A034870 -id:A034870 - cp's OEIS frontend

A091043 Normalized triangle of odd numbered entries of even numbered rows of Pascal's triangle A007318.

1, 1, 1, 3, 10, 3, 1, 7, 7, 1, 5, 60, 126, 60, 5, 3, 55, 198, 198, 55, 3, 7, 182, 1001, 1716, 1001, 182, 7, 1, 35, 273, 715, 715, 273, 35, 1, 9, 408, 4284, 15912, 24310, 15912, 4284, 408, 9, 5, 285, 3876, 19380, 41990, 41990, 19380, 3876, 285, 5, 11, 770, 13167, 85272

Offset: 1

Wolfdieter Lang, Jan 23 2004

b(n)= A006519(n), with b(n) defined in the formula. For every odd n b(n)=1.

The row polynomials Po(n,x) := 2*b(n)*sum(a(n,m)*x^m,m=0..n-1), n>=1, appear as numerators of the generating functions for the odd numbered column sequences of array A034870. b(n) is defined in the formula below.

			[1];[1,1];[3,10,3];[1,7,7,1];[5,60,126,60,5];...
n=3: GCD(3,10,3)=GCD(3,10)=1=b(3)=A006519(3); n=4: GCD(4,28,28,4)=GCD(4,28)=4=b(4)=A006519(4).

W. Lang, First 9 rows.

a(n, m)= binomial(2*n, 2*m+1)/(2*b(n)), n>=m+1>=1, else 0, with b(n) := GCD(seq(binomial(2*n, 2*m+1)/2, m=0..n-1)), where GCD denotes the greatest common divisor of a set of numbers (here one half of the odd numbered entries in the even numbered rows of Pascal's triangle). It suffices to consider m=0..floor((n-1)/2) due to symmetry.

A159041 Triangle read by rows: row n (n>=0) gives the coefficients of the polynomial p(n,x) of degree n defined in comments.

Original entry on oeis.org

1, 1, 1, 1, -10, 1, 1, -25, -25, 1, 1, -56, 246, -56, 1, 1, -119, 1072, 1072, -119, 1, 1, -246, 4047, -11572, 4047, -246, 1, 1, -501, 14107, -74127, -74127, 14107, -501, 1, 1, -1012, 46828, -408364, 901990, -408364, 46828, -1012, 1, 1, -2035, 150602, -2052886, 7685228, 7685228, -2052886, 150602, -2035, 1

Offset: 0

Roger L. Bagula, Apr 03 2009

Let E(n,k) (1 <= k <= n) denote the Eulerian numbers as defined in A008292. Then we define polynomials p(n,x) for n >= 0 as follows.
p(n,x) = (1/(1-x)) * ( Sum_{k=0..floor(n/2)} (-1)^k*E(n+2,k+1)*x^k + Sum_{k=ceiling((n+2)/2)..n+1} (-1)^(n+k)*E(n+2,k+1)*x^k ).
For example,
p(0,x) = (1-x)/(1-x) = 1,
p(1,x) = (1-x^2)/(1-x) = 1 + x,
p(2,x) = (1 - 11*x + 11*x^2 - x^3)/(1-x) = 1 - 10*x + x^2,
p(3,x) = (1 - 26*x + 26*x^3 - x^4)/(1-x) = 1 - 25*x - 25*x^2 + x^3,
p(4,x) = (1 - 57*x + 302*x^2 - 302*x^3 + 57*x^3 + x^5)/(1-x)
       = 1 - 56*x + 246*x^2 - 56*x^3 + x^4.
More generally, there is a triangle-to-triangle transformation U -> T defined as follows.
Let U(n,k) (1 <= k <= n) be a triangle of nonnegative numbers in which the rows are symmetric about the middle. Define polynomials p(n,x) for n >= 0 by
p(n,x) = (1/(1-x)) * ( Sum_{k=0..floor(n/2)} (-1)^k*U(n+2,k+1)*x^k + Sum_{k=ceiling((n+2)/2)..n+1} (-1)^(n+k)*U(n+2,k+1)*x^k ).
The n-th row of the new triangle T(n,k) (0 <= k <= n) gives the coefficients in the expansion of p(n+2).
The new triangle may be defined recursively by: T(n,0)=1; T(n,k) = T(n,k-1) + (-1)^k*U(n+2,k) for 1 <= k <= floor(n/2); T(n,k) = T(n,n-k).
Note that the central terms in the odd-numbered rows of U(n,k) do not get used.
The following table lists various sequences constructed using this transform:
Parameter Triangle  Triangle  Odd-numbered
m             U         T        rows
0          A007312  A007312   A034870
1          A008292  A159041   A171692
2          A060187  A225356   A225076
3          A142458  A225433   A225398
4          A142459  A225434   A225415

