A053121 -id:A053121 - cp's OEIS frontend

A054456 Convolution triangle of A000129(n) (Pell numbers).

1, 2, 1, 5, 4, 1, 12, 14, 6, 1, 29, 44, 27, 8, 1, 70, 131, 104, 44, 10, 1, 169, 376, 366, 200, 65, 12, 1, 408, 1052, 1212, 810, 340, 90, 14, 1, 985, 2888, 3842, 3032, 1555, 532, 119, 16, 1, 2378, 7813, 11784, 10716, 6482, 2709, 784, 152, 18, 1, 5741, 20892, 35223

Offset: 0

Wolfdieter Lang, Apr 27 2000 and May 08 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The G.f. for the row polynomials p(n,x) (increasing powers of x) is Pell(z)/(1-x*z*Pell(z)) with Pell(x)=1/(1-2*x-x^2) = g.f. for A000129(n+1) (Pell numbers without 0).
Column sequences are A000129(n+1), A006645(n+1), A054457(n) for m=0..2.
Riordan array (1/(1-2x-x^2),x/(1-2x-x^2)). - Paul Barry, Mar 15 2005
As a Riordan array, this factors as (1/(1-x^2),x/(1-x^2))*(1/(1-2x),x/(1-2x)), [abs(A049310) times square of A007318, or A038207]. - Paul Barry, Jul 28 2005
Coefficients of polynomials defined by P(x, 0) = 1; P(x, 1) = 2 - x; P(x, n) = (2 - x)*P(x, n - 1) + P(x, n - 2). - Roger L. Bagula, Mar 24 2008
Subtriangle (obtained by dropping the first column) of the triangle given by (0, 2, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 19 2013
T(n,k) is the number of words of length n over {0,1,2,3} having k letters 3 and avoiding runs of odd length of the letter 0. - Milan Janjic, Jan 14 2017

			Fourth row polynomial (n=3): p(3,x)= 12+14*x+6*x^2+x^3
Triangle begins:
{1},
{2, 1},
{5, 4, 1},
{12, 14, 6, 1},
{29, 44, 27, 8, 1},
{70, 131,104, 44, 10, 1},
{169, 376, 366, 200, 65, 12, 1},
{408, 1052, 1212, 810, 340, 90, 14, 1},
{985, 2888, 3842, 3032, 1555, 532, 119, 16, 1},
{2378, 7813, 11784, 10716, 6482, 2709, 784, 152, 18, 1},
{5741, 20892, 35223, 36248, 25235, 12432, 4396, 1104, 189, 20, 1},
The triangle (0, 2, 1/2, -1/2, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, ...) begins:
1
0, 1
0, 2, 1
0, 5, 4, 1
0, 12, 14, 6, 1
0, 29, 44, 27, 8, 1 - _Philippe Deléham_, Feb 19 2013

Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A000129. Row sums: A006190(n+1).

Cf. A129844.

G := 1/(1-(x+2)*z-z^2): Gser := simplify(series(G, z = 0, 18)): for n from 0 to 15 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 15 do seq(coeff(P[n], x, j), j = 0 .. n) end do; # yields sequence in triangular form - Emeric Deutsch, Aug 30 2015
T := (n,k) -> `if`(n=0,1,2^(n-k)*binomial(n,k)*hypergeom([(k-n)/2,(k-n+1)/2],[-n], -1)): seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Apr 25 2016
# Uses function PMatrix from A357368. Adds column 1,0,0,0,... to the left.
PMatrix(10, A000129); # Peter Luschny, Oct 19 2022

P[x_, 0] := 1; P[x_, 1] := 2 - x; P[x_, n_] := P[x, n] = (2 - x) P[x, n - 1] + P[x, n - 2]; Table[Abs@ CoefficientList[P[x, n], x], {n, 0, 10}] // Flatten (* Roger L. Bagula, Mar 24 2008, edited by Michael De Vlieger, Apr 25 2018 *)

a(n, m) := ((n-m+1)*a(n, m-1) + (n+m)*a(n-1, m-1))/(4*m), n >= m >= 1, a(n, 0)= P(n+1)= A000129(n+1) (Pell numbers without P(0)), a(n, m) := 0 if n

G.f. for column m: Pell(x)*(x*Pell(x))^m, m >= 0, with Pell(x) G.f. for A000129(n+1).

