A126156 -id:A126156 - cp's OEIS frontend

A241885 Write the coefficient of x^n/n! in the expansion of (x/(exp(x)-1))^(1/2) as f(n)/g(n); sequence gives f(n).

1, -1, 1, 1, -3, -19, 79, 275, -2339, -11813, 14217, 95265, -4634445, -193814931, 131301607, 1315505395, -3890947599, -136146236611, 46949081169401, 124889801445461, -10635113572583999, -158812278992229461, 56918172351554857, 8484151253958927197

Offset: 0

Jitender Singh, May 01 2014

For g(n) see A242225(n).
The old definition was "Numerator of (B_n)^(1/2) in the Cauchy type product (sometimes known as binomial transform) where B_n is the n-th Bernoulli number".
The Nørlund polynomials N(a, n, x) with parameter a = 1/2 evaluated at x = 0 give the rational values. - Peter Luschny, Feb 18 2024

			For n=1, B_1=-1/2 and B_1^(1/2)=-1/4 so a(1)=-1.
For n=6, B_6=1/6 and B_6^(1/2)=79/86016 so a(6)=79.
1/1, -1/4, 1/48, 1/64, -3/1280, -19/3072, 79/86016, 275/49152, -2339/2949120, -11813/1310720, 14217/11534336 = A241885 / A242225.

Robert Israel, Table of n, a(n) for n = 0..479
David Broadhurst, Relations between A241885/A242225, A222411/A222412, and A350194/A350154.
Jitender Singh, On an arithmetic convolution, arXiv:1402.0065 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.6.7.

Cf. A242225 (denominators), A126156, A242233.
Cf. also A222411/A222412, A350194/A350154.
Cf. A370416/A370417.

g := proc(f, n) option remember; local g0, m; g0 := sqrt(f(0));
if n=0 then g0 else if n=1 then 0 else add(binomial(n, m)*g(f,m)*g(f,n-m), m=1..n-1) fi; (f(n)-%)/(2*g0) fi end:
a := n -> numer(g(bernoulli, n));
seq(a(n), n = 0..23); # Peter Luschny, May 07 2014

a := 1
g[0] := Sqrt[f[0]]
f[k_] := BernoulliB[k]
g[1] := f[1]/(2 g[0]^1);
g[k_] := (f[k] - Sum[Binomial[k, m] g[m] g[k - m], {m, 1, k - 1}])/(2 g[0])
Table[Factor[g[k]], {k, 0, 15}] // TableForm
(* Alternative: *)
Table[Numerator@NorlundB[n, 1/2, 0], {n, 0, 23}]  (* Peter Luschny, Feb 18 2024 *)

a(n)=numerator(sum(k=0,n,binomial(-1/2,k)*binomial(n+1/2,n-k)*stirling(n+k,k,2)/binomial(n+k,k))) \\ Tani Akinari, Oct 08 2024

Theorem: a(n)/A242225(n) = n!*A222411(n)/(A222412(n)*(-1)^n/(1-2*n)) = n!*A350194(n)/(A350154(n)*(2*n+1)). - David Broadhurst, Apr 23 2022 (see Link).
For any arithmetic function f and a positive integer k > 1, define the k-th root of f to be the arithmetic function g such that g*g*...*g(k times)=f and is determined by the following recursive formula:
g(0) = f(0)^(1/m);
g(1) = f(1)/(m*g(0)^(m-1));
g(k) = 1/(m*g(0)^(m-1))*(f(k) - Sum_{k_1+...+k_m=k,k_i= 2.
This formula is applicable for any rational root of an arithmetic function with respect to the Cauchy type product.
E.g.f: sqrt(x/(exp(x)-1)); take numerators. - Peter Luschny, May 08 2014
a(n) = numerator(Sum_{k=0..n} binomial(-1/2,k)*binomial(n+1/2,n-k)*Stirling2(n+k,k)/binomial(n+k,k)). - Tani Akinari, Oct 08 2024

Simpler definition from N. J. A. Sloane, Apr 24 2022 at the suggestion of David Broadhurst

A126155 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 5 times first term.