			Triangle begins as follows:
  1;
  1,     1;
  1,   -10,      1;
  1,   -25,    -25,        1;
  1,   -56,    246,      -56,       1;
  1,  -119,   1072,     1072,    -119,       1;
  1,  -246,   4047,   -11572,    4047,    -246,        1;
  1,  -501,  14107,   -74127,  -74127,   14107,     -501,      1;
  1, -1012,  46828,  -408364,  901990, -408364,    46828,  -1012,     1;
  1, -2035, 150602, -2052886, 7685228, 7685228, -2052886, 150602, -2035, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Roger L. Bagula, Another Mathematica program for A159041.

Cf. A007312, A008292, A034870, A060187, A142458, A142459, A159041, A171692, A225076, A225356, A225398, A225415, A225433, A225434.

A008292 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 or k = n then 1; else k*procname(n-1, k)+(n-k+1)*procname(n-1, k-1) ; end if; end proc:
# row n of new triangle T(n,k) in terms of old triangle U(n,k):
p:=proc(n) local k; global U;
simplify( (1/(1-x)) * ( add((-1)^k*U(n+2,k+1)*x^k,k=0..floor(n/2)) + add((-1)^(n+k)*U(n+2,k+1)*x^k, k=ceil((n+2)/2)..n+1 )) );
end;
U:=A008292;
for n from 0 to 6 do lprint(simplify(p(n))); od: # N. J. A. Sloane, May 11 2013
A159041 := proc(n, k)
    if k = 0 then
        1;
    elif k <= floor(n/2) then
        A159041(n, k-1)+(-1)^k*A008292(n+2, k+1) ;
    else
        A159041(n, n-k) ;
    end if;
end proc: # R. J. Mathar, May 08 2013

A[n_, 1] := 1;
A[n_, n_] := 1;
A[n_, k_] := (n - k + 1)A[n - 1, k - 1] + k A[n - 1, k];
p[x_, n_] = Sum[x^i*If[i == Floor[n/2] && Mod[n, 2] == 0, 0, If[i <= Floor[n/2], (-1)^i*A[n, i], -(-1)^(n - i)*A[n, i]]], {i, 0, n}]/(1 - x);
Table[CoefficientList[FullSimplify[p[x, n]], x], {n, 1, 11}];
Flatten[%]

def A008292(n,k): return sum( (-1)^j*(k-j)^n*binomial(n+1,j) for j in (0..k) )
@CachedFunction
def T(n,k):
    if (k==0 or k==n): return 1
    elif (k <= (n//2)): return T(n,k-1) + (-1)^k*A008292(n+2,k+1)
    else: return T(n,n-k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 18 2022

T(n, k) = T(n, k-1) + (-1)^k*A008292(n+2, k+1) if k <= floor(n/2), otherwise T(n, n-k), with T(n, 0) = T(n, n) = 1. - R. J. Mathar, May 08 2013

Edited by N. J. A. Sloane, May 07 2013, May 11 2013

A260492 Pascal's triangle aerated with columns of zeros.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 2, 0, 1, 1, 0, 3, 0, 3, 0, 1, 1, 0, 4, 0, 6, 0, 4, 0, 1, 1, 0, 5, 0, 10, 0, 10, 0, 5, 0, 1, 1, 0, 6, 0, 15, 0, 20, 0, 15, 0, 6, 0, 1, 1, 0, 7, 0, 21, 0, 35, 0, 35, 0, 21, 0, 7, 0, 1, 1, 0, 8, 0, 28, 0, 56, 0, 70, 0, 56, 0, 28, 0, 8, 0, 1, 1, 0, 9, 0, 36, 0, 84, 0, 126, 0, 126, 0, 84, 0, 36, 0, 9, 0, 1, 1, 0, 10, 0, 45, 0, 120, 0, 210, 0, 252, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1

