A111959 -id:A111959 - cp's OEIS frontend

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310

Offset: 0

Wolfdieter Lang

Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.
As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005
The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007
As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010
From Wolfdieter Lang, Aug 10 2017: (Start)
The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):
  (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).
This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).
For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)
From Peter Bala, Mar 04 2018: (Start)
The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....
2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).
Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)

			Array begins:
  1,  2,   6,  20,   70, ...
  1,  6,  30, 140,  630, ...
  1, 10,  70, 420, 2310, ...
  1, 14, 126, 924, 6006, ...
Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
From _Paul Barry_, Apr 14 2010: (Start)
As a number triangle, T(n, m) begins:
n\k       0      1       2       3      4      5     6    7   8  9 10 ...
0:        1
1:        2      1
2:        6      6       1
3:       20     30      10       1
4:       70    140      70      14      1
5:      252    630     420     126     18      1
6:      924   2772    2310     924    198     22     1
7:     3432  12012   12012    6006   1716    286    26    1
8:    12870  51480   60060   36036  12870   2860   390   30   1
9:    48620 218790  291720  204204  87516  24310  4420  510  34  1
10:  184756 923780 1385670 1108536 554268 184756 41990 6460 646 38  1
... [Reformatted and extended by _Wolfdieter Lang_, Aug 10 2017]
Production matrix begins
      2, 1,
      2, 4, 1,
     -4, 0, 4, 1,
     10, 0, 0, 4, 1,
    -28, 0, 0, 0, 4, 1,
     84, 0, 0, 0, 0, 4, 1,
   -264, 0, 0, 0, 0, 0, 4, 1,
    858, 0, 0, 0, 0, 0, 0, 4, 1,
  -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - _Wolfdieter Lang_, Aug 10 2017
From _Peter Bala_, Feb 15 2018: (Start)
With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
-x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ).
x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End)

Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Muniru A Asiru, Antidiagonals n = 0..50, flattened
Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
Wolfdieter Lang, First 10 rows.
Wolfdieter Lang, On polynomials related to derivatives of the generating function of Catalan numbers, Fib. Quart. 40,4 (2002) 299-313; T(n,m) is called B(n,m) there.
Benjamin Lovitz and Nathaniel Johnston, A hierarchy of eigencomputations for polynomial optimization on the sphere, arXiv:2310.17827 [math.OC], 2023. See pp. 35, 39. See also Math. Prog. (2025) Series A.
Helmut Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.

Columns include: A000984 (m=0), A002457 (m=1), A002802 (m=2), A020918 (m=3), A020920 (m=4), A020922 (m=5), A020924 (m=6), A020926 (m=7), A020928 (m=8), A020930 (m=9), A020932 (m=10).
Row sums: A046748.
Cf. A001147, A006882, A007318, A068555, A111959, A283150, A283151.

Flat(List([0..9],n->List([0..n],m->Binomial(2*n,n)*Binomial(n,m)/Binomial(2*m,m)))); # Muniru A Asiru, Jul 19 2018

[Binomial(n+1,k+1)*Catalan(n)/Catalan(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 28 2024

t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *)

T(i,j)=if(i<0 || j<0,0,(2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)

def A046521(n,k): return binomial(n+1, k+1)*catalan_number(n)/catalan_number(k)
flatten([[A046521(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 28 2024

T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1.
As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010
From Peter Bala, Apr 11 2012: (Start):
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555.
The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
T(n,k) = 4^(n-k)*A006882(2*n - 1)/(A006882(2*n - 2*k)*A006882(2*k - 1)) = 4^(n-k)*(2*n - 1)!!/((2*n - 2*k)!*(2*k - 1)!!). - Peter Bala, Nov 07 2016
Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
From Peter Bala, Aug 13 2021: (Start)
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of  formal power series then
G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
The m-th power of this array has entries m^(n-k)*T(n,k). (End)

A012150 Expansion of e.g.f. exp(tan(arcsin(x))).

Original entry on oeis.org

1, 1, 1, 4, 13, 76, 421, 3256, 25369, 245008, 2449801, 28441216, 346065061, 4700478784, 67243537453, 1047088053376, 17192488230961, 302112622479616, 5593309059948049, 109527844826856448, 2255588021494237501

Offset: 0

Patrick Demichel (patrick.demichel(AT)hp.com)

nonn

			exp(tan(arcsin(x))) = 1+x+1/2!*x^2+4/3!*x^3+13/4!*x^4+76/5!*x^5...

Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.

Cf. A012194, A088009, A111959, A190863.

A012150 := proc(n) if n = 0 then 1; else add( (1+(-1)^(n-k)) *binomial((n-2)/2,(n-k)/2)/(2*k!), k=1..n) ; %*n! ; end if; end proc: # R. J. Mathar, Mar 20 2011

Range[0, 20]! CoefficientList[Series[Exp[Tan[ArcSin[x]]], {x, 0, 20}], x] (* Or *)
f[n_] := n! Sum[(1 + (-1)^(n - k)) Binomial[(n - 2)/2, (n - k)/2]/2/k!, {k, n}]; f[0] = 1; Array[f, 21, 0] (* Robert G. Wilson v, Feb 19 2011 *)

my(x='x+O('x^30)); Vec(serlaplace(exp(tan(asin(x))))) \\ Michel Marcus, Oct 30 2022

From Vladimir Kruchinin, Feb 17 2011: (Start)
a(n) = n!*Sum_{k=1..n} A111959(n-1,k-1)*2^(k-n)/k!.
a(n) = n!*Sum_{k=1..n} (1+(-1)^(n-k))*C((n-2)/2,(n-k)/2)/(2*k!), n>0.
E.g.f.: exp(x/sqrt(1-x^2)). (End)
E.g.f.: S(x) = exp(x/sqrt(1-x^2)) = 1 + 2*(x/sqrt(1-x^2))/(G(0) - x/sqrt(1-x^2)), G(k) = 8*k + 2 + (x^2)/((1-x^2)*(8*k+6) + x^2/G(k+1)); (continued fraction). - Sergei N. Gladkovskii, Dec 16 2011
a(n) = (3*n^2 - 12*n + 13)*a(n-2) - 3*(n-4)*(n-3)^2*(n-2)*a(n-4) + (n-6)*(n-5)*(n-4)^2*(n-3)*(n-2)*a(n-6). - Vaclav Kotesovec, Nov 08 2013
a(n) ~ n^(n-1/3) * exp(3/2*n^(1/3)-n) / sqrt(3) * (1 - 19/(36*n^(1/3)) + 553/(2592*n^(2/3))). - Vaclav Kotesovec, Nov 08 2013
a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n/2-1,k)/(n-2*k)!. - Seiichi Manyama, Jun 08 2024

Name edited by Michel Marcus, Oct 30 2022

A111960 Renewal array for central trinomial numbers A002426.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 7, 7, 3, 1, 19, 20, 12, 4, 1, 51, 61, 40, 18, 5, 1, 141, 182, 135, 68, 25, 6, 1, 393, 547, 441, 251, 105, 33, 7, 1, 1107, 1640, 1428, 888, 420, 152, 42, 8, 1, 3139, 4921, 4572, 3076, 1596, 654, 210, 52, 9, 1, 8953, 14762, 14535, 10456, 5880, 2652, 966, 280, 63, 10, 1

Offset: 0

Paul Barry, Aug 23 2005

Also the convolution triangle of A002426. - Peter Luschny, Oct 06 2022

			Triangle T(n,k) begins:
   1;
   1,  1;
   3,  2,  1;
   7,  7,  3,  1;
  19, 20, 12,  4, 1;
  51, 61, 40, 18, 5, 1;
  ...
From _Paul Barry_, May 12 2009: (Start)
Production matrix is
  1, 1,
  2, 1, 1,
  0, 2, 1, 1,
  -2, 0, 2, 1, 1,
  0, -2, 0, 2, 1, 1,
  4, 0, -2, 0, 2, 1, 1. (End)

Row sums are A111961.
Diagonal sums are A111962.
Inverse is A111963.
Factors as A007318*A111959.
Column k=0 gives A002426.
Cf. A026325.

# Uses function PMatrix from A357368. Adds a row and column above and to the left.
PMatrix(10, n -> A002426(n - 1)); # Peter Luschny, Oct 06 2022

Factors as (1/(1-x), x/(1-x))*(1/sqrt(1-4x^2), x/sqrt(1-4x^2)).
From Paul Barry, May 12 2009: (Start)
Equals ((1-x^2)/(1+x+x^2),x/(1+x+x^2))^{-1}*(1,x/(1-x^2))=A094531*(1,x/(1-x^2)).
Riordan array (1/sqrt(1-2x-3x^2), x/sqrt(1-2x-3x^2));
T(n,k) = Sum_{j=0..n} C(n,j)*C((j-1)/2,(j-k)/2)*2^(j-k)*(1+(-1)^(j-k))/2.
G.f.: 1/(1-xy-x-2x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). (End)