Original entry on oeis.org

1, 1, 5, 1, 7, 35, 55, 35, 7, 139, 695, 1195, 1415, 1195, 695, 139, 5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473, 357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721

Offset: 0

Paul D. Hanna, Dec 20 2006

			The triangle begins:
  1;
  1, 5, 1;
  7, 35, 55, 35, 7;
  139, 695, 1195, 1415, 1195, 695, 139;
  5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, 5473;
  357721, 1788605, 3175705, 4343885, 5126905, 5403005, 5126905, 4343885, 3175705, 1788605, 357721; ...
If we write the triangle like this:
  .......................... ....1;
  ................... ....1, ....5, ....1;
  ............ ....7, ...35, ...55, ...35, ....7;
  ..... ..139, ..695, .1195, .1415, .1195, ..695, ..139;
  .5473, 27365, 48145, 63365, 69025, 63365, 48145, 27365, .5473;
then the first term in each row is the sum of the previous row:
  5473 = 139 + 695 + 1195 + 1415 + 1195 + 695 + 139
the next term is 5 times the first:
  27365 = 5*5473,
and the remaining terms in each row are obtained by the rule illustrated by:
  48145 = 2*27365 - 5473 - 8*139;
  63365 = 2*48145 - 27365 - 8*695;
  69025 = 2*63365 - 48145 - 8*1195;
  63365 = 2*69025 - 63365 - 8*1415;
  48145 = 2*63365 - 69025 - 8*1195;
  27365 = 2*48145 - 63365 - 8*695;
  5473 = 2*27365 - 48145 - 8*139.
An alternate recurrence is illustrated by:
  27365 = 5473 + 4*(139 + 695 + 1195 + 1415 + 1195 + 695 + 139);
  48145 = 27365 + 4*(695 + 1195 + 1415 + 1195 + 695);
  63365 = 48145 + 4*(1195 + 1415 + 1195);
  69025 = 63365 + 4*(1415);
and then for k>n, T(n,k) = T(n,2n-k).

Cf. A126156 (column 0); diagonals: A126157, A126158; A126159; variants: A008301 (p=1), A125053 (p=2), A126150 (p=3).

T := proc(n,k) option remember; local j;
  if n = 1 then 1
elif k = 1 then add(T(n-1, j), j=1..2*n-3)
elif k = 2 then 5*T(n, 1)
elif k > n then T(n, 2*n-k)
else 2*T(n, k-1)-T(n, k-2)-8*T(n-1, k-2)
  fi end:
seq(print(seq(T(n,k), k=1..2*n-1)), n=1..6); # Peter Luschny, May 12 2014

T[n_, k_] := T[n, k] = Which[n==1, 1, k==1, Sum[T[n-1, j], {j, 1, 2n-3}], k==2, 5 T[n, 1], k>n, T[n, 2n-k], True, 2 T[n, k-1] - T[n, k-2] - 8 T[n-1, k-2]];
Table[T[n, k], {n, 1, 6}, {k, 1, 2n-1}] (* Jean-François Alcover, Jun 15 2019, from Maple *)

PARI
```
{T(n,k) = local(p=4);if(2*n
				
```

/* Alternate Recurrence: */
{T(n,k) = local(p=4);if(2*n

from functools import cache
@cache
def R(n, k):
      return (1 if n == 1 else sum(R(n-1, j) for j in range(1, 2*n-2))
                if k == 1 else 5*R(n, 1) if k == 2 else R(n, 2*n-k)
                if k > n else 2*R(n, k-1) - R(n, k-2) - 8*R(n-1, k-2))
def A126155(n, k): return R(n+1, k+1)
for n in range(5): print([A126155(n, k) for k in range(2*n+1)])
# Peter Luschny, Dec 14 2023

Sum_{k=0..2n} (-1)^k*C(2n,k)*T(n,k) = (-8)^n.