Offset: 0

Peter Bala, Aug 15 2015

To obtain this array we take Pascal's triangle A007318 and insert a column of zeros at columns 1, 3, 5, .... Thus the n-th row of this array gives the coefficients in the expansion of (1 + x^2)^n.
The transpose of this array is a generalized Riordan array (1, 1 + x^2) as defined by Wilson in 2005 (see the Bala link for details). Note that these are not the same as the generalized Riordan arrays introduced by Wang and Wang in 2008.
Call this array C and let p(x) = 1 + x^2. Then C^2 gives the coefficients in the expansion of the polynomials ( p(p(x)) )^n, C^3 gives the coefficients in the expansion of the polynomials ( p(p(p(x))) )^n and so on.
See A204293 for Pascal's triangle aerated by both row and columns.
The triangle is read by rows of lengths equal to the odd numbers 2*n + 1 = A005408(n), n >= 0; although it is then (from the 3rd line of comments & formula on) considered as an infinite square matrix C, the upper right being filled with zeros. - M. F. Hasler, Aug 19 2015

			Table begins
n\k 0  1  2  3  4  5  6  7  8  9 10 11 12
- - - - - - - - - - - - - - - - - - - - -
0   1
1   1  0  1
2   1  0  2  0  1
3   1  0  3  0  3  0  1
4   1  0  4  0  6  0  4  0  1
5   1  0  5  0 10  0 10  0  5  0  1
6   1  0  6  0 15  0 20  0 15  0  6  0  1
...

P. Bala, Notes on generalized Riordan arrays
Evgeniy Burlachenko, Riordan arrays and generalized Euler polynomials, arXiv:1709.02229 [math.NT], 2017.
Evgeniy Burlachenko, Riordan arrays, Chebyshev polynomials, Fibonacci bases, arXiv:1903.10435 [math.NT], 2019.
Alan D. Sokal, How to generalize (and not to generalize) the Chu-Vandermonde identity, arXiv:1804.08919 [math.CO], 2018.
W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Cf. A003095, A027826, A007318, A034870, A072191, A161869, A204293.

#A260492
#define the aerated Pascal matrix (with indexing starting at 1)
C := Matrix(30, 30, (i,j) -> (1 - mod(j-1, 2))*binomial(i-1, floor((j-1)/2))):
for n from 1 to 12 do seq(C(n, k), k = 1 .. 2*n-1) end do;
# alternative program using row polynomials
rowpoly:= proc(n) option remember; expand((1 + x^2)^n) end:
T := (n, k)-> coeff(rowpoly(n), x, k):
seq(seq(T(n, k), k = 0..2*n), n = 0..11);

T[n_, k_] := If[OddQ[k], 0, Binomial[n, k/2]];
Table[T[n, k], {n, 0, 10}, {k, 0, 2n}] // Flatten (* Jean-François Alcover, Mar 23 2018 *)

C=matrix(20,20,m,n,if(n%2,binomial(m-1,n\2))); a=concat(vector(5,i,vector(i*2-1,j,C[i,j]))) \\ (C is read by rows, not antidiagonals.) - M. F. Hasler, Aug 19 2015