A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2xA(k,x) + (4k+1)x^2*(A(k,x))^2), k >= 0.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 1, 7, 9, 13, 1, 1, 1, 9, 13, 37, 25, 1, 1, 1, 11, 17, 73, 81, 61, 1, 1, 1, 13, 21, 121, 169, 301, 125, 1, 1, 1, 15, 25, 181, 289, 841, 729, 295, 1, 1, 1, 17, 29, 253, 441, 1801, 2197, 2549, 625, 1, 1, 1, 19, 33, 337, 625, 3301, 4913, 10123, 6561, 1447, 1

Offset: 0

Werner Schulte, Oct 29 2016

For k = 0 see A000012, for k = 1 see A098615, and for k = 2 see A200376.
It will be interesting using the formulae for k < 0 (attention: signed terms!). Especially for k = -1 see A157674.
If G is the g.f. of central binomial coefficients (see A000984) and B(k,x) = G(k*x^2), then B(k,x) = A(k,x)/(1+x*A(k,x)) and A(k,x) = B(k,x) / (1-x*B(k,x)) for k >= 0. - Werner Schulte, Aug 07 2017

			The terms define the array T(k,n) for k >= 0 and n >= 0, i.e.,
k\n  0  1   2   3    4     5      6      7       8        9  . . .
0:   1  1   1   1    1     1      1      1       1        1  . . .
1:   1  1   3   5   13    25     61    125     295      625  . . .
2:   1  1   5   9   37    81    301    729    2549     6561  . . .
3:   1  1   7  13   73   169    841   2197   10123    28561  . . .
4:   1  1   9  17  121   289   1801   4913   28057    83521  . . .
5:   1  1  11  21  181   441   3301   9261   63071   194481  . . .
6:   1  1  13  25  253   625   5461  15625  123565   390625  . . .
7:   1  1  15  29  337   841   8401  24389  219619   707281  . . .
8:   1  1  17  33  433  1089  12241  35937  362993  1185921  . . .
9:   1  1  19  37  541  1369  17101  50653  567127  1874161  . . .
etc.

Cf. A000012, A000045, A000108, A000984, A015441, A046521, A098615, A111959, A200376.

A(k,x) = (x + sqrt(1 - 4*k*x^2))/(1 - (4*k+1)*x^2) for k >= 0.
T(k,0) = 1 and T(k,2*n+2) = (4*k+1)^(n+1)-2*(Sum_{i=0..n} A000108(i)*k^(i+1)* (4*k+1)^(n-i)), and T(k,2*n+1) = (4*k+1)^n for k >= 0 and n >= 0.
A(k,x) = 1/(1 - x - 2*k*x^2*C(k*x^2)), k >= 0, where C is the g.f. of A000108.
Conjecture: If B(k,n) satisfy B(k,0) = B(k,1) = 1 and B(k,n+2) = B(k,n+1) + k*B(k,n) for k >= 0 and n >= 0 (generalized Fibonacci numbers, see A015441) and G(k,x) = Sum_{n>=0} A000108(n)*B(k,n)*x^n for k >= 0, then you will have (1): A(k,x*G(k,x)) = G(k,x) and (2): G(k,x/A(k,x)) = A(k,x) for k >= 0. Especially for k = 1 see A098615 and for k = 2 see A200376.
Conjecture: T(k,2*n) = Sum_{i=0..n} A046521(n,i)*k^(n-i) for k, n >= 0. - Werner Schulte, Aug 02 2017
Recurrence: T(k,2*n+2) = (4*k+1)*T(k,2*n)-2*k^(n+1)*A000108(n) with initial value T(k,0) = 1 for k >= 0 and n >= 0. - Werner Schulte, Aug 09 2017
T(k,n) = Sum_{i=0..n} A111959(n,i)*k^((n-i)/2) for k >= 0 and n >= 0. - Werner Schulte, Aug 09 2017

cp's OEIS Frontend

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

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A012150 Expansion of e.g.f. exp(tan(arcsin(x))).

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A111960 Renewal array for central trinomial numbers A002426.

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A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2*x*A(k,x) + (4*k+1)*x^2*(A(k,x))^2), k >= 0.

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A277604 Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2xA(k,x) + (4k+1)x^2*(A(k,x))^2), k >= 0.