers A008292, see A155585. - Tom Copeland, Sep 30 2015
Absolute values of the alternating sums of the odd-numbered rows (where the single 1 at the apex of the triangle is counted as row #1) of the Eulerian triangle, A008292. The actual alternating sums alternate in sign, e.g., 1, -2, 16, -272, etc. (Even-numbered rows have alternating sums always 0.) - Gregory Gerard Wojnar, Sep 28 2018
The sequence is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 1]. - Allen Stenger, Aug 03 2020
From Peter Bala, Dec 24 2021: (Start)
Conjectures:
1) The sequence taken modulo any integer k eventually becomes periodic with period dividing phi(k).
2) The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k, except when p = 2, n = 1 and k = 1 or 2.
3) For i >= 1 define a_i(n) = a(n+i). The Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i >= 1 the expansion of exp(Sum_{n >= 1} a_i(n)*x^n/n) has integer coefficients. For an example, see A262145.(End)

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + 1/3*x^3 + 2/15*x^5 + 17/315*x^7 + 62/2835*x^9 + O(x^11).
tanh(x) = x - 1/3*x^3 + 2/15*x^5 - 17/315*x^7 + 62/2835*x^9 - 1382/155925*x^11 + ...
(sec x)^2 = 1 + x^2 + 2/3*x^4 + 17/45*x^6 + ...
a(3)=16 because we have: {1, 3, 2, 5, 4}, {1, 4, 2, 5, 3}, {1, 4, 3, 5, 2},
  {1, 5, 2, 4, 3}, {1, 5, 3, 4, 2}, {2, 3, 1, 5, 4}, {2, 4, 1, 5, 3},
  {2, 4, 3, 5, 1}, {2, 5, 1, 4, 3}, {2, 5, 3, 4, 1}, {3, 4, 1, 5, 2},
  {3, 4, 2, 5, 1}, {3, 5, 1, 4, 2}, {3, 5, 2, 4, 1}, {4, 5, 1, 3, 2},
  {4, 5, 2, 3, 1}. - _Geoffrey Critzer_, May 19 2013

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Index entries for "core" sequences
Index entries for sequences related to boustrophedon transform
Index entries for sequences related to Bernoulli numbers.

A350972 is essentially the same sequence.
a(n)=2^(n-1)*A002105(n). Apart from signs, 2^(2n-2)*A001469(n) = n*a(n).
Cf. A001469, A002430, A036279, A000364 (secant numbers), A000111 (secant-tangent numbers), A024283, A009764. First diagonal of A059419 and of A064190.
Cf. A009006, A009725, A029584, A012509, A009123, A009567.
Equals A002425(n) * 2^A101921(n).
Equals leftmost column of A162005. - Johannes W. Meijer, Jun 27 2009
Cf. A258880, A258901. Cf. A002105, A080635, A234797.
Cf. A019538, A110501.

series(tan(x),x,40);
with(numtheory): a := n-> abs(2^(2*n)*(2^(2*n)-1)*bernoulli(2*n)/(2*n));
A000182_list := proc(n) local T,k,j; T[1] := 1;
for k from 2 to n do T[k] := (k-1)*T[k-1] od;
   for k from 2 to n do
       for j from k to n do
           T[j] := (j-k)*T[j-1]+(j-k+2)*T[j] od od;
seq(T[j], j=1..n)  end:
A000182_list(15);  # Peter Luschny, Apr 02 2012

Table[ Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}], {n, 0, 7}] (* Victor Adamchik, Oct 05 2005 *)
v[1] = 2; v[n_] /; n >= 2 := v[n] = Sum[ Binomial[2 n - 3, 2 k - 2] v[k] v[n - k], {k, n - 1}]; Table[ v[n]/2, {n, 15}] (* Zerinvary Lajos, Jul 08 2009 *)
Rest@ Union[ Range[0, 29]! CoefficientList[ Series[ Tan[x], {x, 0, 30}], x]] (* Harvey P. Dale, Oct 19 2011; modified by Robert G. Wilson v, Apr 02 2012 *)
t[1, 1] = 1; t[1, 0] = 0; t[n_ /; n > 1, m_] := t[n, m] = m*(m+1)*Sum[t[n-1, k], {k, m-1, n-1}]; a[n_] := t[n, 1]; Table[a[n], {n, 1, 15}]  (* Jean-François Alcover, Jan 02 2013, after A064190 *)
a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, m! SeriesCoefficient[ Tan[x], {x, 0, m}]]]; (* Michael Somos, Mar 12 2015 *)
a[ n_] := If[ n < 1, 0, ((-16)^n - (-4)^n) Zeta[1 - 2 n]]; (* Michael Somos, Mar 12 2015 *)
Table[2 PolyGamma[2n - 1, 1/2]/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
a[ n_] := a[n] = If[ n < 2, Boole[n == 1], Sum[Binomial[2 n - 2, 2 k - 1] a[k] a[n - k], {k, n - 1}]]; (* Michael Somos, Aug 02 2018 *)
a[n_] := (2^(2*n)*(2^(2*n) - 1)*Abs[BernoulliB[2*n]])/(2*n); a /@  Range[20]  (* Stan Wagon, Nov 21 2022 *)

a(n):=sum(sum(binomial(k,r)*sum(sum(binomial(l,j)/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(j-2*i)^(2*n),i,0,floor((j-1)/2))*(-1)^(l-j),j,1,l)*(-1)^l*binomial(r+l-1,r-1),l,1,2*n)*(-1)^(1-r),r,1,k)/k,k,1,2*n); /* Vladimir Kruchinin, Aug 23 2010 */

a[n]:=if n=1 then 1 else 2*sum(sum(binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j),k,1,j),j,1,n-1);
makelist(a[n],n,1,30); /* Tani Akinari, Sep 20 2023 */