T(n,k) = (1 - k mod 2) binomial(n,floor(k/2)).
O.g.f.: 1/(1 - (1 + x^2)*t) = 1 + (1 + x^2)*t + (1 + 2*x^2 + x^4)*t^2 + ....
Let C denote this array.
Row sums of C = 2^n; Row sums of C^2 = 5^n; Row sums of C^3 = 26^n; Row of sums C^4 = 677^n. In general the row sums of C^m = A003095(m)^n.
First column of C^2 = 2^n; first column of C^3 = 5^n, first column of C^4 = 26^n and so on.
Let P denote Pascal's triangle A007318. Then C * transpose(C) = P * transpose(P) = the square symmetric Pascal matrix.
For n >= 0, (P^n)*C is the array P^(n+1) aerated by columns.
First column of P*C = 2^n; first column of (P*C)^2 = 6^n; first column of (P*C)^3 = 38^n, and so on, where [2, 6, 38, ...] is A072191.
Let R equal Pascal's triangle aerated with rows of zeros. Then C*R = P^2.
R*P*C is P^3 aerated by both rows and columns
Conjecturally, the limit of R^n * C^n as n -> oo has as its first column an aerated version of A027826, with zeros elsewhere in the array.
Aeration by rows/columns amounts to multiplication to the left/right by the identity matrix aerated by rows/column: R = J*P, C = P*J'. This makes obvious that P^n*C = P^(n+1)*J', R*P^n = J*P^(n+1), R*P^n*C = J*P^(n+2)*J' (aerated by both rows and columns). - M. F. Hasler, Aug 19 2015

A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2x + 1)^n + (x^2 - 1)(x + 1)^n.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 1, 0, 2, 6, 6, 2, 0, 3, 13, 22, 18, 7, 1, 0, 4, 23, 56, 75, 60, 29, 8, 1, 0, 5, 36, 115, 215, 261, 215, 121, 45, 10, 1, 0, 6, 52, 206, 495, 806, 938, 798, 496, 220, 66, 12, 1, 0, 7, 71, 336, 987, 2016, 3031, 3452, 3010, 2003, 1001, 364, 91

Offset: 0

Franck Maminirina Ramaharo, Feb 28 2018

			The triangle T(n, k) begins:
n\k  0  1   2    3    4     5     6     7     8     9    10   11  12  13 14
0:   0  0   1
1:   0  1   2    1
2:   0  2   6    6    2
3:   0  3  13   22   18     7     1
4:   0  4  23   56   75    60    29     8     1
5:   0  5  36  115  215   261   215   121    45    10     1
6:   0  6  52  206  495   806   938   798   496   220    66   12   1
7:   0  7  71  336  987  2016  3031  3452  3010  2003  1001  364  91  14  1

M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972.
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996.

Paul Barry, On the Connection Coefficients of the Chebyshev-Boubaker polynomials, The Scientific World Journal, Volume 2013 (2013), Article ID 657806, 10 pages.
Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
A. M. Mathai and P. N. Rathie, Enumeration of almost cubic maps, Journal of Combinatorial Theory, Series B, Vol 13 (1972), 83-90.
Franck Ramaharo, A one-variable bracket polynomial for some Turk's head knots, arXiv:1807.05256 [math.CO], 2018.
Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.

Row sums: A000302 (powers of 4).

Cf. A034870, A034871, A032443.

T := (n, k) -> binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k);
for n from 0 to 10 do seq(T(n, k), k = 0 .. max(2*n, n + 2)) od;

T(n, k) := binomial(2*n, k) + binomial(n, k - 2) - binomial(n, k)$
a : []$
for n:0 thru 10 do
  a : append(a, makelist(T(n, k), k, 0, max(2*n, n + 2)))$
a;

row(n) = Vecrev((x^2 + 2*x + 1)^n + (x^2 - 1)*(x + 1)^n); \\ Michel Marcus, Nov 12 2022

T(n,k) = binomial(2*n,k) + binomial(n,k-2) - binomial(n,k).
T(n,k) = T(n-1,k-1)+ T(n-1,k) + A034871(n-1,k-1), with T(n,0) = T(0,1) = 0 and T(0,2) = 1
T(n,1) = A001477(n).
T(n,2) = A143689(n).
T(n,3) = n + A002492(n-1) - A000292(n-2).
T(n,n) = A247493(n+1,n).
T(n,n+1) = n + A001791(n).
T(n,n+2) = 1 + A002694(n), n >= 2.
T(n,n+k) = binomial(2*n, n-k) = A094527(n,k), for k >= 3 and n>=k.
G.f.: 1/(1 - y*(x^2 + 2*x + 1)) + (x^2 - 1)/(1 - y*(x + 1)).

A099510 Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + 2*z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.