Number triangle T(n, k) with T(n, 0)=A000129(n), T(1, 1)=1, T(n, k)=0 if k>n, T(n, k)=T(n-1, k-1)+T(n-2, k)+2T(n-1, k) otherwise; T(n, k)=if(k<=n, sum{j=0..floor((n-k)/2), C(n-j, k)C(n-k-j, j)2^(n-2j-k)}; - Paul Barry, Mar 15 2005

Bivariate g.f. G(x,z) = 1/[1 - (2 + x)z - z^2]. G.f. for column k = z^k/(1 - 2z - z^2)^{k+1} (k>=0). - Emeric Deutsch, Aug 30 2015

T(n,k) = 2^(n-k)*C(n,k)*hypergeom([(k-n)/2,(k-n+1)/2],[-n],-1)) for n>=1. - Peter Luschny, Apr 25 2016

A095666 Pascal (1,4) triangle.

Original entry on oeis.org

4, 1, 4, 1, 5, 4, 1, 6, 9, 4, 1, 7, 15, 13, 4, 1, 8, 22, 28, 17, 4, 1, 9, 30, 50, 45, 21, 4, 1, 10, 39, 80, 95, 66, 25, 4, 1, 11, 49, 119, 175, 161, 91, 29, 4, 1, 12, 60, 168, 294, 336, 252, 120, 33, 4, 1, 13, 72, 228, 462, 630, 588, 372, 153, 37, 4, 1, 14, 85, 300, 690, 1092

Offset: 0

Wolfdieter Lang, Jun 11 2004

This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A096940 (q=5), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m is G(z,x) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.]
T(2*n,n) = A029609(n) for n > 0, A029609 are the central terms of the Pascal (2,3) triangle A029600. - Reinhard Zumkeller, Apr 08 2012

			Triangle begins:
  [4];
  [1,4];
  [1,5,4];
  [1,6,9,4];
  [1,7,15,13,4];
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Wolfdieter Lang, First 10 rows.

Row sums: A020714(n-1), n >= 1, 4 if n=0.
Alternating row sums are [4, -3, followed by 0's].
Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+4), A055999(n+1), A060488(n+3), A095667-71, A095819.

a095666 n k = a095666_tabl !! n !! k
a095666_row n = a095666_tabl !! n
a095666_tabl = [4] : iterate
   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1,4]
-- Reinhard Zumkeller, Apr 08 2012

a(n,k):=(1+3*k/n)*binomial(n,k) # Mircea Merca, Apr 08 2012

A095666[n_, k_] := If[n == k,  4, (3*k/n + 1)*Binomial[n, k]];
Table[A095666[n, k], {n, 0, 12}, {k, 0, n}] (* Paolo Xausa, Apr 14 2025 *)

Recursion: a(n, m) = 0 if m > n, a(0, 0) = 4; a(n, 0) = 1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1 + 3*k/n)*binomial(n,k). - Mircea Merca, Apr 08 2012

A073387 Convolution triangle of A002605(n) (generalized (2,2)-Fibonacci), n>=0.

Original entry on oeis.org

1, 2, 1, 6, 4, 1, 16, 16, 6, 1, 44, 56, 30, 8, 1, 120, 188, 128, 48, 10, 1, 328, 608, 504, 240, 70, 12, 1, 896, 1920, 1872, 1080, 400, 96, 14, 1, 2448, 5952, 6672, 4512, 2020, 616, 126, 16, 1, 6688, 18192, 23040, 17856, 9352, 3444, 896, 160, 18, 1

Offset: 0

Wolfdieter Lang, Aug 02 2002

The g.f. for the row polynomials P(n,x) = Sum_{m=0..n} T(n,m)*x^m is 1/(1-(2+x+2*z)*z). See Shapiro et al. reference and comment under A053121 for such convolution triangles.