{a(n) = if( n<1, 0, ((-4)^n - (-16)^n) * bernfrac(2*n) / (2*n))};

{a(n) = my(an); if( n<2, n==1, an = vector(n, m, 1); for( m=2, n, an[m] = sum( k=1, m-1, binomial(2*m - 2, 2*k - 1) * an[k] * an[m-k])); an[n])}; /* Michael Somos */

{a(n) = if( n<1, 0, (2*n - 1)! * polcoeff( tan(x + O(x^(2*n + 2))), 2*n - 1))}; /* Michael Somos */

{a(n) = my(X=x+x*O(x^n),Egf); Egf = x*sum(m=0,n, prod(k=1,m, tanh(2*k*X))); (n-1)!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */

/* Continued Fraction for the e.g.f. tan(x), from Paul D. Hanna: */
{a(n)=local(CF=1+O(x)); for(i=1, n, CF=1/(2*(n-i+1)-1-x^2*CF)); (2*n-1)!*polcoeff(x*CF, 2*n-1)}

/* O.g.f. Sum_{n>=1} a(n)*x^n, from Paul D. Hanna Feb 05 2013: */
{a(n)=polcoeff( x+2*x*sum(m=1, n, x^m*prod(k=1, m, (2*k-1)^2/(1+(2*k-1)^2*x +x*O(x^n))) ), n)}

# The objective of this implementation is efficiency.
# n -> [0, a(1), a(2), ..., a(n)] for n > 0.
def A000182_list(n):
    T = [0 for i in range(1, n+2)]
    T[1] = 1
    for k in range(2, n+1):
        T[k] = (k-1)*T[k-1]
    for k in range(2, n+1):
        for j in range(k, n+1):
            T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
    return T
print(A000182_list(100)) # Peter Luschny, Aug 07 2011

from sympy import bernoulli
def A000182(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023

# Algorithm of L. Seidel (1877)
# n -> [a(1), ..., a(n)] for n >= 1.
def A000182_list(len) :
    R = []; A = {-1:0, 0:1}; k = 0; e = 1
    for i in (0..2*len-1) :
        Am = 0; A[k + e] = 0; e = -e
        for j in (0..i) : Am += A[k]; A[k] = Am; k += e
        if e > 0 : R.append(A[i//2])
    return R
A000182_list(15) # Peter Luschny, Mar 31 2012