Original entry on oeis.org

1, 1, 2, 1, 4, 1, 1, 6, 6, 4, 1, 8, 15, 20, 1, 1, 10, 28, 56, 15, 6, 1, 12, 45, 120, 70, 56, 1, 1, 14, 66, 220, 210, 252, 28, 8, 1, 16, 91, 364, 495, 792, 210, 120, 1, 1, 18, 120, 560, 1001, 2002, 924, 792, 45, 10, 1, 20, 153, 816, 1820, 4368, 3003, 3432, 495, 220, 1, 1, 22, 190

Offset: 0

Paul D. Hanna, Oct 20 2004

Row sums form A099511. In general if T(n,k) = coefficient of z^k in (a + b*z + c*z^2)^(n-[k/2]), then the resulting number triangle will have the o.g.f.: ((1-a*x-c*x^2*y^2) + b*x*y)/((1-a*x-c*x^2*y^2)^2 - x*(b*x*y)^2).

			Rows begin:
[1],
[1,2],
[1,4,1],
[1,6,6,4],
[1,8,15,20,1],
[1,10,28,56,15,6],
[1,12,45,120,70,56,1],
[1,14,66,220,210,252,28,8],
[1,16,91,364,495,792,210,120,1],
[1,18,120,560,1001,2002,924,792,45,10],...
and can be derived from coefficients of (1+2*z+z^2)^n:
[1],
[1,2,1],
[1,4,6,4,1],
[1,6,15,20,15,6,1],
[1,8,28,56,70,56,28,8,1],
[1,10,45,120,210,252,210,120,45,10,1],...
by shifting each column k down by [k/2] rows.

Cf. A034870, A099509, A099511, A099512.

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

G.f.: (1-x+2*x*y-x^2*y^2)/((1-x)^2-2*x^2*y^2-2*x^3*y^2+x^4*y^4). T(n, k) = binomial(2*n-2*(k\2), k).

A225076 Triangle read by rows: absolute values of odd-numbered rows of A225356.

Original entry on oeis.org

1, 1, 22, 1, 1, 236, 1446, 236, 1, 1, 2178, 58479, 201244, 58479, 2178, 1, 1, 19672, 1736668, 19971304, 49441990, 19971304, 1736668, 19672, 1, 1, 177134, 46525293, 1356555432, 9480267666, 19107752148, 9480267666, 1356555432, 46525293, 177134, 1

Offset: 1

Roger L. Bagula, Apr 26 2013

An equivalent definition: take the polynomials corresponding to rows 2, 4, 6, 8, ... of A060187, divide by x+1, and extract the coefficients. [Corrected by Petros Hadjicostas, Apr 17 2020]

			Triangle T(n,m) (for n >= 1 and 0 <= m <= 2*n - 2) begins as follows:
  1;
  1,    22,       1;
  1,   236,    1446,      236,        1;
  1,  2178,   58479,   201244,    58479,     2178,       1;
  1, 19672, 1736668, 19971304, 49441990, 19971304, 1736668, 19672, 1;
  ...

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened

Cf. A002671 (row sums), A034870, A060187, A171692, A225398.

(* Power series via an infinite sum *)
p[x_,n_] = (x-1)^(2*n)*Sum[(2*k+1)^(2*n-1)*x^k,{k,0,Infinity}];
Table[CoefficientList[p[x,n]/(1+x),x],{n,1,10}]//Flatten
(* First alternative method: recurrence *)
t[n_, k_, m_]:= t[n,k,m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1,k-1,m] + (m*k - (m-1))*t[n-1,k,m]];
T[n_, k_]:= T[n, k]= t[n+1,k+1,2]; (* t(n,k,2) = A060187 *)
Flatten[Table[CoefficientList[Sum[T[n, k]*x^k, {k,0,n}]/(x+1), x], {n,14,2}]]
(* Second alternative method: polynomial expansion *)
p[t_] = Exp[t]*x/(-Exp[2*t] + x);
Flatten[Table[CoefficientList[(n!*(-1+x)^(n+1)/(x*(x+1)))*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 1, 13, 2}]]

def A060187(n, k): return sum( (-1)^(k-j)*(2*j-1)^(n-1)*binomial(n, k-j) for j in (1..k) )
def A225076(n,k): return sum( (-1)^(k-j-1)*A060187(2*n, j+1) for j in (0..k-1) )
flatten([[A225076(n, k) for k in (1..2*n-1)] for n in (1..12)]) # G. C. Greubel, Mar 19 2022