T(n, k) is the number of words of length n over {0,1,2,3} having k letters 3 and avoiding runs of odd length for the letters 0,1. - Milan Janjic, Jan 14 2017

			Lower triangular matrix, T(n,k), n >= k >= 0, else 0:
    1;
    2,    1;
    6,    4,    1;
   16,   16,    6,    1;
   44,   56,   30,    8,   1;
  120,  188,  128,   48,  10,   1;
  328,  608,  504,  240,  70,  12,   1;
  896, 1920, 1872, 1080, 400,  96,  14,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Wolfdieter Lang, First 10 rows.

Cf. A002605, A007482 (row sums), A053121, A073403, A073404.
Cf. A001045, A005563, A015518, A159920.
Columns: A002605 (k=0), A073388 (k=1), A073389 (k=2), A073390 (k=3), A073391 (k=4), A073392 (k=5), A073393 (k=6), A073394 (k=7), A073397 (k=8), A073398 (k=9).

A073387:= func< n,k | (&+[2^(n-k-j)*Binomial(n-j,k)*Binomial(n-k-j,j): j in [0..Floor((n-k)/2)]]) >;
[A073387(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 03 2022

T := (n,k) -> `if`(n=0,1,2^(n-k)*binomial(n,k)*hypergeom([(k-n)/2, (k-n+1)/2], [-n], -2)): seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Apr 25 2016

T[n_, k_]:=T[n,k]=Sum[2^(n-k-j)*Binomial[n-j,k]*Binomial[n-k-j,j], {j,0,(n-k)/2}];
Table[T[n,k], {n,0,10}, {k,0,n}]//Flatten (* Jean-François Alcover, Jun 04 2019 *)

def A073387(n,k): return sum(2^(n-k-j)*binomial(n-j,k)*binomial(n-k-j,j) for j in range(((n-k+2)//2)))
flatten([[A073387(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Oct 03 2022

T(n, k) = 2*(p(k-1, n-k)*(n-k+1)*T(n-k+1) + q(k-1, n-k)*(n-k+2)*T(n-k))/(k!*12^k), n >= k >= 1, with T(n) = T(n, k=0) = A002605(n), else 0; p(m, n) = Sum_{j=0..m} A(m, j)*n^(m-j) and q(m, n) = Sum_{j=0..m} B(m, j)*n^(m-j) with the number triangles A(k, m) = A073403(k, m) and B(k, m) = A073404(k, m).
T(n, k) = Sum_{j=0..floor((n-k)/2)} 2^(n-k-j)*binomial(n-j, k)*binomial(n-k-j, j) if n > k, else 0.
T(n, k) = ((n-k+1)*T(n, k-1) + 2*(n+k)*T(n-1, k-1))/(6*k), n >= k >= 1, T(n, 0) = A002605(n+1), else 0.
Sum_{k=0..n} T(n, k) = A007482(n).
G.f. for column m (without leading zeros): 1/(1-2*x*(1+x))^(m+1), m>=0.
T(n,k) = 2^(n-k)*binomial(n,k)*hypergeom([(k-n)/2, (k-n+1)/2], [-n], -2) for n>=1. - Peter Luschny, Apr 25 2016
From G. C. Greubel, Oct 03 2022: (Start)
T(n, n-1) = A005843(n), n >= 1.
T(n, n-2) = 2*A005563(n-1), n >= 2.
T(n, n-3) = 4*A159920(n-1), n >= 2.
Sum_{k=0..n} (-1)^k*T(n, k) = A001045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A015518(n+1). (End)

A093561 (4,1) Pascal triangle.

