A008301 -id:A008301 - cp's OEIS frontend

A125053 Variant of triangle A008301, read by rows of 2*n+1 terms, such that the first column is the secant numbers (A000364).

1, 1, 3, 1, 5, 15, 21, 15, 5, 61, 183, 285, 327, 285, 183, 61, 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385, 50521, 151563, 247065, 325947, 378105, 396363, 378105, 325947, 247065, 151563, 50521, 2702765, 8108295, 13311741, 17908935

Offset: 0

Paul D. Hanna, Nov 21 2006, Dec 20 2006

Foata and Han refer to this as the triangle of Poupard numbers h_n(k). - N. J. A. Sloane, Feb 17 2014

Central terms (A125054) equal the binomial transform of the tangent numbers (A000182).

			If we write the triangle like this:
......................... ...1;
................... ...1, ...3, ...1;
............. ...5, ..15, ..21, ..15, ...5;
....... ..61, .183, .285, .327, .285, .183, ..61;
. 1385, 4155, 6681, 8475, 9129, 8475, 6681, 4155, 1385;
then the first nonzero term is the sum of the previous row:
1385 = 61 + 183 + 285 + 327 + 285 + 183 + 61,
the next term is 3 times the first:
4155 = 3*1385,
and the remaining terms in each row are obtained by the rule illustrated by:
6681 = 2*4155 - 1385 - 4*61;
8475 = 2*6681 - 4155 - 4*183;
9129 = 2*8475 - 6681 - 4*285;
8475 = 2*9129 - 8475 - 4*327;
6681 = 2*8475 - 9129 - 4*285;
4155 = 2*6681 - 8475 - 4*183;
1385 = 2*4155 - 6681 - 4*61.
An alternate recurrence is illustrated by:
4155 = 1385 + 2*(61 + 183 + 285 + 327 + 285 + 183 + 61);
6681 = 4155 + 2*(183 + 285 + 327 + 285 + 183);
8475 = 6681 + 2*(285 + 327 + 285);
9129 = 8475 + 2*(327);
and then for k>n, T(n,k) = T(n,2*n-k).

Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened
Dominique Foata and Guo-Niu Han, Seidel Triangle Sequences and Bi-Entringer Numbers, Nov 20 2013;
Foata, Dominique; Han, Guo-Niu; Strehl, Volker The Entringer-Poupard matrix sequence. Linear Algebra Appl. 512, 71-96 (2017).

Cf. A008301, A000364 (secant numbers, which are the row sums), A125054 (central terms), A125055, A000182, A008282.

Cf. A210111 (left half).

a125053 n k = a125053_tabf !! n !! k
a125053_row n = a125053_tabf !! n
a125053_tabf = iterate f [1] where
f zs = zs' ++ reverse (init zs') where
zs' = (sum zs) : g (map (* 2) zs) (sum zs)
g [x] y = [x + y]
g xs y = y' : g (tail $ init xs) y' where y' = sum xs + y
-- Reinhard Zumkeller, Mar 17 2012

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 3*T(n, 1)
elif k > n then T(n, 2*n-k)
else 2*T(n, k-1) - T(n, k-2) - 4*T(n-1, k-2)
  fi end:
seq(print(seq(T(n,k), k=1..2*n-1)), n=1..5); # Peter Luschny, May 11 2014

t[n_, k_] := t[n, k] = If[2*n < k || k < 0, 0, If[n == 0 && k == 0, 1, If[k == 0, Sum[t[n-1, j], {j, 0, 2*n-2}], If[k <= n, t[n, k-1] + 2*Sum[t[n-1, j], {j, k-1, 2*n-1-k}], t[n, 2*n-k]]]]]; Table[t[n, k], {n, 0, 6}, {k, 0, 2*n}] // Flatten (* Jean-François Alcover, Dec 06 2012, translated from Pari *)

PARI
```
T(n,k)=if(2*n
				
```

Sum_{k=0..2n} C(2n,k)*T(n,k) = 4^n * A000182(n), where A000182 are the tangent numbers.