A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 11, 26, 12, 8, 57, 120, 136, 32, 16, 361, 970, 760, 560, 80, 32, 2763, 7052, 8860, 3680, 2000, 192, 64, 24611, 72530, 72884, 58520, 15120, 6496, 448, 128, 250737, 716528, 976464, 538048, 314720, 55552, 19712, 1024, 256

Offset: 1

Peter Bala, Jan 28 2011

Define a polynomial sequence S(n,x) recursively by
(1)... S(n+1,x) = x*S(n,x-1)+(x+1)*S(n,x+1) with S(0,x) = 1.
This table lists the coefficients of these polynomials (for n>=1) in ascending powers of x.
The first few polynomials are
S(0,x) = 1
S(1,x) = 2*x+1
S(2,x) = 4*x^2+4*x+3
S(3,x) = 8*x^3+12*x^2+26*x+11.
The sequence [1,1,3,11,57,...] of constant terms of the polynomials is the sequence of Springer numbers A001586. The zeros of the polynomials S(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
Compare the recurrence (1) with the recurrence relation satisfied by the coefficients T(n,k) of the polynomials of A104035, namely
(2)... T(n+1,k) = k*T(n,k-1)+(k+1)*T(n,k+1).

			Table begin
n\k|.....0.....1.....2.....3.....4.....5......6
===============================================
0..|.....1
1..|.....1.....2
2..|.....3.....4.....4
3..|....11....26....12.....8
4..|....57...120...136....32...16
5..|...361...970...760...560...80.....32
6..|..2763..7052..8860..3680..2000...192....64
...

Peter Bala, The zeros of the row polynomials of A185417

Cf A001586 (1st column and row sums), A104035, A126156, A147309, A185415, A185418, A185419

#A185417
S := proc(n,x) option remember;
description 'polynomials S(n,x)'
if n = 0 return 1 else return x*S(n-1,x-1)+(x+1)*S(n-1,x+1)
end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(S(n,x),x); end do;

S[0, ] = 1; S[n, x_] := S[n, x] = x*S[n-1, x-1] + (x+1)*S[n-1, x+1]; Table[ CoefficientList[S[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 15 2015 *)

E.g.f: F(x,t) = 1/(cos(t)-sin(t))*(tan(2*t)+sec(2*t))^x
= (cos(t)+sin(t))^x/(cos(t)-sin(t))^(x+1)
= 1 + (2*x+1)*t + (4*x^2+4*x+3)*t^2/2! + ....
Note that (tan(t)+sec(t))^x is the e.g.f for table A147309.
ROW POLYNOMIALS
The easily checked identity d/dt F(x,t) = x*F(x-1,t)+(x+1)*F(x+1,t) shows that the row generating polynomials of this table are the polynomials S(n,x) described in the Comments section above.
The polynomials S(n,-x) satisfy a Riemann hypothesis: that is, the zeros of S(n,-x) lie on the vertical line Re(x) = 1/2 in the complex plane - see the link.
RELATION WITH OTHER SEQUENCES
1st column [1,1,3,11,57,...] is A001586.
Row sums sequence [1,3,11,57,...] is also A001586.
For n>=1, the values 1/2^n*P(2*n,-1/2) = [1,7,139,5473,...] appear to be A126156.

A365673 Array A(n, k) read by ascending antidiagonals. Polygonal number weighted generalized Catalan sequences.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 4, 1, 1, 1, 4, 15, 8, 1, 1, 1, 5, 34, 105, 16, 1, 1, 1, 6, 61, 496, 945, 32, 1, 1, 1, 7, 96, 1385, 11056, 10395, 64, 1, 1, 1, 8, 139, 2976, 50521, 349504, 135135, 128, 1, 1, 1, 9, 190, 5473, 151416, 2702765, 14873104, 2027025, 256, 1

Offset: 0

Peter Luschny, Sep 30 2023

Using polygonal numbers as weights, a recursion for triangles is defined, whose main diagonals represents a family of sequences, which include, among others, the powers of 2, the double factorial of odd numbers, the reduced tangent numbers, and the Euler numbers.
Apart from the edge cases k = 0 and k = n the recursion is T(n, k) = w(n, k) * T(n, k - 1) + T(n - 1, k). T(n, 0) = 1 and T(n, n) = T(n, n-1) if n > 0.
The weights w(n, k) identical to 1 yield the recursion of the Catalan triangle A009766 (with main diagonal the Catalan numbers). Here the polygonal numbers are used as weights in the form w(n, k) = p(s, n - k + 1), where the parameter s is the number of sides of the polygon and p(s, n) = ((s-2) * n^2 - (s-4) * n) / 2, see A317302.