Original entry on oeis.org

1, 4, 1, 4, 5, 1, 4, 9, 6, 1, 4, 13, 15, 7, 1, 4, 17, 28, 22, 8, 1, 4, 21, 45, 50, 30, 9, 1, 4, 25, 66, 95, 80, 39, 10, 1, 4, 29, 91, 161, 175, 119, 49, 11, 1, 4, 33, 120, 252, 336, 294, 168, 60, 12, 1, 4, 37, 153, 372, 588, 630, 462, 228, 72, 13, 1, 4, 41, 190, 525, 960, 1218

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).
This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x) = (1+3*z)/(1-(1+x)*z).
The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

			Triangle begins
  [1];
  [4, 1];
  [4, 5, 1];
  [4, 9, 6, 1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, >Rows n = 0..125 of triangle, flattened
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A020714(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 3 for n=2 and 0 otherwise.
Columns m=1..9: A016813, A000384 (hexagonal), A002412, A002417, A034263, A051947, A050483, A052181, A055843.
Cf. A007318, A093562 (d=5), A228196, A228576.

a093561 n k = a093561_tabl !! n !! k
a093561_row n = a093561_tabl !! n
a093561_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [4, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093561(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+3*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(4;n-m, m) for 0<= m <= n, otherwise 0, with F(4;0, 0)=1, F(4;n, 0)=4 if n>=1 and F(4;n, m) = (4*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=4 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. row m (without leading zeros): (1+3*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 3*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(4 + 9*x + 6*x^2/2! + x^3/3!) = 4 + 13*x + 28*x^2/2! + 50*x^3/3! + 80*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A093563 (6,1)-Pascal triangle.

Original entry on oeis.org

1, 6, 1, 6, 7, 1, 6, 13, 8, 1, 6, 19, 21, 9, 1, 6, 25, 40, 30, 10, 1, 6, 31, 65, 70, 40, 11, 1, 6, 37, 96, 135, 110, 51, 12, 1, 6, 43, 133, 231, 245, 161, 63, 13, 1, 6, 49, 176, 364, 476, 406, 224, 76, 14, 1, 6, 55, 225, 540, 840, 882, 630, 300, 90, 15, 1, 6, 61, 280, 765, 1380

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+5*z)/(1-(1+x)*z).
The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins
  1;
  6,  1;
  6,  7,  1;
  6, 13,  8,  1;
  6, 19, 21,  9,  1;
  6, 25, 40, 30, 10,  1;
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Wolfdieter Lang, First 10 rows and array of figurate numbers

Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
Cf. A007318, A093564 (d=7), A228196, A228576.
The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.

a093563 n k = a093563_tabl !! n !! k
a093563_row n = a093563_tabl !! n
a093563_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
-- Reinhard Zumkeller, Aug 31 2014

lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* Jean-François Alcover, Sep 16 2011, after g.f. *)

from math import comb, isqrt
def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 5*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A093562 (5,1) Pascal triangle.

Original entry on oeis.org

1, 5, 1, 5, 6, 1, 5, 11, 7, 1, 5, 16, 18, 8, 1, 5, 21, 34, 26, 9, 1, 5, 26, 55, 60, 35, 10, 1, 5, 31, 81, 115, 95, 45, 11, 1, 5, 36, 112, 196, 210, 140, 56, 12, 1, 5, 41, 148, 308, 406, 350, 196, 68, 13, 1, 5, 46, 189, 456, 714, 756, 546, 264, 81, 14, 1, 5, 51, 235, 645, 1170

Offset: 0

Wolfdieter Lang, Apr 22 2004

This is the fifth member, d=5, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-1, for d=1..4.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+4*z)/(1-(1+x)*z).
The SW-NE diagonals give A022095(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 4. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
The array F(5;n,m) gives in the columns m >= 1 the figurate numbers based on A016861, including the heptagonal numbers A000566 (see the W. Lang link).
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013
The n-th row polynomial is (4 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Mar 02 2018

