A125055 -id:A125055 - 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.

A125054 Central terms of triangle A125053.

Original entry on oeis.org

1, 3, 21, 327, 9129, 396363, 24615741, 2068052367, 225742096209, 31048132997523, 5252064083753061, 1071525520294178007, 259439870666594250489, 73542221109962636293083, 24125551094579137082039181, 9068240688454120376775401247, 3871645204706420218816959159969

Offset: 0

Paul D. Hanna, Nov 21 2006

nonn

Triangle A125053 is a variant of triangle A008301 (enumeration of binary trees) such that the leftmost column is the secant numbers (A000364).
Right edge of triangle A210108.
Apparently all terms (except the initial 1) have 3-valuation 1. - F. Chapoton, Aug 02 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)

Cf. A125053, A125055, A000182.

b[n_]:=n!*SeriesCoefficient[Tan[x],{x,0,n}]; Table[Sum[Binomial[n,k]*b[2*k+1],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, May 30 2015 *)

Binomial transform of A000182 (e.g.f. tan(x)).
a(n) = Sum_{k=0..n} A130847(n,k)*2^k. - Philippe Deléham, Jul 22 2007
G.f.: 1/(1-sqrt(x))/Q(0), where Q(k)= 1 + sqrt(x) - x*(2*k+1)*(2*k+2)/(1 - sqrt(x) - x*(2*k+2)*(2*k+3)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Apr 27 2013
G.f.: Q(0)/(1-3*x), where Q(k) = 1 - 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2/( 4*x^2*(2*k+1)*(2*k+3)*(k+1)^2 - (1 - 8*x*k^2 - 8*x*k -3*x)*(1 - 8*x*k^2 - 24*x*k -19*x)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 23 2013
G.f.: Q(0)/(1-1*x), where Q(k) = 1 - (2*k+1)*(2*k+2)*x/(x*(2*k+1)*(2*k+2) - (1-x)/(1 - (2*k+2)*(2*k+3)*x/(x*(2*k+2)*(2*k+3) - (1-x)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2013
a(n) ~ 2^(4*n+5) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+3/2)). - Vaclav Kotesovec, May 30 2015
From Peter Bala, May 11 2017: (Start)
O.g.f. as an S-fraction: A(x) = 1/(1 - 3*x/(1 - 4*x/(1 - 15*x/(1 - 16*x/(1 - 35*x/(1 - 36*x/(1 - ...))))))), 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,....
A(x) = 1/(1 + 3*x - 6*x/(1 - 2*x/(1 + 3*x - 20*x/(1 - 12*x/(1 + 3*x - 42*x/(1 - 30*x/(1 + 3*x - ...))))))), , where the unsigned coefficients in the partial numerators [6, 2, 20, 12, 42, 30, ...] are obtained from the sequence [2, 6, 12, 20, ..., n*(n + 1), ...] by swapping adjacent terms. (End)

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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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