E.g.f.: log(sec x) = Sum_{n > 0} a(n)*x^(2*n)/(2*n)!.
E.g.f.: tan x = Sum_{n >= 0} a(n+1)*x^(2*n+1)/(2*n+1)!.
E.g.f.: (sec x)^2 = Sum_{n >= 0} a(n+1)*x^(2*n)/(2*n)!.
2/(exp(2x)+1) = 1 + Sum_{n>=1} (-1)^(n+1) a(n) x^(2n-1)/(2n-1)! = 1 - x + x^3/3 - 2*x^5/15 + 17*x^7/315 - 62*x^9/2835 + ...
a(n) = 2^(2*n) (2^(2*n) - 1) |B_(2*n)| / (2*n) where B_n are the Bernoulli numbers (A000367/A002445 or A027641/A027642).
Asymptotics: a(n) ~ 2^(2*n+1)*(2*n-1)!/Pi^(2*n).
Sum[2^(2*n + 1 - k)*(-1)^(n + k + 1)*k!*StirlingS2[2*n + 1, k], {k, 1, 2*n + 1}]. - Victor Adamchik, Oct 05 2005
a(n) = abs[c(2*n-1)] where c(n)= 2^(n+1) * (1-2^(n+1)) * Ber(n+1)/(n+1) = 2^(n+1) * (1-2^(n+1)) * (-1)^n * Zeta(-n) = [ -(1+EN(.))]^n = 2^n * GN(n+1)/(n+1) = 2^n * EP(n,0) = (-1)^n * E(n,-1) = (-2)^n * n! * Lag[n,-P(.,-1)/2] umbrally = (-2)^n * n! * C{T[.,P(.,-1)/2] + n, n} umbrally for the signed Euler numbers EN(n), the Bernoulli numbers Ber(n), the Genocchi numbers GN(n), the Euler polynomials EP(n,t), the Eulerian polynomials E(n,t), the Touchard / Bell polynomials T(n,t), the binomial function C(x,y) = x!/[(x-y)!*y! ] and the polynomials P(j,t) of A131758. - Tom Copeland, Oct 05 2007
a(1) = A094665(0,0)*A156919(0,0) and a(n) = Sum_{k=1..n-1} 2^(n-k-1)*A094665(n-1, k)*A156919(k,0) for n = 2, 3, .., see A162005. - Johannes W. Meijer, Jun 27 2009
G.f.: 1/(1-1*2*x/(1-2*3*x/(1-3*4*x/(1-4*5*x/(1-5*6*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-2x-12x^2/(1-18x-240x^2/(1-50x-1260x^2/(1-98x-4032x^2/(1-162x-9900x^2/(1-... (continued fraction);
coefficient sequences given by 4*(n+1)^2*(2n+1)*(2n+3) and 2(2n+1)^2 (see Van Fossen Conrad reference). (End)
E.g.f.: x*Sum_{n>=0} Product_{k=1..n} tanh(2*k*x) = Sum_{n>=1} a(n)*x^n/(n-1)!. - Paul D. Hanna, May 11 2010 [corrected by Paul D. Hanna, Sep 28 2023]
a(n) = (-1)^(n+1)*Sum_{j=1..2*n+1} j!*Stirling2(2*n+1,j)*2^(2*n+1-j)*(-1)^j for n >= 0. Vladimir Kruchinin, Aug 23 2010: (Start)
If n is odd such that 2*n-1 is prime, then a(n) == 1 (mod (2*n-1)); if n is even such that 2*n-1 is prime, then a(n) == -1 (mod (2*n-1)). - Vladimir Shevelev, Sep 01 2010
Recursion: a(n) = (-1)^(n-1) + Sum_{i=1..n-1} (-1)^(n-i+1)*C(2*n-1,2*i-1)* a(i). - Vladimir Shevelev, Aug 08 2011
E.g.f.: tan(x) = Sum_{n>=1} a(n)*x^(2*n-1)/(2*n-1)! = x/(1 - x^2/(3 - x^2/(5 - x^2/(7 - x^2/(9 - x^2/(11 - x^2/(13 -...))))))) (continued fraction from J. H. Lambert - 1761). - Paul D. Hanna, Sep 21 2011
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 09 2013: (Start)
Continued fractions:
E.g.f.: (sec(x))^2 = 1+x^2/(x^2+U(0)) where U(k) = (k+1)*(2k+1) - 2x^2 + 2x^2*(k+1)*(2k+1)/U(k+1).
E.g.f.: tan(x) = x*T(0) where T(k) = 1-x^2/(x^2-(2k+1)*(2k+3)/T(k+1)).
E.g.f.: tan(x) = x/(G(0)+x) where G(k) = 2*k+1 - 2*x + x/(1 + x/G(k+1)).
E.g.f.: tanh(x) = x/(G(0)-x) where G(k) = k+1 + 2*x - 2*x*(k+1)/G(k+1).
E.g.f.: tan(x) = 2*x - x/W(0) where W(k) = 1 + x^2*(4*k+5)/((4*k+1)*(4*k+3)*(4*k+5) - 4*x^2*(4*k+3) + x^2*(4*k+1)/W(k+1)).
E.g.f.: tan(x) = x/T(0) where T(k) = 1 - 4*k^2 + x^2*(1 - 4*k^2)/T(k+1).
E.g.f.: tan(x) = -3*x/(T(0)+3*x^2) where T(k)= 64*k^3 + 48*k^2 - 4*k*(2*x^2 + 1) - 2*x^2 - 3 - x^4*(4*k -1)*(4*k+7)/T(k+1).
G.f.: 1/G(0) where G(k) = 1 - 2*x*(2*k+1)^2 - x^2*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1).
G.f.: 2*Q(0) - 1 where Q(k) = 1 + x^2*(4*k + 1)^2/(x + x^2*(4*k + 1)^2 - x^2*(4*k + 3)^2*(x + x^2*(4*k + 1)^2)/(x^2*(4*k + 3)^2 + (x + x^2*(4*k + 3)^2)/Q(k+1) )).
G.f.: (1 - 1/G(0))*sqrt(-x), where G(k) = 1 + sqrt(-x) - x*(k+1)^2/G(k+1).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/Q(k+1)). (End)
O.g.f.: x + 2*x*Sum_{n>=1} x^n * Product_{k=1..n} (2*k-1)^2 / (1 + (2*k-1)^2*x). - Paul D. Hanna, Feb 05 2013
a(n) = (-4)^n*Li_{1-2*n}(-1). - Peter Luschny, Jun 28 2012
a(n) = (-4)^n*(4^n-1)*Zeta(1-2*n). - Jean-François Alcover, Dec 05 2013
Asymptotic expansion: 4*((2*(2*n-1))/(Pi*e))^(2*n-1/2)*exp(1/2+1/(12*(2*n-1))-1/(360*(2*n-1)^3)+1/(1260*(2*n-1)^5)-...). (See Luschny link.) - Peter Luschny, Jul 14 2015
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = tan(x) satisfies the differential equation A''(x) = 2*A(x)*A'(x) with A(0) = 0 and A'(0) = 1, leading to the recurrence a(0) = 0, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0, 1, 0, 2, 0, 16, 0, 272, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 1/2 and a(1) = 1 produces A080635. Cf. A002105, A234797. (End)
a(n) = 2*polygamma(2*n-1, 1/2)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
a(n) = 2^(2n-2)*|p(2n-1,-1/2)|, where p_n(x) are the shifted row polynomials of A019538. E.g., a(2) = 2 = 2^2 * |1 + 6(-1/2) + 6(-1/2)^2|. - Tom Copeland, Oct 19 2016
From Peter Bala, May 05 2017: (Start)
With offset 0, the o.g.f. A(x) = 1 + 2*x + 16*x^2 + 272*x^3 + ... has the property that its 4th binomial transform 1/(1 - 4*x) A(x/(1 - 4*x)) has the S-fraction representation 1/(1 - 6*x/(1 - 2*x/(1 - 20*x/(1 - 12*x/(1 - 42*x/(1 - 30*x/(1 - ...))))))), where the coefficients [6, 2, 20, 12, 42, 30, ...] in the partial numerators of the continued fraction are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. Compare with the S-fraction associated with A(x) given above by Paul Barry.
A(x) = 1/(1 + x - 3*x/(1 - 4*x/(1 + x - 15*x/(1 - 16*x/(1 + x - 35*x/(1 - 36*x/(1 + x - ...))))))), where the unsigned coefficients in the partial numerators [3, 4, 15, 16, 35, 36,...] come in pairs of the form 4*n^2 - 1, 4*n^2 for n = 1,2,.... (End)
a(n) = Sum_{k = 1..n-1} binomial(2*n-2, 2*k-1) * a(k) * a(n-k), with a(1) = 1. - Michael Somos, Aug 02 2018
a(n) = 2^(2*n-1) * |Euler(2*n-1, 0)|, where Euler(n,x) are the Euler polynomials. - Daniel Suteu, Nov 21 2018 (restatement of one of Copeland's 2007 formulas.)
x - Sum_{n >= 1} (-1)^n*a(n)*x^(2*n)/(2*n)! = x - log(cosh(x)). The series reversion of x - log(cosh(x)) is (1/2)*x - (1/2)*log(2 - exp(x)) = Sum_{n >= 0} A000670(n)*x^(n+1)/(n+1)!. - Peter Bala, Jul 11 2022
For n > 1, a(n) = 2*Sum_{j=1..n-1} Sum_{k=1..j} binomial(2*j,j+k)*(-4*k^2)^(n-1)*(-1)^k/(4^j). - Tani Akinari, Sep 20 2023
a(n) = A110501(n) * 4^(n-1) / n (Han and Liu, 2018). - Amiram Eldar, May 17 2024