			Array A(n, k) starts:                            (polygon|diagonal|triangle)
[0] 1, 1, 1,   1,     1,       1,         1, ...  A258837  A000012
[1] 1, 1, 2,   4,     8,      16,        32, ...  A080956  A011782
[2] 1, 1, 3,  15,   105,     945,     10395, ...  A001477  A001147  A001498
[3] 1, 1, 4,  34,   496,   11056,    349504, ...  A000217  A002105  A365674
[4] 1, 1, 5,  61,  1385,   50521,   2702765, ...  A000290  A000364  A060058
[5] 1, 1, 6,  96,  2976,  151416,  11449296, ...  A000326  A126151  A366138
[6] 1, 1, 7, 139,  5473,  357721,  34988647, ...  A000384  A126156  A365672
[7] 1, 1, 8, 190,  9080,  725320,  87067520, ...  A000566  A366150  A366149
[8] 1, 1, 9, 249, 14001, 1322001, 188106489, ...  A000567
           A054556                         A366137

Wikipedia, Polygonal number.

Poly weights: A258837, A080956, A001477, A000217, A000290, A000326, A000384.
Rows: A000012, A011782, A001147, A002105, A000364, A126151, A126156, A366150.
Triangles: A001498, A365674, A060058, A366138, A365672, A366149.
Cf. A009766, A366137 (central diagonal), A317302 (table of polygonal numbers).
Cf. A112934, A303943, A305532, A305533.

poly := (s, n) -> ((s - 2) * n^2 - (s - 4) * n) / 2:
T := proc(s, n, k) option remember; if k = 0 then 1 else if k = n then T(s, n, k-1) else poly(s, n - k + 1) * T(s, n, k - 1) + T(s, n - 1, k) fi fi end:
for n from 0 to 8 do A := (n, k) -> T(n, k, k): seq(A(n, k), k = 0..9) od;
# Alternative, using continued fractions:
A := proc(p, L) local CF, poly, k, m, P, ser;
   poly := (s, n) -> ((s - 2)*n^2 - (s - 4)*n)/2;
   CF := 1 + x;
   for k from 1 to L do
       m := L - k + 1;
       P := poly(p, m);
       CF := 1/(1 - P*x*CF)
   od;
   ser := series(CF, x, L);
   seq(coeff(ser, x, m), m = 0..L-1)
end:
for p from 0 to 8 do lprint(A(p, 8)) od;

poly[s_, n_] := ((s - 2) * n^2 - (s - 4) * n) / 2;
T[s_, n_, k_] := T[s, n, k] = If[k == 0, 1, If[k == n, T[s, n, k - 1], poly[s, n - k + 1] * T[s, n, k - 1] + T[s, n - 1, k]]];
A[n_, k_] := T[n, k, k];
Table[A[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 27 2023, from first Maple program *)

A(p, n) = {
       my(CF = 1 + x,
           poly(s, n) = ((s - 2)*n^2 - (s - 4)*n)/2,
           m, P
       );
       for(k = 1, n,
           m = n - k + 1;
           P = poly(p, m);
           CF = 1/(1 - P*x*CF)
        );
        Vec(CF + O(x^(n)))
}
for(p = 0, 8, print(A(p, 8)))
\\  Michel Marcus and Peter Luschny, Oct 02 2023

from functools import cache
@cache
def T(s, n, k):
    if k == 0: return 1
    if k == n: return T(s, n, k - 1)
    p = (n - k + 1) * ((s - 2) * (n - k + 1) - (s - 4)) // 2
    return p * T(s, n, k - 1) + T(s, n - 1, k)
def A(n, k): return T(n, k, k)
for n in range(9): print([A(n, k) for k in range(9)])

A126157 Main diagonal and central terms of symmetric triangle A126155.