Sum_{k=0..2n} (-1)^n*C(2n,k)*T(n,k) = (-4)^n.

A210108 Left half of Poupard's triangle, A008301.

Original entry on oeis.org

1, 1, 2, 4, 8, 10, 34, 68, 94, 104, 496, 992, 1420, 1712, 1816, 11056, 22112, 32176, 40256, 45496, 47312, 349504, 699008, 1026400, 1309568, 1528384, 1666688, 1714000, 14873104, 29746208, 43920304, 56696384, 67419664, 75523808, 80571184, 82285184, 819786496

Offset: 0

Reinhard Zumkeller, Mar 17 2012

Reinhard Zumkeller, Rows n=0..120 of triangle, flattened

Cf. A002105 (left edge), A005799 (right edge); A210111.

a210108 n k = a210108_tabl !! n !! k
a210108_row n = a210108_tabl !! n
a210108_tabl = zipWith take [1..] a008301_tabf

T(n,k) = A008301(n,k), 0 <= k <= n.

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.

A126150 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 4 times first term.

Original entry on oeis.org

1, 1, 4, 1, 6, 24, 36, 24, 6, 96, 384, 636, 744, 636, 384, 96, 2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976, 151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416, 11449296, 45797184

Offset: 0

Paul D. Hanna, Dec 19 2006

			Triangle begins:
  1;
  1, 4, 1;
  6, 24, 36, 24, 6;
  96, 384, 636, 744, 636, 384, 96;
  2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976;
  151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416; ...
If we write the triangle like this:
  .......................... ....1;
  ................... ....1, ....4, ....1;
  ............ ....6, ...24, ...36, ...24, ....6;
  ..... ...96, ..384, ..636, ..744, ..636, ..384, ...96;
  .2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, .2976;
then the first term in each row is the sum of the previous row:
  2976 = 96 + 384 + 636 + 744 + 636 + 384 + 96
the next term is 4 times the first:
  11904 = 4*2976,
and the remaining terms in each row are obtained by the rule illustrated by:
  20256 = 2*11904 - 2976 - 6*96;
  26304 = 2*20256 - 11904 - 6*384;
  28536 = 2*26304 - 20256 - 6*636;
  26304 = 2*28536 - 26304 - 6*744;
  20256 = 2*26304 - 28536 - 6*636;
  11904 = 2*20256 - 26304 - 6*384;
  2976 = 2*11904 - 20256 - 6*96.
An alternate recurrence is illustrated by:
  11904 = 2976 + 3*(96 + 384 + 636 + 744 + 636 + 384 + 96);
  20256 = 11904 + 3*(384 + 636 + 744 + 636 + 384);
  26304 = 20256 + 3*(636 + 744 + 636);
  28536 = 26304 + 3*(744);
and then for k>n, T(n,k) = T(n,2n-k).

Cf. A126151 (column 0); diagonals: A126152, A126153; A126154; variants: A008301, A125053, A126155.

PARI
```
T(n,k)=local(p=3);if(2*n
				
```

/* Alternate Recurrence: */ T(n,k)=local(p=3);if(2*n

Sum_{k=0..2n} (-1)^k*C(2n,k)*T(n,k) = (-6)^n.

A002105 Reduced tangent numbers: 2^n(2^{2n} - 1)|B_{2n}|/n, where B_n = Bernoulli numbers.