A000340 a(0)=1, a(n) = 3*a(n-1) + n + 1.

Original entry on oeis.org

1, 5, 18, 58, 179, 543, 1636, 4916, 14757, 44281, 132854, 398574, 1195735, 3587219, 10761672, 32285032, 96855113, 290565357, 871696090, 2615088290, 7845264891, 23535794695, 70607384108, 211822152348, 635466457069

Offset: 0

N. J. A. Sloane, Simon Plouffe

From Johannes W. Meijer, Feb 20 2009: (Start)
Second right hand column (n-m=1) of the A156920 triangle.
The generating function of this sequence enabled the analysis of the polynomials A156921 and A156925.
(End)
Partial sums of A003462, and thus the second partial sums of A000244 (3^n). Also column k=2 of A106516. - John Keith, Jan 04 2022

			G.f. = 1 + 5*x + 18*x^2 + 58*x^3 + 179*x^4 + 543*x^5 + 1636*x^6 + ...

F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
Shaoshi Chen, Hanqian Fang, Sergey Kitaev, and Candice X.T. Zhang, Patterns in Multi-dimensional Permutations, arXiv:2411.02897 [math.CO], 2024. See p. 7.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 389
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. Mentions this sequence. See also Proc. Amer. Math. Soc. 148 (2020), 461-469.
Index entries for linear recurrences with constant coefficients, signature (5,-7,3).

From Johannes W. Meijer, Feb 20 2009: (Start)
Cf. A156921, A156925, A156927, A156933. Other columns A156922, A156923, A156924.
Equals A156920 second right hand column.
Equals A142963 second right hand column divided by 2^n.
Equals A156919 second right hand column divided by 2.
(End)
Cf. A014915.
Equals column k=1 of A008971 (shifted). - Jeremy Dover, Jul 11 2021
Cf. A000340, A003462 (first differences), A106516.