			Triangle begins
  [1];
  [5,  1];
  [5,  6,  1];
  [5, 11,  7,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider, Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A007283(n-1), n>=1, 1 for n=0. A082505(n+1), alternating row sums are 1 for n=0, 4 for n=2 and 0 else.
Column sequences give for m=1..9: A016861, A000566 (heptagonal), A002413, A002418, A027800, A051946, A050484, A052255, A055844.
A007318, A093563 (d=6), A228196, A228576.

a093562 n k = a093562_tabl !! n !! k
a093562_row n = a093562_tabl !! n
a093562_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [5, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093562(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+(r-a<<2))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(5;n-m, m) for 0<= m <= n, otherwise 0, with F(5;0, 0)=1, F(5;n, 0)=5 if n>=1 and F(5;n, m):=(5*n+m)*binomial(n+m-1, m-1)/m if m>=1.
G.f. column m (without leading zeros): (1+4*x)/(1-x)^(m+1), m>=0.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=5 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
T(n, k) = C(n, k) + 4*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(5 + 11*x + 7*x^2/2! + x^3/3!) = 5 + 16*x + 34*x^2/2! + 60*x^3/3! + 95*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A093565 (8,1) Pascal triangle.

Original entry on oeis.org

1, 8, 1, 8, 9, 1, 8, 17, 10, 1, 8, 25, 27, 11, 1, 8, 33, 52, 38, 12, 1, 8, 41, 85, 90, 50, 13, 1, 8, 49, 126, 175, 140, 63, 14, 1, 8, 57, 175, 301, 315, 203, 77, 15, 1, 8, 65, 232, 476, 616, 518, 280, 92, 16, 1, 8, 73, 297, 708, 1092, 1134, 798, 372, 108, 17, 1, 8, 81, 370, 1005

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(8;n,m) gives in the columns m>=1 the figurate numbers based on A017077, including the decagonal numbers A001107,(see the W. Lang link).
This is the eighth member, d=8, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-4, for d=1..7.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x):=Sum_{m=0..n} a(n,m)*x^m is G(z,x)=(1+7*z)/(1-(1+x)*z).
The SW-NE diagonals give A022098(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 7. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
  [1];
  [8,  1];
  [8,  9,  1];
  [8, 17, 10,  1];
  ...

Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
W. Lang, First 10 rows and array of figurate numbers .

Row sums: A005010(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 7 for n=2 and 0 else.
The column sequences give for m=1..9: A017077, A001107 (decagonal), A007585, A051797, A051878, A050404, A052226, A056001, A056122.
Cf. A093644 (d=9).

a093565 n k = a093565_tabl !! n !! k
a093565_row n = a093565_tabl !! n
a093565_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [8, 1]
-- Reinhard Zumkeller, Aug 31 2014

a(n, m)=F(8;n-m, m) for 0<= m <= n, otherwise 0, with F(8;0, 0)=1, F(8;n, 0)=8 if n>=1 and F(8;n, m):=(8*n+m)*binomial(n+m-1, m-1)/m if m>=1.
Recursion: a(n, m)=0 if m>n, a(0, 0)= 1; a(n, 0)=8 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+7*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 7*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(8 + 17*x + 10*x^2/2! + x^3/3!) = 8 + 25*x + 52*x^2/2! + 90*x^3/3! + 140*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014

A096940 Pascal (1,5) triangle.

Original entry on oeis.org

5, 1, 5, 1, 6, 5, 1, 7, 11, 5, 1, 8, 18, 16, 5, 1, 9, 26, 34, 21, 5, 1, 10, 35, 60, 55, 26, 5, 1, 11, 45, 95, 115, 81, 31, 5, 1, 12, 56, 140, 210, 196, 112, 36, 5, 1, 13, 68, 196, 350, 406, 308, 148, 41, 5, 1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5, 1, 15, 95, 345, 810, 1302

Offset: 0

Wolfdieter Lang, Jul 16 2004

This is the fifth member, q=5, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1)), A029635 (q=2) (but with a(0,0)=2, not 1), A095660 (q=3), A095666 (q=4), A096956 (q=6).
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column no. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is G(z,x)=g(z)/(1-x*z*f(z)). Here: g(x)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins:
  5;
  1,  5;
  1,  6,  5;
  1,  7, 11,   5;
  1,  8, 18,  16,   5;
  1,  9, 26,  34,  21,   5;
  1, 10, 35,  60,  55,  26,   5;
  1, 11, 45,  95, 115,  81,  31,   5;
  1, 12, 56, 140, 210, 196, 112,  36,   5;
  1, 13, 68, 196, 350, 406, 308, 148,  41,  5;
  1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.