Triangle read by rows: row n gives coefficients in the expansion of the polynomial ((x - 1)^(2*n)/(x + 1)) * Sum_{k >= 0} (2*k + 1)^(2*n-1)*x^k. The infinite sum simplifies to a polynomial.
Sum_{m=0..2*n-2} T(n,m)*t^m = 2^(2*n-1) * (1-t)^(2*n) * LerchPhi(t, 1-2*n, 1/2)/(1 + t).
Sum_{k=1..n} T(n, k) = A002671(n-1).
T(n,m) = Sum_{k=0..m-1} (-1)^(m-k-1)*A060187(2*n,k+1) for n >= 1 and 1 <= m <= 2*n-1. - Petros Hadjicostas, Apr 17 2020

Edited by N. J. A. Sloane, May 06 2013, May 11 2013

A225398 Triangle read by rows: absolute values of odd-numbered rows of A225433.

Original entry on oeis.org

1, 1, 38, 1, 1, 676, 4806, 676, 1, 1, 10914, 362895, 1346780, 362895, 10914, 1, 1, 174752, 20554588, 263879264, 683233990, 263879264, 20554588, 174752, 1, 1, 2796190, 1063096365, 35677598760, 267248150610, 554291429748, 267248150610, 35677598760, 1063096365, 2796190, 1

Offset: 1

Roger L. Bagula, Apr 26 2013 (Entered by N. J. A. Sloane, May 06 2013)

			Triangle begins:
  1;
  1,     38,        1;
  1,    676,     4806,       676,         1;
  1,  10914,   362895,   1346780,    362895,     10914,        1;
  1, 174752, 20554588, 263879264, 683233990, 263879264, 20554588, 174752, 1;

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened

Cf. A034870, A171692, A225076.

(* First program *)
t[n_, k_, m_]:= t[n,k,m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1,k-1,m] + (m*k-(m- 1))*t[n-1,k,m]];
T[n_, k_]:= T[n, k]= t[n+1, k+1, 3]; (* t(n,k,3) = A142458 *)
Flatten[Table[CoefficientList[Sum[T[n, k]*x^k, {k,0,n}]/(1+x), x], {n, 1, 14, 2}]]
(* Second program *)
t[n_, k_, m_]:= t[n, k, m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1,k-1,m] + (m*k-m +1)*t[n-1,k,m]]; (* t(n,k,3) = A142458 *)
A225398[n_, k_]:= A225398[n, k]= Sum[(-1)^(k-j-1)*t[2*n,j+1,3], {j,0,k-1}];
Table[A225398[n, k], {n,12}, {k,2*n-1}] //Flatten (* G. C. Greubel, Mar 19 2022 *)

@CachedFunction
def T(n, k, m):
    if (k==1 or k==n): return 1
    else: return (m*(n-k)+1)*T(n-1, k-1, m) + (m*k-m+1)*T(n-1, k, m)
def A142458(n, k): return T(n, k, 3)
def A225398(n,k): return sum( (-1)^(k-j-1)*A142458(2*n, j+1) for j in (0..k-1) )
flatten([[A225398(n, k) for k in (1..2*n-1)] for n in (1..12)]) # G. C. Greubel, Mar 19 2022

From G. C. Greubel, Mar 19 2022: (Start)
T(n, k) = Sum_{j=0..k-1} (-1)^(k-j-1)*A142458(2*n, j+1).
T(n, n-k) = T(n, k). (End)

Edited by N. J. A. Sloane, May 11 2013

A225415 Triangle read by rows: absolute values of odd-numbered rows of A225434.