[(3^(n+2)-2*n-5)/4: n in [0..30]]; // Vincenzo Librandi, Aug 15 2011

a[ -1]:=0:a[0]:=1:for n from 1 to 50 do a[n]:=4*a[n-1]-3*a[n-2]+1 od: seq(a[n],n=0..50); # Miklos Kristof, Mar 09 2005
A000340:=-1/(3*z-1)/(z-1)**2; # conjectured by Simon Plouffe in his 1992 dissertation

a[ n_] := MatrixPower[ {{1, 0, 0}, {1, 1, 0}, {1, 1, 3}}, n + 1][[3, 1]]; (* Michael Somos, May 28 2014 *)
RecurrenceTable[{a[0]==1,a[n]==3a[n-1]+n+1},a,{n,30}] (* or *) LinearRecurrence[{5,-7,3},{1,5,18},30] (* Harvey P. Dale, Jan 31 2017 *)

G.f.: 1/((1-3*x)*(1-x)^2).
a(n) = (3^(n+2) - 2*n - 5)/4.
a(n) = Sum_{k=0..n+1} (n-k+1)*3^k = Sum_{k=0..n+1} k*3^(n-k+1). - Paul Barry, Jul 30 2004
a(n) = Sum_{k=0..n} binomial(n+2, k+2)*2^k. - Paul Barry, Jul 30 2004
a(-1)=0, a(0)=1, a(n) = 4*a(n-1) - 3*a(n-2) + 1. - Miklos Kristof, Mar 09 2005
a(n) = 5*a(n-1) - 7*a(n-2) + 3*a(n-3). - Johannes W. Meijer, Feb 20 2009
a(-2 - n) = 3^-n * A014915(n). - Michael Somos, May 28 2014
E.g.f.: exp(x)*(9*exp(2*x) - 2*x - 5)/4. - Stefano Spezia, Nov 09 2024

A050488 a(n) = 3(2^n-1) - 2n.

Original entry on oeis.org

0, 1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821, 25769803707

Offset: 0

James Sellers, Dec 26 1999

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).
Number of subgroups of C_(2^n) X C_(2^n) (see A060724).
Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008
Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010
(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012
The partial sums of A033484. - J. M. Bergot, Oct 03 2012
Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012
Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013
Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015
Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, Flattened Stirling Permutations, arXiv:2306.13034 [math.CO], 2023. See p. 14.
Tamas Lengyel, On p-adic properties of the Stirling numbers of the first kind, Journal of Number Theory, 148 (2015) 73-94.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

A050487(2^m-1).
Equals (1/2) A051667.
Cf. A000079, A000225, A000295, A005408, A008292, A033484, A054852, A060724, A066810, A125165, A126277, A142963, A156919, A156920, A156925.

List([0..30],n->3*(2^n-1)-2*n); # Muniru A Asiru, Oct 26 2018

a050488 n = sum $ zipWith (*) a000079_list (reverse $ take n a005408_list)
-- Reinhard Zumkeller, Jul 24 2015

[3*(2^n-1) - 2*n: n in [0..30]]; // G. C. Greubel, Oct 23 2018

seq(coeff(series(x*(x+1)/((1-x)^2*(1-2*x)),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 26 2018

Table[3(2^n-1)-2n,{n,0,30}] (* or *) LinearRecurrence[{4,-5,2}, {0,1,5}, 40] (* Harvey P. Dale, Apr 09 2018 *)

a(n)=3*(2^n-1)-2*n \\ Charles R Greathouse IV, Sep 24 2015

for n in range(0, 30): print(3*(2**n-1) - 2*n, end=', ') # Stefano Spezia, Oct 27 2018

Row sums of A125165: (1, 5, 15, 37, ...). Binomial transform of [1, 4, 6, 6, 6, ...] = [1, 5, 15, 37, ...]. 4th diagonal from the right of A126777 = (1, 5, 15, ...). - Gary W. Adamson, Dec 23 2006
a(n) = 2*a(n-1) + (2n-1). - Gary W. Adamson, Sep 30 2007
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n+1) = A156920(n+1,1).
a(n+1) = A156919(n+1,1)/2^n.
a(n+1) = A142963(n+2,1)/2.
a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.
G.f.: z*(1+z)/((1-z)^2*(1-2*z)).
(End)
a(n) = 2*n + 2*a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n+1) = Sum_{k=0..n} A000079(k) * A005408(n-k), convolution of the powers of 2 with the odd numbers. - Reinhard Zumkeller, Mar 08 2012
E.g.f.: exp(x)*(3*exp(x) - 2*x - 3). - Stefano Spezia, May 15 2023

A162005 The EG1 triangle.

Original entry on oeis.org

1, 2, 1, 16, 28, 1, 272, 1032, 270, 1, 7936, 52736, 36096, 2456, 1, 353792, 3646208, 4766048, 1035088, 22138, 1, 22368256, 330545664, 704357760, 319830400, 27426960, 199284, 1, 1903757312, 38188155904, 120536980224, 93989648000