Original entry on oeis.org

1, 1, 4, 34, 496, 11056, 349504, 14873104, 819786496, 56814228736, 4835447317504, 495812444583424, 60283564499562496, 8575634961418940416, 1411083019275488149504, 265929039218907754399744, 56906245479134057176170496, 13722623393637762299131396096, 3704005473270641755597685653504

Offset: 1

N. J. A. Sloane

Comments from R. L. Graham, Apr 25 2006 and Jun 08 2006: "This sequence also gives the number of ways of arranging 2n tokens in a row, with 2 copies of each token from 1 through n, such that the first token is a 1 and between every pair of tokens labeled i (i=1..n-1) there is exactly one token labeled i+1.
"For example, for n=3, there are 4 possibilities: 123123, 121323, 132312 and 132132 and indeed a(3) = 4. This is the work of my Ph. D. student Nan Zang. See also A117513, A117514, A117515.
"Develin and Sullivant give another occurrence of this sequence and show that their numbers have the same generating function, although they were unable to find a 1-1-mapping between their problem and Poupard's."
The sequence 1,0,1,0,4,0,34,0,496,0,11056, ... counts increasing complete binary trees with e.g.f. sec^2(x/sqrt 2). - Wenjin Woan, Oct 03 2007
a(n) = number of increasing full binary trees on vertex set [2n-1] with the left-largest property: the largest descendant of each non-leaf vertex occurs in its left subtree (Poupard). The first Mathematica recurrence below counts these trees by number 2k-1 of vertices in the left subtree of the root: the root is necessarily labeled 1 and n necessarily occurs in the left subtree and so there are Binomial[2n-3,2k-2] ways to choose the remaining labels for the left subtree. - David Callan, Nov 29 2007
Number of bilabeled unordered increasing trees with 2n labels. - Markus Kuba, Nov 18 2014
Conjecture: taking the sequence modulo an integer k gives an eventually purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 10 begins [1, 1, 4, 4, 6, 6, 4, 4, 6, 6, 4, 4, 6, 6, ...] with an apparent period [4, 4, 6, 6] of length 4 = phi(10) beginning at a(3). - Peter Bala, May 08 2023
Let c(1), c(2), c(3), ... be a geometric progression and s = (2*c(1)/c(2))^(1/2). Then c(1)*s*tan(x/s) = Sum_{n>0} a(n) * c(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Jan 15 2025

			G.f. = x + x^2 + 4*x^3 + 34*x^4 + 496*x^5 + 11056*x^6 + 349504*x^7 + ...

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).

T. D. Noe, Table of n, a(n) for n = 1..100
G. E. Andrews, J. Jimenez-Urroz, and K. Ono, q-series identities and values of certain L-functions, Duke Math J. Volume 108, Number 3 (2001), 395-419.
R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
Peter Bala, A continued fraction expansion related to A002105
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
M. P. Develin and S. P. Sullivant, Markov Bases of Binary Graph Models, Annals of Combinatorics 7 (2003) 441-466.
Dominique Foata and Guo-Niu Han, The doubloon polynomial triangle, Ram. J. 23 (2010), 107-126
Dominique Foata and Guo-Niu Han, Doubloons and new q-tangent numbers, Quart. J. Math. 62 (2) (2011) 417-432
Dominique Foata and Guo-Niu Han, Tree Calculus for Bivariable Difference Equations, 2012. - From _N. J. A. Sloane_, Feb 02 2013
R. L. Graham and Nan Zang, Enumerating split-pair arrangements, J. Combin. Theory Ser. A 115 (2008), no. 2, 293-303.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Markus Kuba and Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 2014.
B. A. Kupershmidt, On skew-symmetric and general deformations of Lax pseudodifferential operators, J. Math. Sci. 151 (4) (2008) 3139, eq. (1.13).
Zhicong Lin, Shi-Mei Ma, David G. L. Wang, and Liuquan Wang, Positivity and divisibility of alternating descent polynomials, arXiv:2011.02685 [math.CO], 2020.
E. Norton, Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions, arXiv preprint arXiv:1302.5411 [math.RA], 2013.
Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
C. Poupard, Deux proprietes des arbres binaires ordonnes stricts, European J. Combin., 10 (1989), 369-374.
M. Safaryan, Problem 11725, Amer. Math. Monthly, 120 (2013), 661.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 66-67.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 64-67. [Annotated scanned copy]
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011.
Index entries for sequences related to Bernoulli numbers.