David A. Corneth, Table of n, a(n) for n = 0..9999
Wolfdieter Lang, First 10 rows.

Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].

Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.

a(n,k):=piecewise(n=0,5,0Mircea Merca, Apr 08 2012

a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m,t)} \\ David A. Corneth, Aug 28 2019

Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; a(n, 0)=1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (5-4*x)/(1-x)^(m+1), m>=0.
a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012

A098615 G.f. A(x) satisfies: A(xG(x)) = G(x), where G(x) is the g.f. for A098614(n) = Fibonacci(n+1)Catalan(n).

Original entry on oeis.org

1, 1, 3, 5, 13, 25, 61, 125, 295, 625, 1447, 3125, 7151, 15625, 35491, 78125, 176597, 390625, 880125, 1953125, 4390901, 9765625, 21920913, 48828125, 109486993, 244140625, 547018941, 1220703125, 2733608905, 6103515625, 13662695645, 30517578125, 68294088535, 152587890625, 341399727335, 762939453125, 1706739347095, 3814697265625, 8532741458075, 19073486328125, 42660172763995, 95367431640625

Offset: 0

Paul D. Hanna, Oct 14 2004

nonn

G.f. satisfies: A(x) = x/(series reversion of x*G098614(x)), where G098614 is the g.f. for A098614 = {1*1, 1*1, 2*2, 3*5, 5*14, 8*42, 13*132, ...}.
Hankel transform is 2^n. Image of F(n+1) under the Riordan array (c(x^2),xc(x^2)), c(x) the g.f. of A000108. The sequence 0,1,1,3,5,... has general term Sum_{k=0..floor(n/2)} (C(n-1,k) - C(n-1,k-1))*F(n-2k). It is the image of the Fibonacci numbers under the transform of generating functions g(x)-> g(xc(x^2)), c(x) the g.f. of A000108. This sequence has Hankel transform -(-4)^((n-1)/2)(1-(-1)^n)/2. - Paul Barry, Oct 01 2007
The sequence of fractions 1, 1/2, 3/4, 5/8, 13/16, 25/32, ... or a(n)/2^n is the image of F(n+1) under the Chebyshev related (rational) Riordan array c((x/2)^2),(x/2)c((x/2)^2)) where c(x) is the g.f. of A000108. The Hankel transform of this fraction sequence is 1/(2^(n^2)). - Paul Barry, Jun 17 2008

Paul D. Hanna, Table of n, a(n) for n = 0..300
Paul Barry and A. Hennessy, Meixner-Type Results for Riordan Arrays and Associated Integer Sequences, J. Int. Seq. 13 (2010) # 10.9.4, example 30.
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.

Cf. A000045, A000108, A046748, A053121, A054335, A098614.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!( (x+Sqrt(1-4*x^2))/(1-5*x^2) )); // G. C. Greubel, Jul 31 2024

Array[Sum[Binomial[(# - 1)/2, (# - k)/2]*2^(# - k - 1)*((-1)^(# - k) + 1), {k, 0, #}] &, 42, 0] (* or *)
CoefficientList[Series[(Sqrt[1 - 4 x^2] + x)/(1 - 5 x^2), {x, 0, 41}], x] (* Michael De Vlieger, May 20 2021 *)

a(n):=sum(binomial((n-1)/2,(n-k)/2)*2^(n-k-1)*((-1)^(n-k)+1),k,0,n); /* Vladimir Kruchinin, Apr 16 2011 */