Offset: 1

Johannes W. Meijer, Jun 27 2009, Jul 02 2009, Aug 31 2009

We define the EG1 matrix by EG1[2m-1,1] = 2*eta(2m-1) and the recurrence relation EG1[2m-1,n] = EG1[2m-1,n-1] - EG1[2m-3,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, .. and n = 2, 3, .., with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. For the EG2[2m,n] coefficients see A008955.
The n-th term of the row coefficients EG1[1-2*m,n] for m = 1, 2, .., can be generated with REG1(1-2*m,n) = (-1)^(m+1)*2^(1-m)*ECGP(1-2*m, n)*(1/n)*4^(-n)*(2*n)!/((n-1)!)^2 . For information about the ECGP polynomials see A094665 and the examples below.
We define the o.g.f.s. of the REG1(1-2*m,n) by GFREG1(z,1-2*m) = sum(REG1(1-2*m,n)* z^(n-1), n=1..infinity) for m = 1, 2, .., with GFREG1(z,1-2*m) = (-1)^(m+1)* RG(z,1-2*m)/ (2^(2*m-1)*(1-z)^((2*m+1)/2)). The RG(z,1-2m) polynomials led to the EG1 triangle.
We used the coefficients of the A156919 and A094665 triangles to determine those of the EG1 triangle, see the Maple program. The A156919 triangle gives information about the sums SF(p) = sum(n^(p-1)*4^(-n)*z^(n-1)*(2*n)!/((n-1)!)^2, n=1..infinity) for p= 0, 1, 2, .. .
Contribution from Johannes W. Meijer, Nov 23 2009: (Start)
The EG1 matrix is related to the ED2 array A167560 because sum(EG1(2*m-1,n)*z^(2*m-1), m=1..infinity) = ((2*n-1)!/(4^(n-1)*(n-1)!^2))*int(sinh(y*(2*z))/cosh(y)^(2*n),y=0..infinity).
(End)
Appears to equal triangle A322230 with rows read in reverse order. Triangle A322230 describes the e.g.f. S(x,k) = Integral C(x,k)*D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2*S(x,k)^2 = 1. - Paul D. Hanna, Dec 22 2018
Appears to equal triangle A325220, which has e.g.f. S(x,k) = -i * sn( i * Integral C(x,k) dx, k) such that C(x,k) = cn( i * Integral C(x,k) dx, k), where sn(x,k) and cn(x,k) are Jacobi Elliptic functions. - Paul D. Hanna, Apr 13 2019

			The first few rows of the EG1 triangle are :
[1]
[2, 1]
[16, 28, 1]
[272, 1032, 270, 1]
The first few RG(z,1-2*m) polynomials are:
RG(z,-1) = 1
RG(z,-3) = 2+z
RG(z,-5) = 16+28*z+z^2
RG(z,-7) = 272+1032*z+270*z^2+z^3
The first few GFREG1(z,1-2*m) are:
GFREG1(z,-1) = (1)*(1)/(2*(1-z)^(3/2))
GFREG1(z,-3) = (-1)*(2+z)/(2^3*(1-z)^(5/2))
GFREG1(z,-5) = (1)*(16+28*z+z^2)/( 2^5*(1-z)^(7/2))
GFREG1(z,-7) = (-1)*(272+1032*z+270*z^2+z^3)/(2^7*(1-z)^(9/2))
The first few REG1(1-2*m,n) are:
REG1(-1,n) = (1/1)*(1)*(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-3,n) = (-1/2)*(n) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-5,n) = (1/4) *(n+3*n^2) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-7,n) = (-1/8)*(4*n+15*n^2+15*n^3) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
The first few ECGP(1-2*m,n) polynomials are:
ECGP(-1,n) = 1
ECGP(-3,n) = n
ECGP(-5,n) = n+3*n^2
ECGP(-7,n) = 4*n+15*n^2+15*n^3

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.

A079484 equals the row sums.
A000182 (ZAG numbers), A162006 and A162007 equal the first three left hand columns.
A000012, A004004 (2x), A162008, A162009 and A162010 equal the first five right hand columns.
Related to A094665, A083061 and A156919 (DEF triangle).
Cf. A161198 [(1-x)^((-1-2*n)/2)], A008955 (EG2[2m, n])
Cf. A167560 (ED2 array).
Cf. A322230 (reversed rows), A325220.

nmax:=7; mmax := nmax: imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1) * (x+1)*T1(i-1, x+1)-2*x^2*T1(i-1, x)): dx := degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1=0..dx): od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1)*A156919(n-1, m-1) end do end do: for n from 0 to nmax do SF(n) := sum(A156919(n, k1)*z^k1, k1=0..n)/(2^(n+1)*(1-z)^((2*n+3)/2)) od: GFREG1(z, -1) := A156919(0, 0)*A094665 (0, 0) / (2*(1-z)^(3/2)): for m from 2 to nmax do GFREG1(z, 1-2*m) := simplify((-1)^(m+1)*2^(1-m)* sum(A094665(m-1, k2)*SF(k2), k2=1..m-1)) od: for m from 1 to mmax do g(m) := sort((numer ((-1)^(m+1)* GFREG1(z, 1-2*m))), ascending) od: for n from 1 to nmax do for m from 1 to n do a(n, m) := abs(coeff(g(n), z, m-1)) od: od: seq(seq(a(n, m), m=1..n), n=1..nmax);
# Maple program edited by Johannes W. Meijer, Sep 25 2012

A different form of the recurrence relation is EG1[1-2*m,n] = (EG1[3-2*m,n]-EG1[3-2*m,n+1])* (n^2) for m = 2, 3, .., with EG1[ -1,n] = (1/n)*4^(-n)*((2*n)!/(n-1)!^2).