Row sums of A008301.
Left edge of triangle A210108.
Cf. A000111, A000217, A000364, A000464, A002439, A008281, A079144, A158690.

A002105:= func< n | (-1)^(n+1)*2^n*(4^n - 1)*Bernoulli(2*n)/n >;
[A002105(n): n in [1..30]]; // G. C. Greubel, Sep 20 2024

S := proc(n, k) option remember; if k=0 then `if`(n=0, 1, 0) else S(n, k-1) + S(n-1, n-k) fi end: A002105 := n -> S(2*n-1, 2*n-1)/2^(n-1): seq(A002105(i),i=1..16); # Peter Luschny, Jul 08 2012
# The above function written as a formula: a(n) = A008281(2*n-1, 2*n-1)/2^(n-1).
# Alternatively, based on the triangular numbers A000217:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else
A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Sep 30 2023

u[1] = 1; u[n_]/;n>=2 := u[n] = Sum[Binomial[2n-3,2k-2]u[k]u[n-k],{k,n-1}]; Table[u[n],{n,8}] (* Poupard and also Develin and Sullivant, give a different recurrence that involves a symmetric sum: v[1] = 1; v[n_]/;n>=2 := v[n] = 1/2 Sum[Binomial[2n-2,2k-1]v[k]v[n-k],{k,n-1}] *) (*David Callan, Nov 29 2007 *)
a[n_] := (-1)^n 2^(n+1) PolyLog[1-2n, -1]; Array[a, 10] (* Vladimir Reshetnikov, Jan 23 2011 *)
Table[(-1)^(n+1)*2^n*(2^(2n)-1)*BernoulliB[2n]/n,{n,1,20}] (* Vaclav Kotesovec, Nov 03 2014 *)
eulerCF[f_, len_] := Module[{g}, g[len-1]=1; g[k_]:=g[k]=1-f[k]/(f[k]-1/g[k+1]); CoefficientList[g[0] + O[x]^len, x]]; A002105List[len_] := eulerCF[(1/2) x (#+1) (#+2)&, len]; A002105List[19] (* Peter Luschny, Aug 08 2015 after Sergei N. Gladkovskii *)
Table[PolyGamma[2n-1, 1/2] 2^(2-n)/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
Table[EulerE[2n-1, 0] (-2)^n, {n, 1, 10}] (* Vladimir Reshetnikov, Oct 21 2015 *)

{a(n) = if( n<1, 0, ((-2)^n - (-8)^n) * bernfrac(2*n) / n)}; /* Michael Somos, Jun 22 2002 */

{a(n) = if( n<0, 0, (2*n)! * polcoeff( -2 * log( cos(x / quadgen(8) + O(x^(2*n + 1)))), 2*n))}; /* Michael Somos, Jul 17 2003 */

{a(n) = if( n<0, 0, -(-2)^(n+1) * sum(i=1, 2*n, 2^-i * sum(j=1, i, (-1)^j * binomial( i-1, j-1) * j^(2*n - 1))))}; /* Michael Somos, Sep 07 2013 */

{a(n)=local(CF=1+x*O(x^n));if(n<1,return(0), for(k=1,n,CF=1/(1-(n-k+1)*(n-k+2)/2*x*CF));return(Vec(CF)[n]))}  /* Paul D. Hanna */

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

from sympy import bernoulli
def A002105(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 A002105_list(n) :
    D = [0]*(n+2); D[1] = 1
    R = []; z = 1/2; b = True
    for i in(0..2*n-1) :
        h = i//2 + 1
        if b :
            for k in range(h-1, 0, -1) : D[k] += D[k+1]
            z *= 2
        else :
            for k in range(1, h+1, 1) :  D[k] += D[k-1]
        b = not b
        if b : R.append(D[h]*z/h)
    return R
A002105_list(16) # Peter Luschny, Jun 29 2012

def A002105(n): return (-1)^(n+1)*2^n*(4^n -1)*bernoulli(2*n)/n
[A002105(n) for n in range(1,31)] # G. C. Greubel, Sep 20 2024

