A078812 -id:A078812 - cp's OEIS frontend

A106195 Riordan array (1/(1-2x), x(1-x)/(1-2*x)).

1, 2, 1, 4, 3, 1, 8, 8, 4, 1, 16, 20, 13, 5, 1, 32, 48, 38, 19, 6, 1, 64, 112, 104, 63, 26, 7, 1, 128, 256, 272, 192, 96, 34, 8, 1, 256, 576, 688, 552, 321, 138, 43, 9, 1, 512, 1280, 1696, 1520, 1002, 501, 190, 53, 10, 1, 1024, 2816, 4096, 4048, 2972, 1683, 743, 253, 64, 11

Offset: 0

Gary W. Adamson, Apr 24 2005; Paul Barry, May 21 2006

Extract antidiagonals from the product P * A, where P = the infinite lower triangular Pascal's triangle matrix; and A = the Pascal's triangle array:
  1, 1,  1,  1, ...
  1, 2,  3,  4, ...
  1, 3,  6, 10, ...
  1, 4, 10, 20, ...
  ...
Row sums are Fibonacci(2n+2). Diagonal sums are A006054(n+2). Row sums of inverse are A105523. Product of Pascal triangle A007318 and A046854.
A106195 with an appended column of ones = A055587. Alternatively, k-th column (k=0, 1, 2) is the binomial transform of bin(n, k).
T(n,k) is the number of ideals in the fence Z(2n) having k elements of rank 1. - Emanuele Munarini, Mar 22 2011
Subtriangle of the triangle given by (0, 2, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 22 2012

			Triangle begins
   1;
   2,   1;
   4,   3,   1;
   8,   8,   4,  1;
  16,  20,  13,  5,  1;
  32,  48,  38, 19,  6, 1;
  64, 112, 104, 63, 26, 7, 1;
(0, 2, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, ...) begins :
  1;
  0,  1;
  0,  2,   1;
  0,  4,   3,   1;
  0,  8,   8,   4,  1;
  0, 16,  20,  13,  5,  1;
  0, 32,  48,  38, 19,  6, 1;
  0, 64, 112, 104, 63, 26, 7, 1. - _Philippe Deléham_, Mar 22 2012

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
E. Munarini, N. Zagaglia Salvi, On the Rank Polynomial of the Lattice of Order Ideals of Fences and Crowns, Discrete Mathematics 259 (2002), 163-177.

Column 0 = 1, 2, 4...; (binomial transform of 1, 1, 1...); column 1 = 1, 3, 8, 20...(binomial transform of 1, 2, 3...); column 2: 1, 4, 13, 38...= binomial transform of bin(n, 2): 1, 3, 6...

Cf. A001792, A001906, A002620, A007318, A029653, A049612, A055587, A078812, A208341.

a106195 n k = a106195_tabl !! n !! k
a106195_row n = a106195_tabl !! n
a106195_tabl = [1] : [2, 1] : f [1] [2, 1] where
   f us vs = ws : f vs ws where
     ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
                      ([0] ++ us ++ [0])
-- Reinhard Zumkeller, Dec 16 2013

[ (&+[Binomial(n-k, n-j)*Binomial(j, k): j in [0..n]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Mar 15 2020

T := (n, k) -> hypergeom([-n+k, k+1],[1],-1):
seq(lprint(seq(simplify(T(n, k)), k=0..n)), n=0..7); # Peter Luschny, May 20 2015

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := u[n - 1, x] + (x + 1) v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207605 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A106195 *)
(* Clark Kimberling, Feb 19 2012 *)
Table[Hypergeometric2F1[-n+k, k+1, 1, -1], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 15 2020 *)

create_list(sum(binomial(i,k)*binomial(n-k,n-i),i,0,n),n,0,8,k,0,n); /* Emanuele Munarini, Mar 22 2011 */

from sympy import Poly, symbols
x = symbols('x')
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

from mpmath import hyp2f1, nprint
def T(n, k): return hyp2f1(k - n, k + 1, 1, -1)
for n in range(13): nprint([int(T(n, k)) for k in range(n + 1)]) # Indranil Ghosh, May 28 2017, after formula from Peter Luschny

[[sum(binomial(n-k,n-j)*binomial(j,k) for j in (0..n)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 15 2020

T(n,k) = Sum_{j=0..n} C(n-k,n-j)*C(j,k).
From Emanuele Munarini, Mar 22 2011: (Start)
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-k,i)*2^(n-k-i).
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-i,k)*(-1)^i*2^(n-k-i).
Recurrence: T(n+2,k+1) = 2*T(n+1,k+1)+T(n+1,k)-T(n,k). (End)
From Clark Kimberling, Feb 19 2012: (Start)
Define u(n,x) = u(n-1,x)+v(n-1,x), v(n,x) = u(n-1,x)+(x+1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1. Then v matches A106195 and u matches A207605. (End)
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k-1). - Philippe Deléham, Mar 22 2012
T(n+k,k) is the coefficient of x^n y^k in 1/(1-2x-y+xy). - Ira M. Gessel, Oct 30 2012
T(n, k) = A208341(n+1,n-k+1), k = 0..n. - Reinhard Zumkeller, Dec 16 2013
T(n, k) = hypergeometric_2F1(-n+k, k+1, 1 , -1). - Peter Luschny, May 20 2015
G.f. 1/(1-2*x+x^2*y-x*y). - R. J. Mathar, Aug 11 2015
Sum_{k=0..n} T(n, k) = Fibonacci(2*n+2) = A088305(n+1). - G. C. Greubel, Mar 15 2020