{ a(n) = polcoeff((sqrt(1-4*x^2+x^2*O(x^n))+x)/(1-5*x^2),n) }
for(n=0,50,print1(a(n),", "))

def A098615_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (x+sqrt(1-4*x^2))/(1-5*x^2) ).list()
A098615_list(30) # G. C. Greubel, Jul 31 2024

G.f.: (x + sqrt(1-4*x^2)) / (1-5*x^2).
G.f. satisfies: A(x) = sqrt(1 + 2*x*A(x) + 5*x^2*A(x)^2). - Paul D. Hanna, Nov 18 2014
a(2*n) = A046748(n).
a(2*n+1) = 5^n.
a(n) = Sum_{k=0..floor((n+1)/2)} (C(n,k) - C(n,k-1))*Fibonacci(n-2k+1). - Paul Barry, Oct 01 2007
G.f.: 1/(1-x-2x^2/(1-x^2/(1-x^2/(1-x^2/(1-x^2/(1-.... (continued fraction). - Paul Barry, Feb 09 2009
a(n) = Sum_{k=0..n} binomial((n-1)/2,(n-k)/2)*2^(n-k-1)*(1+(-1)^(n-k)). - Vladimir Kruchinin, Apr 16 2011
From Gary W. Adamson, Sep 22 2011: (Start)
a(n) is the upper left term in M^n, M = an infinite square production matrix as follows:
   1, 1, 1, 0, 0, 0, ...
   1, 0, 0, 1, 0, 0, ...
   1, 0, 0, 0, 1, 0, ...
   0, 1, 0, 0, 0, 1, ...
   0, 0, 1, 0, 0, 0, ...
   0, 0, 0, 1, 0, 0, ...
   0, 0, 0, 0, 1, 0, ...
   0, 0, 0, 0, 0, 1, ...
   ... (End)
a(n) = Sum_{k=0..floor(n/2)} A054335(n-k,n-2k). - Philippe Deléham, Feb 01 2012
a(n) = Sum_{k=0..n} A053121(n,k)*A000045(k+1). - Philippe Deléham, Feb 03 2012
n*a(n) +(n-1)*a(n-1) -3*(3*n-4)*a(n-2) -3*(3*n-7)*a(n-3) +20*(n-3)*a(n-4) +20*(n-4)*a(n-5) = 0. - R. J. Mathar, Jul 21 2017

A126930 Inverse binomial transform of A005043.

Original entry on oeis.org

1, -1, 2, -3, 6, -10, 20, -35, 70, -126, 252, -462, 924, -1716, 3432, -6435, 12870, -24310, 48620, -92378, 184756, -352716, 705432, -1352078, 2704156, -5200300, 10400600, -20058300, 40116600, -77558760, 155117520, -300540195, 601080390, -1166803110

Offset: 0

Philippe Deléham, Mar 17 2007

Successive binomial transforms are A005043, A000108, A007317, A064613, A104455. Hankel transform is A000012.
Moment sequence of the trace of the square of a random matrix in USp(2)=SU(2). If X=tr(A^2) is a random variable (a distributed with Haar measure) then a(n) = E[X^n]. - Andrew V. Sutherland, Feb 29 2008
From Tom Copeland, Nov 08 2014: (Start)
This array is one of a family of Catalan arrays related by compositions of the special fractional linear (Mobius) transformation P(x,t) = x/(1-t*x); its inverse Pinv(x,t) = P(x,-t); an o.g.f. of the Catalan numbers A000108, C(x) = [1-sqrt(1-4x)]/2; and its inverse Cinv(x) = x*(1-x). The Motzkin sums, or Riordan numbers, A005043 are generated by Mot(x)=C[P(x,1)]. One could, of course, choose the Riordan numbers as the parent sequence.
O.g.f.: G(x) = C[P[P(x,1),1]1] = C[P(x,2)] = (1-sqrt(1-4*x/(1+2*x)))/2 = x - x^2 + 2 x^3 - ... = Mot[P(x,1)].
Ginv(x) =  Pinv[Cinv(x),2] = P[Cinv(x),-2] = x(1-x)/[1-2x(1-x)] = (x-x^2)/[1-2(x-x^2)] = x*A146559(x).
Cf. A091867 and A210736 for an unsigned version with a leading 1. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Francesc Fite, Kiran S. Kedlaya, Victor Rotger and Andrew V. Sutherland, Sato-Tate distributions and Galois endomorphism modules in genus 2, arXiv preprint arXiv:1110.6638 [math.NT], 2011.
Kiran S. Kedlaya and Andrew V. Sutherland, Hyperelliptic curves, L-polynomials and random matrices, arXiv:0803.4462 [math.NT], 2008-2010.
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 100.