E.g.f.: 2*log(sec(x / sqrt(2))) = Sum_{n>0} a(n) * x^(2*n) / (2*n)!. - Michael Somos, Jun 22 2002
A000182(n) = 2^(n-1) * a(n). - Michael Somos, Jun 22 2002
a(n) = 2^(n-1)/n * A110501(n). - Don Knuth, Jan 16 2007
a(n+1) = Sum_{k = 0..n} A094665(n, k). - Philippe Deléham, Jun 11 2004
O.g.f.: A(x) = x/(1-x/(1-3*x/(1-6*x/(1-10*x/(1-15*x/(... -n*(n+1)/2*x/(1 - ...))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
sqrt(2) tan( x/sqrt(2)) = Sum_(n>=0) (x^(2n+1)/(2n+1)!) a_n. - Dominique Foata and Guo-Niu Han, Oct 24 2008
Basic hypergeometric generating function: Sum_{n>=0} Product {k = 1..n} (1-exp(-2*k*t))/Product {k = 1..n} (1+exp(-2*k*t)) = 1 + t + 4*t^2/2! + 34*t^3/3! + 496*t^4/4! + ... [Andrews et al., Theorem 4]. For other sequences with generating functions of a similar type see A000364, A000464, A002439, A079144 and A158690. - Peter Bala, Mar 24 2009
E.g.f.: Sum_{n>=0} Product_{k=1..n} tanh(k*x) = Sum_{n>=0} a(n)*x^n/n!. - Paul D. Hanna, May 11 2010
a(n) = (-1)^(n+1)*sum(j!*stirling2(2*n+1,j)*2^(n+1-j)*(-1)^(j),j,1,2*n+1), n>=0. - Vladimir Kruchinin, Aug 23 2010
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) = upper left term in M^n, a(n+1) = sum of top row terms in M^n; where M = the infinite square production matrix:
   1, 3, 0,  0,  0, 0, 0, ...
   1, 3, 6,  0,  0, 0, 0, ...
   1, 3, 6, 10,  0, 0, 0, ...
   1, 3, 6, 10, 15, 0, 0, ... (End)
E.g.f. A(x) satisfies differential equation A''(x)=exp(A(x)). - Vladimir Kruchinin, Nov 18 2011
E.g.f.: For E(x)=sqrt(2)* tan( x/sqrt(2))=x/G(0); G(k)= 4*k + 1 - x^2/(8*k + 6 - x^2/G(k+1)); (from continued fraction Lambert's, 2-step). - Sergei N. Gladkovskii, Jan 14 2012
a(n) = (-1)^n*2^(n+1)*Li_{1-2*n}(-1). (See also the Mathematica prog. by Vladimir Reshetnikov.) - Peter Luschny, Jun 28 2012
G.f.: 1/G(0) where G(k) = 1 - x*( 4*k^2 + 4*k + 1 ) - x^2*(k+1)^2*( 4*k^2 + 8*k + 3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 14 2013
G.f.: 1/Q(0), where Q(k) = 1 - (k+1)*(k+2)/2*x/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: (1/G(0))/sqrt(x) - 1/sqrt(x), where G(k) = 1 - sqrt(x)*(2*k+1)/(1 + sqrt(x)*(2*k+1)/(1 + sqrt(x)*(k+1)/(1 - sqrt(x)*(k+1)/G(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Jul 07 2013
log(2) - 1/1 + 1/2 - 1/3 + ... + (-1)^n / n = (-1)^n / 2 * (1/n - 1 / (2*n^2) + 1 / (2*n^2)^2 - 4 / (2*n^2)^3 + ... + (-1)^k * a(k) / (2*n^2)^k + ...) asymptotic expansion. - Michael Somos, Sep 07 2013
G.f.: T(0), where T(k) = 1-x*(k+1)*(k+2)/(x*(k+1)*(k+2)-2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 24 2013
a(n) ~ 2^(3*n+2) * n^(2*n-1/2) / (exp(2*n) * Pi^(2*n-1/2)). - Vaclav Kotesovec, Nov 03 2014
From Peter Bala, Sep 10 2015: (Start)
The e.g.f. A(x) = sqrt(2)*tan(x/sqrt(2)) satisfies A''(x) = A(x)*A'(x), hence the recurrence a(0) = 0, a(1) = 1, else a(n) = Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0,1,0,1,0,4,0,34,0,496,...].
Note that the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence [1,1,1,2,5,16,61,272,...] = A000111. (End)
a(n) = polygamma(2*n-1, 1/2)*2^(2-n)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
E.g.f.: sqrt(2)*tan(x/sqrt(2)) = Sum_{n>0} a(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Mar 05 2017
From Peter Bala, May 05 2017: (Start)
Let B(x) = A(x)/x = 1 + x + 4*x^2 + 34*x^3 + ... denote the shifted o.g.f. Then B(x) = 1/(1 + 2*x - 3*x/(1 - x/(1 + 2*x - 10*x/(1 - 6*x/(1 + 2*x - 21*x/(1 - 15*x/(1 + 2*x - 36*x/(1 - 28*x/(1 + 2*x - ...))))))))), where the coefficient sequence [3, 1, 10, 6, 21, 15, 36, 28, ...] in the partial numerators of the continued fraction is obtained by swapping adjacent triangular numbers. Cf. A079144.
It follows (by means of an equivalence transformation) that the second binomial transform of B(x), with g.f. equal to 1/(1 - 2*x)*B(x/(1 - 2*x)), has the S-fraction representation 1/(1 - 3*x/(1 - x/(1 - 10*x/(1 - 6*x/(1 - 21*x/(1 - 15*x/(1 - 36*x/(1 - 28*x/(1 - ...))))))))). Compare with the S-fraction representation of the g.f. A(x) given above by Hanna, dated Oct 07 2005. (End)
The computation can be based on the triangular numbers, a(n) = T(n, n) where T(n, k) = A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) for 0 < k < n, and T(n, 0) = 1, T(n, n) = T(n, k-1) if k > 0. This is equivalent to Paul D. Hanna's continued fraction 2005. - Peter Luschny, Sep 30 2023