Original entry on oeis.org

1, 5, 55, 1415, 69025, 5403005, 616437655, 96365988815, 19756766836225, 5140259013390005, 1654190282113104055, 645005323804145184215, 299613583126435893179425, 163464903833408195554809005

Offset: 0

Paul D. Hanna, Dec 20 2006

nonn

Cf. A126155; A126156 (column 0), A126158 (diagonal).

a(n)=Sum_{k, 0<=k<=n}A130847(n,k)*4^k. - Philippe Deléham, Jul 22 2007

A126158 Secondary diagonal of symmetric triangle A126155: a(n) = A126155(n+1,n).

Original entry on oeis.org

1, 35, 1195, 63365, 5126905, 594825635, 93900238195, 19371302880965, 5061231946045105, 1633629246059544035, 638388562675692767995, 297033561831219312442565, 162266449500902451982091305

Offset: 0

Paul D. Hanna, Dec 20 2006

nonn

Cf. A126155; A126156 (column 0), A126157 (main diagonal).

a(n)=A126157(n+1)-4*A126157(n). - Philippe Deléham, Jul 22 2007

A261042 Generating function g(0) where g(k) = 1 - x2(k+1)(k+2)/(x2(k+1)(k+2) - 1/g(k+1)).

Original entry on oeis.org

1, 4, 64, 2176, 126976, 11321344, 1431568384, 243680935936, 53725527801856, 14893509177769984, 5070334006399074304, 2079588119566033616896, 1011390382859091900891136, 575501120339508919401447424, 378784713733072451034702413824, 285539131625477547496925147693056

Offset: 0

Peter Luschny, Aug 08 2015

More generally let G(y) defined by the Taylor expansion of the continued fraction
g(y,k) = 1 - (y*x*(k+1)*(k+2)) / ((y*x*(k+1)*(k+2)) - 1/g(y,k+1)). Then
G(1/2) -> A002105, G(1) -> A000182, G(2) -> A261042, G(4) -> A253165 and G(1/8)(n) *2^(n-1+padic(n,2)) -> A002425.

Cf. A000182, A002105, A002425, A126156 (example section), A253165.

eulerCF := proc(f, len) local g, k; g := 1;
for k from len-2 by -1 to 0 do g := 1 - f(k)/(f(k)-1/g) od;
PolynomialTools:-CoefficientList(convert(series(g, x, len), polynom), x) end:
A261042_list := len -> eulerCF(k -> x*2*(k+1)*(k+2), len): A261042_list(16);
# Alternative:
ser := series(cos(x/sqrt(2))^(-2), x, 32):
seq(2^(2*n)*(2*n)!*coeff(ser, x, 2*n), n = 0..15); # Peter Luschny, Sep 03 2022

fracGen[f_, len_] := Module[{g, k}, g[len] = 1; For[k = len-1, k >= 0, k--, g[k] = 1-f[k]/(f[k]-1/g[k+1])]; CoefficientList[g[0] + O[x]^(len+1), x] ]; A261042list[len_] := fracGen[x*2*(#+1)*(#+2)&, len-1]; A261042list[16] (* Jean-François Alcover, Aug 08 2015, after Peter Luschny *)

def A261042_list(len):
    f = lambda k: x*2*(k+1)*(k+2)
    g = 1
    for k in range(len-2,-1,-1):
        g = (1-f(k)/(f(k)-1/g)).simplify_rational()
    return taylor(g, x, 0, len-1).list()
A261042_list(16)

a(n) = 2^(2*n)*(2*n)!*[x^(2*n)] cos(x/sqrt(2))^(-2). - Peter Luschny, Sep 03 2022

A087736 Triangle T(n,k) read by rows given by [0, 1, 3, 6, 10, 15, 21, ...] DELTA [1, 3, 6, 10, 15, 21, 28,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 4, 0, 4, 23, 34, 0, 34, 249, 606, 496, 0, 496, 4354, 14181, 20434, 11056, 0, 11056, 112238, 450097, 894838, 885032, 349504, 0, 349504, 4008024, 18911670, 47136095, 65613780, 48468804, 14873104

Offset: 0

Philippe Deléham, Sep 30 2003, Jul 17 2007

			Triangle begins:
1;
0, 1;
0, 1, 4;
0, 4, 23, 34;
0, 34, 249, 606, 496;
0, 496, 4354, 14181, 20434, 11056;
0, 11056, 112238, 450097, 894838, 885032, 349504;
0, 349504, 4008024, 18911670, 47136095, 65613780, 48468804, 14873104 ;...