A142963 Triangle read by rows, coefficients of the polynomials P(k, x) = (1/2) Sum_{p=0..k-1} Stirling2(k, p+1)x^p(1-4x)^(k-1-p)(2*p+2)!/(p+1)!.

Original entry on oeis.org

1, 1, 2, 1, 10, 4, 1, 30, 72, 8, 1, 74, 516, 464, 16, 1, 166, 2584, 7016, 2864, 32, 1, 354, 10740, 64240, 84480, 17376, 64, 1, 734, 40008, 450280, 1321760, 949056, 104704, 128, 1, 1498, 139108, 2681296, 14713840, 24198976, 10223488, 629248, 256, 1, 3030, 462264, 14341992

Offset: 1

Wolfdieter Lang, Sep 15 2008

Previous name: Table of coefficients of row polynomials of certain o.g.f.s.
The o.g.f.s G(k, x) for the k-family of sequences S(k, n):= Sum_{p=0..n} p^k*binomial(2*p, p)*binomial(2*(n-p), n-p), k=0,1,... (convolution of two sequences involving the central binomial coefficients) are 1/(1-4*x) for k=0 and 2*x*P(k, x)/(1-4*x)^(k+1) for k=1,2,..., with the row polynomials P(k, x) = Sum_{m=0..k-1} a(n,m)*x^m).
The author was led to compute the sums S(k, n) by a question asked by M. Greiter, Jun 27 2008.
In order to keep the index k>=1 of Sigma(k, n) also for the polynomials P(k, x), their degree is then k-1.

			Triangle starts:
[1]
[1,   2]
[1,  10,     4]
[1,  30,    72,      8]
[1,  74,   516,    464,      16]
[1, 166,  2584,   7016,    2864,     32]
[1, 354, 10740,  64240,   84480,  17376,     64]
[1, 734, 40008, 450280, 1321760, 949056, 104704, 128]
...
P(3,x) = 1+10*x+4*x^2.
G(3,x) = 2*x*(1+10*x+4*x^2)/(1-4*x)^4.

Vincenzo Librandi, Table of n, a(n) for n = 1..210
Wolfdieter Lang, First 10 rows and more.
L. Liu, Y. Wang, A unified approach to polynomial sequences with only real zeros, arXiv:math/0509207v5 [math.CO], 2005-2006.

Left hand column sequences 2*A142964, 4*A142965, 8*A142966, 16*A142968.
Row sums A142967.
From Johannes W. Meijer, Feb 20 2009: (Start)
A156919 and this sequence can be mapped onto A156920.
Cf. A156921, A156925, A156927, A156933.
Right hand column sequences 2^n*A000340, 2^n*A156922, 2^n*A156923, 2^n*A156924. (End)
Cf. A142961, A142962.

A142963 := proc(n,m): if n=m+1 then 2^(n-1); elif m=0 then 1 ; elif m<0 or m>n-1 then 0; else (m+1)*procname(n-1, m)+(4*n-4*m-2)*procname(n-1, m-1); end if; end proc: seq(seq(A142963(n,m), m=0..n-1), n=1..9); # Johannes W. Meijer, Sep 28 2011
# Alternatively (assumes offset 0):
p := (n,x) -> (1/2)*add(Stirling2(n+1,k+1)*x^k*(1-4*x)^(n-k)*(2*k+2)!/(k+1)!, k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od;
# Peter Luschny, Jun 18 2017

t[, 0] = 1; t[n, m_] /; m == n-1 := 2^m; t[n_, m_] := (m+1)*t[n-1, m] + (4*n-4*m-2)*t[n-1, m-1]; Table[t[n, m], {n, 1, 10}, {m, 0, n-1}] // Flatten (* Jean-François Alcover, Jun 21 2013, after Johannes W. Meijer *)

G(k, x) = Sum_{p=0..k} S2(k, p)*((2*p)!/p!)*x^p/(1-4*x)^(p+1), k >= 0 (here k >= 1), with the Stirling2 triangle S2(k, p):=A048993(k, p). (Proof from the product of the o.g.f.s of the two convoluted sequences and the normal ordering (x^d_x)^k = Sum_{p=0..k} S2(k, p)*x^p*d_x^p, with the derivative operator d_x.)
a(k,m) = [x^m]P(k, x) = [x^m] ((1-4*x)^(k+1))*G(k,x)/(2*x), k>=1, m=0,1,...,k-1.
For the triangle coefficients the following relation holds: T(n,m) = (m+1)*T(n-1,m) + (4*n-4*m-2)*T(n-1,m-1) with T(n,m=0) = 1 and T(n,m=n-1) = 2^(n-1), n >= 1 and 0 <= m <= n-1. - Johannes W. Meijer, Feb 20 2009
From Peter Bala, Jan 18 2018: (Start)
(x*d/dx)^n (1/(sqrt(1 - 4*x)) = 2*x*P(n,x)/sqrt(1 - 4*x)^(n+1/2) for n >= 1.
x*P(n,x)/(1 - 4*x)^(n+1/2) = (1/2)*Sum_{k >= 1} binomial(2*k,k)* k^n*x^k for n >= 1.
P(n+1,x) = ((4*n - 2)*x + 1)*P(n,x) - x*(4*x - 1)*d/dx(P(n,x)).
Hence the polynomial P(n,x) has all real zeros by Liu et al., Theorem 1.1, Corollary 1.2. (End)

Minor edits by Johannes W. Meijer, Sep 28 2011
A more precise name by Peter Luschny, Jun 18 2017
Name reformulated with offset corrected, edited by Wolfdieter Lang, Aug 23 2019

A142965 One fourth of third column (m=2) of triangle A142963.