Original entry on oeis.org

1, 1, 58, 1, 1, 1556, 12006, 1556, 1, 1, 39054, 1461615, 5647300, 1461615, 39054, 1, 1, 976552, 135028828, 1838120344, 4873361350, 1838120344, 135028828, 976552, 1, 1, 24414050, 11462721645, 414730580760, 3221733789330, 6783391017228, 3221733789330, 414730580760, 11462721645, 24414050, 1

Offset: 1

Roger L. Bagula, May 07 2013

			Triangle begins:
  1;
  1,     58,         1;
  1,   1556,     12006,       1556,          1;
  1,  39054,   1461615,    5647300,    1461615,      39054,         1;
  1, 976552, 135028828, 1838120344, 4873361350, 1838120344, 135028828, 976552, 1;

G. C. Greubel, Rows n = 1..50 of the irregular triangle, flattened

The m=4 triangle in the sequence A034870 (m=0), A171692 (m=1), A225076 (m=2), A225398 (m=3).

(* First program *)
t[n_, k_, m_]:= t[n, k, m]= If[k==1 || k==n, 1,(m*n-m*k+1)*t[n-1, k-1, m] + (m*k-(m-1))*t[n-1, k, m]];
T[n_, k_]:= T[n, k] = t[n+1, k+1,4]; (* t(n,k,4) = A142459 *)
Flatten[Table[CoefficientList[Sum[T[n, k]*x^k, {k,0,n}]/(1+x), x], {n,1,14,2}]]
(* Second program *)
t[n_, k_, m_]:= t[n, k, m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1,k-1,m] + (m*k-m+1)*t[n-1,k,m]]; (* t(n,k,4) = A142459 *)
T[n_, k_]:= T[n, k]= Sum[ (-1)^(k-j-1)*t[2*n,j+1,4], {j,0,k-1}];
Table[T[n, k], {n,12}, {k,2*n-1}]//Flatten (* G. C. Greubel, Mar 19 2022 *)

@CachedFunction
def T(n, k, m):
    if (k==1 or k==n): return 1
    else: return (m*(n-k)+1)*T(n-1, k-1, m) + (m*k-m+1)*T(n-1, k, m)
def A142459(n, k): return T(n, k, 4)
def A225415(n,k): return sum( (-1)^(k-j-1)*A142459(2*n, j+1) for j in (0..k-1) )
flatten([[A225415(n, k) for k in (1..2*n-1)] for n in (1..12)]) # G. C. Greubel, Mar 19 2022

T(n, k) = Sum_{j=0..k-1} (-1)^(k-j-1)*A142459(2*n, j+1). - G. C. Greubel, Mar 19 2022

Edited by N. J. A. Sloane, May 11 2013

A181145 G.f.: exp( Sum_{n>=1} [Sum_{k=0..2n} C(2n,k)^2y^k]x^n/n ) = Sum_{n>=0,k=0..2n} T(n,k)x^ny^k, as a triangle of coefficients T(n,k) read by rows.

Original entry on oeis.org

1, 1, 4, 1, 1, 12, 27, 12, 1, 1, 24, 134, 236, 134, 24, 1, 1, 40, 410, 1540, 2380, 1540, 410, 40, 1, 1, 60, 975, 6260, 18386, 26216, 18386, 6260, 975, 60, 1, 1, 84, 1981, 19320, 91441, 227052, 306495, 227052, 91441, 19320, 1981, 84, 1, 1, 112, 3612, 49672

Offset: 0

Paul D. Hanna, Oct 16 2010

Compare g.f. to that of the triangle A034870:
* exp( Sum_{n>=1} [Sum_{k=0..2n} C(2n,k)*y^k]*x^n/n )
which consists of the even numbered rows of Pascal's triangle.