Cf. A126120, A126869.

Cf. A000108, A005043, A146559, A091867, A210736.

egf := BesselI(0,2*x) - BesselI(1,2*x):
seq(n!*coeff(series(egf,x,34),x,n),n=0..33); # Peter Luschny, Dec 17 2014

CoefficientList[Series[(1 + 2 x - Sqrt[1 - 4 x^2])/(2 x (1 + 2 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 23 2013 *)
Table[2^n Hypergeometric2F1[3/2, -n, 2, 2], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 02 2015 *)

x='x+O('x^50); Vec((1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x))) \\ Altug Alkan, Nov 03 2015

a(n) = (-1)^n*C(n, floor(n/2)) = (-1)^n*A001405(n).
a(2*n) = A000984(n), a(2*n+1) = -A001700(n).
a(n) = (1/Pi)*Integral_{t=0..Pi}(2cos(2t))^n*2sin^2(t) dt. - Andrew V. Sutherland, Feb 29 2008, Mar 09 2008
a(n) = (-2)^n + Sum_{k=0..n-1} a(k)*a(n-1-k), a(0)=1. - Philippe Deléham, Dec 12 2009
G.f.: (1+2*x-sqrt(1-4*x^2))/(2*x*(1+2*x)). - Philippe Deléham, Mar 01 2013
O.g.f.: (1 + x*c(x^2))/(1 + 2*x), with the o.g.f. c(x) for the Catalan numbers A000108. From the o.g.f. of the Riordan type Catalan triangle A053121. This is the rewritten g.f. given in the previous formula. This is G(-x) with the o.g.f. G(x) of A001405. - Wolfdieter Lang, Sep 22 2013
D-finite with recurrence (n+1)*a(n) +2*a(n-1) +4*(-n+1)*a(n-2)=0. - R. J. Mathar, Dec 04 2013
Recurrence (an alternative): (n+1)*a(n) = 8*(n-2)*a(n-3) + 4*(n-2)*a(n-2) + 2*(-n-1)*a(n-1), n>=3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) ~ (-1)^n*2^(n+1/2)/sqrt(n*Pi). - Fung Lam, Mar 22 2014
E.g.f.: BesselI(0,2*x) - BesselI(1,2*x). - Peter Luschny, Dec 17 2014
a(n) = 2^n*hypergeom([3/2,-n], [2], 2). - Vladimir Reshetnikov, Nov 02 2015
G.f. A(x) satisfies: A(x) = 1/(1 + 2*x) + x*A(x)^2. - Ilya Gutkovskiy, Jul 10 2020

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A054456 Convolution triangle of A000129(n) (Pell numbers).

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A095666 Pascal (1,4) triangle.

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A073387 Convolution triangle of A002605(n) (generalized (2,2)-Fibonacci), n>=0.

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A093561 (4,1) Pascal triangle.

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A093563 (6,1)-Pascal triangle.

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A093562 (5,1) Pascal triangle.

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A098615 G.f. A(x) satisfies: A(xG(x)) = G(x), where G(x) is the g.f. for A098614(n) = Fibonacci(n+1)Catalan(n).