Additional comments from Michael Somos, Jun 25 2002

A005799 Generalized Euler numbers of type 2^n.

Original entry on oeis.org

1, 1, 2, 10, 104, 1816, 47312, 1714000, 82285184, 5052370816, 386051862272, 35917232669440, 3996998043812864, 524203898507631616, 80011968856686405632, 14061403972845412526080, 2818858067801804443910144

Offset: 0

N. J. A. Sloane, Simon Plouffe

Also, a(n) equals the number of alternating permutations (p(1),...,p(2n)) of the multiset {1,1,2,2,...,n,n} satisfying p(1) <= p(2) > p(3) <= p(4) > p(5) <= ... <= p(2n). Hence, A275801(n) <= a(n) <= A275829(n). - Max Alekseyev, Aug 10 2016
This is the BinomialMean transform of A000364 (see A075271 for definition of transform). - John W. Layman, Dec 04 2002
This sequence appears to be middle column in Poupard's triangle A008301.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
Shishuo Fu, Zhicong Lin, and Zhi-Wei Sun, Proof of several conjectures relating permanents to Combinatorial sequences, arXiv:2109.11506v3 [math.CO], 2021-2023.
Shishuo Fu, Zhicong Lin, and Zhi-Wei Sun, Permanent identities, combinatorial sequences, and permutation statistics, Advances in Applied Mathematics, Volume 163, Part A, 102789 (2025).
Ira M. Gessel, Symmetric functions and P-recursiveness, J. Combin. Theory Ser. A 53 (1990), no. 2, 257-285.
H. Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

Right edge of triangle A210108.