			G.f.: A(x,y) = 1 + (1+ 4*y+ y^2)*x + (1 + 12*y+ 27*y^2+ 12*y^3+ y^4)*x^2 + (1+ 24*y+ 134*y^2+ 236*y^3+ 134*y^4+ 24*y^5+ y^6)*x^3 +...
The logarithm of the g.f. begins:
log(A(x,y)) = (1 + 2^2*y + y^2)*x
+ (1 + 4^2*y + 6^2*y^2 + 4^2*y^3 + y^4)*x^2/2
+ (1 + 6^2*y + 15^2*y^2 + 20^2*y^3 + 15^2*y^4 + 6^2*y^5 + y^6)*x^3/3
+ (1 + 8^2*y + 28^2*y^2 + 56^2*y^3 + 70^2*y^4 + 56^2*y^5 + 28^2*y^6 + 8^2*y^7 + y^8)*x^4/4 +...
Triangle begins:
1;
1, 4, 1;
1, 12, 27, 12, 1;
1, 24, 134, 236, 134, 24, 1;
1, 40, 410, 1540, 2380, 1540, 410, 40, 1;
1, 60, 975, 6260, 18386, 26216, 18386, 6260, 975, 60, 1;
1, 84, 1981, 19320, 91441, 227052, 306495, 227052, 91441, 19320, 1981, 84, 1;
1, 112, 3612, 49672, 344260, 1312080, 2883562, 3740572, 2883562, 1312080, 344260, 49672, 3612, 112, 1; ...

Cf. A066357 (row sums), A181146 (main diagonal).

Cf. variants: A181143, A181144, A001263, A034870.

{T(n,k)=polcoeff(polcoeff(exp(sum(m=1,n,sum(j=0,2*m,binomial(2*m,j)^2*y^j)*x^m/m)+O(x^(n+1))),n,x),k,y)}

Row sums form A066357 (with offset), the number of ordered trees on 2n nodes with every subtree at the root having an even number of edges.

Comment and example corrected by Paul D. Hanna, Oct 16 2010

A219462 a(n) = Sum_{k = 1..2n} binomial(2n,k) * Fibonacci(2*k).

Original entry on oeis.org

0, 5, 75, 1000, 13125, 171875, 2250000, 29453125, 385546875, 5046875000, 66064453125, 864794921875, 11320312500000, 148184814453125, 1939764404296875, 25391845703125000, 332383575439453125, 4350957489013671875, 56954772949218750000, 745547657012939453125

Offset: 0

Reinhard Zumkeller, Nov 20 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..500
Doron Zeilbeger, An Enquiry Concerning Human (and Computer!) [Mathematical] Understanding, (2007); page 10.
Index entries for linear recurrences with constant coefficients, signature (15,-25).

Cf. A000045, A000351, A001906, A007318, A034870, A036291, A087453.

a219462 = sum . zipWith (*) a001906_list . a034870_row

Table[Sum[Binomial[2n,k]Fibonacci[2k],{k,2n}],{n,0,20}] (* Harvey P. Dale, Aug 26 2017 *)

a(n) = sum(k = 1, 2*n, binomial(2*n,k) * fibonacci(2*k)); \\ Michel Marcus, Jan 26 2022

a(n) = Sum_{k=1..n} A034870(n,k)*A001906(k).
a(n) = 5^n * Fibonacci(2*n) = A000351(n) * A001906(n).
G.f.: 5*x/(25*x^2-15*x+1). - Colin Barker, Dec 03 2012
E.g.f.: 2*exp(15*x/2)*sinh(5*sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Oct 19 2023

cp's OEIS Frontend

A091043 Normalized triangle of odd numbered entries of even numbered rows of Pascal's triangle A007318.

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A099510 Triangle, read by rows, of trinomial coefficients arranged so that there are n+1 terms in row n by setting T(n,k) equal to the coefficient of z^k in (1 + 2*z + z^2)^(n-[k/2]), for n>=k>=0, where [k/2] is the integer floor of k/2.

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A225076 Triangle read by rows: absolute values of odd-numbered rows of A225356.

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A225398 Triangle read by rows: absolute values of odd-numbered rows of A225433.

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A300192 Triangle read by rows: row n consists of the coefficients of the expansion of the polynomial (x^2 + 2x + 1)^n + (x^2 - 1)(x + 1)^n.

A181145 G.f.: exp( Sum_{n>=1} [Sum_{k=0..2n} C(2n,k)^2y^k]x^n/n ) = Sum_{n>=0,k=0..2n} T(n,k)x^ny^k, as a triangle of coefficients T(n,k) read by rows.

A219462 a(n) = Sum_{k = 1..2n} binomial(2n,k) * Fibonacci(2*k).