Diagonals give A002105: [1, 1, 4, 34, 496, ...] Row sums give A000364 : [1, 1, 5, 61, 1385, ...] Euler numbers.

Cf. A000217 A000384 A002105 A084938.

Sum_{k, 0<=k<=n}T(n,k)*x^(n-k)=A000012(n), A011782(n), A001147(n), A002105(n+1), A000364(n), A126151(n), A126156(n) for n = -3,-2,-1,0,1,2,3 respectively .

Corrected and edited. - Philippe Deléham, Nov 24 2008

A186491 Counts a family of permutations occurring in the study of squeezed states of the simple harmonic oscillator.

Original entry on oeis.org

1, 2, 28, 1112, 87568, 11447072, 2239273408, 612359887232, 223061763490048, 104399900177326592, 61049165415292607488, 43617245341775265585152, 37385513306142843500105728, 37862584188750782065354022912

Offset: 0

Peter Bala, Feb 22 2011

The sequence a(n), with the convention a(0) = 1, enumerates permutations p(1)p(2)...p(4*n) in the symmetric group on 4*n letters having the following properties:
1) The permutation can be written as a product of disjoint two cycles.
2) For i = 1,...,2*n, positions 2*i-1 and 2*i are either both ascents (labeled A) or both descents (labeled D).
The set of permutations satisfying condition (1) forms a subgroup of Symm(4*n) of order A001147(4*n).
Here are some examples of permutations (written in cycle form) in Symm(8), satisfying these conditions, together with their ascent-descent labelings.
... (14)(23)(57)(68)  of type AADDAADD;
... (15)(26)(37)(48)  of type AAAADDDD.
Since the permutations being considered consist of disjoint 2-cycles their ascent-descent labelings must have an equal number of A's and D's.
Further examples can be found in the Example section below.
This family of permutations have arisen in the study of squeezed states
of the simple harmonic oscillator [Sukumar and Hodges].
See A186492 for a recursive triangle to compute this sequence.

			a(1)=2:
The two permutations in Symm(4) satisfying the conditions are
... (13)(24) of type AADD
... (14)(23) of type AADD.
a(2)=28:
Clearly, the ascent-descent structure of one of our permutations must start with an AA and finish with a DD so the two possible types are AAAADDDD and AADDAADD.
There are 4!=24 permutations of type AAAADDDD coming from the bijections of {1,2,3,4} onto {5,6,7,8}.
There are 2*2 = 4 permutations of the remaining type AADDAADD, namely
... (13)(24)(57)(68)
... (13)(24)(58)(67)
... (14)(23)(57)(68)
... (14)(23)(58)(67).

Jitender Singh, On an arithmetic convolution, arXiv:1402.0065 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.6.7.
C. V. Sukumar and A. Hodges, Quantum algebras and parity-dependent spectra, Proc. R. Soc. A (2007) 463.

Cf. A000364, A186492, A126156.

G:= sqrt(sec(2*x)): Gser := series(G, x = 0,32):
seq((2*n)!*coeff(Gser,x^(2*n)), n = 1..15);
# Alternative, using the Singh transformation 'g' from Maple in A126156:
a := n -> (-4)^n*g(euler, 2*n);
seq(a(n), n = 0..13);  # Peter Luschny, Sep 29 2023

a[n]:=if n=0 then 1 else sum(a[n-k]*binomial(2*n,2*k)*(k/(2*n)-1)*(-4)^k,k,1,n);
makelist(a[n],n,0,20); /* Tani Akinari, Sep 19 2023 */