Cf. A000364, A008301, A275801, A275829, A154283, A000657.

T := proc(n, k) option remember;
if n < 0 or k < 0 then 0
elif n = 0 then euler(k, 1)
else T(n-1, k+1) - T(n-1, k) fi end:
a := n -> (-2)^n*T(n, n); seq(a(n), n=0..16); # Peter Luschny, Aug 23 2017

a[n_] := Sum[Binomial[n, i]Abs[EulerE[2i]], {i, 0, n}]/2^n

a(n) = (1/2^n) * Sum_{i=0..n} binomial(n, i) * A000364(i).
From Sergei N. Gladkovskii, Dec 27 2012, Oct 11 2013, Oct 27 2013, Jan 08 2014: (Start) Continued fractions:
G.f.: A(x) = 1/G(0) where G(k) = 1 - x*(k+1)*(2*k+1)/(1 - x*(k+1)*(2*k+1)/G(k+1)).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)^2*(2*k+1)^2/(x^2*(k+1)^2*(2*k+1)^2 - (4*x*k^2 + 2*x*k + x - 1)*( 4*x*k^2 + 10*x*k + 7*x - 1)/Q(k+1)).
G.f.: R(0), where R(k) = 1 - x*(2*k+1)*(k+1)/(x*(2*k+1)*(k+1) - 1/(1 - x*(2*k+1)*(k+1)/(x*(2*k+1)*(k+1) - 1/R(k+1)))).
G.f.: 2/(x*Q(0)), where Q(k) = 2/x - 1 - (2*k+1)^2/(1 - (2*k+2)^2/Q(k+1)). (End)
a(n) ~ 2^(3*n+3) * n^(2*n+1/2) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, May 30 2015
a(n) = 2^n * Sum_{k=0..n} (-1)^k*binomial(n, k)*euler(n+k, 1). - Peter Luschny, Aug 23 2017
From Peter Bala, Dec 21 2019: (Start)
O.g.f. as a continued fraction: 1/(1 - x/(1 - x/(1 - 6*x/(1 - 6*x/(1 - 15*x/(1 - 15*x/(1 - ... - n*(2*n-1)*x/(1 - n*(2*n-1)*x/(1 - ...))))))))) - apply Bala, Proposition 3, with a = 0, b = 1 and replace x with x/2.
Conjectures:
E.g.f. as a continued fraction: 2/(2 - (1-exp(-4*t))/(2 - (1-exp(-8*t))/(2 - (1-exp(-12*t))/(2 - ... )))) = 1 + t + 2*t^2/2! + 10*t^3/3! + 104*t^4/4! + ....
Cf. A000657. [added April 18 2024: for a proof of this conjecture see Fu et al., Section 4.3.]
a(n) = (-2)^(n+1)*Sum_{k = 0..floor((n-1)/2)} binomial(n,2*k+1)*(2^(2*n-2*k) - 1)*Bernoulli(2*n-2*k)/(2*n-2*k) for n >= 1. (End)

Edited by Dean Hickerson, Dec 10 2002

cp's OEIS Frontend

A125053 Variant of triangle A008301, read by rows of 2*n+1 terms, such that the first column is the secant numbers (A000364).

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A210108 Left half of Poupard's triangle, A008301.

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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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A126150 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 4 times first term.

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A002105 Reduced tangent numbers: 2^n*(2^{2n} - 1)*|B_{2n}|/n, where B_n = Bernoulli numbers.

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A005799 Generalized Euler numbers of type 2^n.

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A002105 Reduced tangent numbers: 2^n(2^{2n} - 1)|B_{2n}|/n, where B_n = Bernoulli numbers.