Original entry on oeis.org

1, 18, 129, 646, 2685, 10002, 34777, 115566, 372453, 1175290, 3654369, 11245110, 34349005, 104373282, 315969705, 954002878, 2874983541, 8652474378, 26015617585, 78169534470, 234766551261, 704840716978, 2115654610809, 6349329417486, 19052920751365, 57169029907482

Offset: 0

Wolfdieter Lang, Sep 15 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (10, -40, 82, -91, 52, -12).

Column m=1: 2*A142964; m=3: 8*A142966.
From Johannes W. Meijer, Feb 20 2009: (Start)
Cf. A156925.
Equals A156920(n+2,2).
Equals A156919(n+2,2)/2^n.
(End)

[35/2+2*n^2+12*n-84*2^n-24*2^n*n+135/2*3^n: n in [0..25]]; // Vincenzo Librandi, Jun 18 2017

LinearRecurrence[{10,-40,82,-91,52,-12}, {1,18,129,646,2685,10002},30] (* or *) CoefficientList[Series[(1+8x-11x^2-6x^3)/((x-1)^3 (2x-1)^2 (3x-1)),{x,0,30}],x] (* Harvey P. Dale, Apr 24 2011 *)

a(n) = A142963(n+3,2)/4.
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n) = 10a(n-1) - 40a(n-2) + 82a(n-3) - 91a(n-4) + 52a(n-5) - 12a(n-6).
a(n) = 35/2 + 2*n^2 + 12*n - 84*2^n - 24*2^n*n + 135/2*3^n
G.f.: (1 + 8*z - 11*z^2 - 6*z^3)/((1-z)^3*(1-2*z)^2*(1-3*z)).
(End)

A156922 Third right hand column (n-m=2) of the A156920 triangle.

Original entry on oeis.org

1, 15, 129, 877, 5280, 29658, 159742, 838038, 4323003, 22057825, 111750519, 563535483, 2833221970, 14216330916, 71243079660, 356731958812, 1785306330981, 8931761831331, 44675371382365

Offset: 0

Johannes W. Meijer, Feb 20 2009

Other columns A000340, A156923, A156924.
Equals A156920 third right hand column.
Equals A156919 third right hand column divided by 4.
Equals A142963 third right hand column divided by 2^n.

a(n) = 14*a(n-1)-75*a(n-2)+196*a(n-3)-263*a(n-4)+174*a(n-5)-45*a(n-6)
a(n) = (4*n^2-108*n*3^n+24*n-378*3^n+375*5^n+35)/32
G.f.: GF1(z;RHCnr=3) = (1+z-6*z^2)/((1-5*z)*(1-3*z)^2*(1-z)^3)

A156923 Fourth right hand column (n-m=3) of the A156920 triangle.

Original entry on oeis.org

1, 37, 646, 8030, 82610, 756218, 6411720, 51586344, 400011435, 3020658295, 22373863774, 163379472214, 1180488191108, 8462445970580, 60305767988960, 427848087263712, 3025286818472661

Offset: 0

Johannes W. Meijer, Feb 20 2009

Other columns A000340, A156922, A156924.
Equals A156920 fourth right hand column.
Equals A156919 fourth right hand column divided by 8.
Equals A142963 fourth right hand column divided by 2^n

a(n) = 30*a(n-1)-385*a(n-2)+2776*a(n-3)-12418*a(n-4)+35908*a(n-5)-67818*a(n-6)+82552*a(n-7)-62109*a(n-8)+26190*a(n-9)-4725*a(n-10)
a(n) = (-8*n^3+972*n^2*3^n-84*n^2+7776*n*3^n-11250*n*5^n-286*n+15309*3^n-50625*5^n+36015*7^n-315)/384
G.f.: GF1(z;RHCnr=4) = (1+7*z-79*z^2+119*z^3+126*z^4-270*z^5)/((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)

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A156920 Triangle of the normalized A142963 and A156919 sequences.

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A185412 Triangle T(n,m) read by rows: the matrix product A130595 * A156919.

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A000182 Tangent (or "Zag") numbers: e.g.f. tan(x), also (up to signs) e.g.f. tanh(x).

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A000340 a(0)=1, a(n) = 3*a(n-1) + n + 1.

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A050488 a(n) = 3*(2^n-1) - 2*n.

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A162005 The EG1 triangle.

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A050488 a(n) = 3(2^n-1) - 2n.

A142963 Triangle read by rows, coefficients of the polynomials P(k, x) = (1/2) Sum_{p=0..k-1} Stirling2(k, p+1)x^p(1-4x)^(k-1-p)(2*p+2)!/(p+1)!.