GENERATING FUNCTION
(1)... sqrt(sec(2*x)) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!
= 1 + 2*x^2/2! + 28*x^4/4! + 1112*x^6/6! + ....
Compare with the e.g.f. Of A000364.
O.g.f. as a continued fraction: 1/(1-2*x/(1-12*x/(1-30*x/(...-2*n*(2*n-1)*x/(1-...))))) = 1 + 2*x + 28*x^2 + 1112*x^3 + ....
From Sergei N. Gladkovskii, Oct 23 2012: (Start)
G.f.: 1/U(0) where U(k) = 1 - (4*k+1)*(4*k+2)*x/( 1 - (4*k+3)*(4*k+4)*x/U(k+1)); (continued fraction, 2-step).
G.f.: 1/S(0) where S(k) = 1 - 2*x*(16*k^2 + 4*k + 1) - 8*x^2*(k+1)*(2*k+1)*(4*k+1)*(4*k+3)/S(k+1); (continued fraction, 1-step).
(End)
Let A(x) = Sum_{n>=0} a(n)*x^n = 1/T(0) where T(k)= 1 - (2*k+1)*(2*k+2)*x^2/T(k+1) -(continued fraction, 1-step),- then sqrt(sec(2*x)) = Sum_{n>=0} a(n)*x^n/n!. - Sergei N. Gladkovskii, Oct 25 2012
G.f.: 1/S(0) where S(k)= 1 - (2*k+1)*(2*k+2)*x /S(k+1); (continued fraction, 1-step). - Sergei N. Gladkovskii, Oct 26 2012
G.f.: Q(0), where Q(k) = 1 - x*(2*k+1)*(2*k+2)/(x*(2*k+1)*(2*k+2) - 1/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Oct 09 2013
For n > 0, a(n) = Sum_{k=1..n} a(n-k)*binomial(2*n,2*k)*(k/(2*n)-1)*(-4)^k. - Tani Akinari, Sep 19 2023.

A365672 Triangle read by rows. T(n, k) = 1 if k = 0, equals T(n, k-1) if k = n, and otherwise is (n - k + 1) * (2 * (n - k) + 1) * T(n, k - 1) + T(n - 1, k).

Original entry on oeis.org

1, 1, 1, 1, 7, 7, 1, 22, 139, 139, 1, 50, 889, 5473, 5473, 1, 95, 3549, 58708, 357721, 357721, 1, 161, 10794, 360940, 5771821, 34988647, 34988647, 1, 252, 27426, 1595110, 50434901, 791512162, 4784061619, 4784061619

Offset: 0

Peter Luschny, Sep 29 2023

This triangle is described by Peter Bala (see link).

This a weighted generalized Catalan triangle (A365673) with the hexagonal numbers as weights.

			Triangle T(n, k) starts:
[0] 1;
[1] 1,   1;
[2] 1,   7,     7;
[3] 1,  22,   139,     139;
[4] 1,  50,   889,    5473,     5473;
[5] 1,  95,  3549,   58708,   357721,    357721;
[6] 1, 161, 10794,  360940,  5771821,  34988647,   34988647;
[7] 1, 252, 27426, 1595110, 50434901, 791512162, 4784061619, 4784061619;

Peter Bala, A triangle for calculating A126156.

Cf. A000384, A126156 (main diagonal), A365673 (general case).

T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else (n - k + 1) * (2 * (n - k) + 1) * T(n, k - 1) + T(n - 1, k) fi fi end:

cp's OEIS Frontend

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A126155 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 5 times first term.

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A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

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A365673 Array A(n, k) read by ascending antidiagonals. Polygonal number weighted generalized Catalan sequences.

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A126157 Main diagonal and central terms of symmetric triangle A126155.

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A261042 Generating function g(0) where g(k) = 1 - x*2*(k+1)*(k+2)/(x*2*(k+1)*(k+2) - 1/g(k+1)).

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A087736 Triangle T(n,k) read by rows given by [0, 1, 3, 6, 10, 15, 21, ...] DELTA [1, 3, 6, 10, 15, 21, 28,...] where DELTA is the operator defined in A084938.

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A261042 Generating function g(0) where g(k) = 1 - x2(k+1)(k+2)/(x2(k+1)(k+2) - 1/g(k+1)).