Edited by N. J. A. Sloane, Apr 09 2007, merging two sequences submitted independently by Gary W. Adamson and Paul Barry

A118851 Product of parts in n-th partition in Abramowitz and Stegun order.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 4, 3, 4, 2, 1, 5, 4, 6, 3, 4, 2, 1, 6, 5, 8, 9, 4, 6, 8, 3, 4, 2, 1, 7, 6, 10, 12, 5, 8, 9, 12, 4, 6, 8, 3, 4, 2, 1, 8, 7, 12, 15, 16, 6, 10, 12, 16, 18, 5, 8, 9, 12, 16, 4, 6, 8, 3, 4, 2, 1, 9, 8, 14, 18, 20, 7, 12, 15, 16, 20, 24, 27, 6, 10, 12, 16, 18, 24, 5, 8, 9, 12, 16, 4

Offset: 0

Alford Arnold, May 01 2006

Let Theta(n) denote the set of norm values corresponding to all the partitions of n. The following results hold regarding this set: (i) Theta(n) is a subset of Theta(n+1); (ii) A prime p will appear as a norm only for partitions of n>=p; (iii) There exists a prime p not in Theta(n) for all n>=6; (iv) Let h(k) be the prime floor function which gives the greatest prime less than or equal to the k, then the prime p=h(n+1) does not belong to Theta(n); and (v) The primes not in the set Theta(n) are A000720(A000792(n)) - A000720(n). - Abhimanyu Kumar, Nov 25 2020

			a(9) = 4 because the 9th partition is [2,2] and 2*2 = 4.
Table T(n,k) starts:
  1;
  1;
  2, 1;
  3, 2,  1;
  4, 3,  4,  2,  1;
  5, 4,  6,  3,  4, 2,  1;
  6, 5,  8,  9,  4, 6,  8,  3,  4,  2, 1;
  7, 6, 10, 12,  5, 8,  9, 12,  4,  6, 8, 3, 4,  2,  1;
  8, 7, 12, 15, 16, 6, 10, 12, 16, 18, 5, 8, 9, 12, 16, 4, 6, 8, 3, 4, 2, 1;

Abramowitz and Stegun, Handbook (1964) page 831.

Andrew Howroyd, Table of n, a(n) for n = 0..2713 (rows 0..20)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Walter Bridges and William Craig, On the distribution of the norm of partitions, arXiv:2308.00123 [math.CO], 2023.
Abhimanyu Kumar and Meenakshi Rana, On the treatment of partitions as factorization and further analysis, Journal of the Ramanujan Mathematical Society 35(3), 263-276 (2020).
Wolfdieter Lang, Rows n=1..10.
Robert Schneider and Andrew V. Sills, The Product of Parts or "Norm" of a Partition, INTEGERS, Volume 20A (2020) Paper #A13, 16 pp.
Andrew V. Sills and Robert Schneider, The product of parts or "norm" of a partition, arXiv:1904.08004 [math.NT], 2019-2021.

Cf. A000041 (row lengths), A006906 (row sums).

Cf. A005361, A036035, A036036, A048996, A078812, A085643, A103921, A111786.

C(sig)={vecprod(sig)}
Row(n)={apply(C, [Vecrev(p) | p<-partitions(n)])}
{ for(n=0, 7, print(Row(n))) } \\ Andrew Howroyd, Oct 19 2020

a(n) = A085643(n)/A048996(n).

T(n,k) = A005361(A036035(n,k)). - Andrew Howroyd, Oct 19 2020

Corrected and extended by Franklin T. Adams-Watters, May 26 2006

A207824 Triangle of coefficients of Chebyshev's S(n,x+5) polynomials (exponents of x in increasing order).

Original entry on oeis.org

1, 5, 1, 24, 10, 1, 115, 73, 15, 1, 551, 470, 147, 20, 1, 2640, 2828, 1190, 246, 25, 1, 12649, 16310, 8631, 2400, 370, 30, 1, 60605, 91371, 58275, 20385, 4225, 519, 35, 1, 290376, 501150, 374115, 157800, 41140, 6790, 693, 40, 1

Offset: 0

Philippe Deléham, Feb 20 2012

Riordan array (1/(1-5*x+x^2), x/(1-5*x+x^2)).
Subtriangle of triangle given by (0, 5, -1/5, 1/5, 0, 0, 0, 0, 0, 0, 0, 0...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
Unsigned version of A123967 and A179900.
For 1<=k<=n, T(n,k) equals the number of (n-1)-length  words over {0,1,2,3,4,5} containing k-1 letters equal 5 and avoiding 01. -  Milan Janjic, Dec 20 2016

			Triangle begins :
  1
  5, 1
  24, 10, 1
  115, 73, 15, 1
  551, 470, 147, 20, 1
  2640, 2828, 1190, 246, 25, 1
  12649, 16310, 8631, 2400, 370, 30, 1
  ...
Triangle (0, 5, -1/5, 1/5, 0, 0, 0,...) DELTA (1, 0, 0, 0, ...) begins :
  1
  0, 1
  0, 5, 1
  0, 24, 10, 1
  0, 115, 73, 15, 1
  0, 551, 470, 147, 20, 1
  0, 2640, 2828, 1190, 246, 25, 1
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150)
Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. Triangles of coefficients of Chebyshev's S(n,x+k) polynomials : A207824 (k = 5), A207823 (k = 4), A125662 (k = 3), A078812 (k = 2), A101950 (k = 1), A049310 (k = 0), A104562 (k = -1), A053122 (k = -2), A207815 (k = -3), A159764 (k = -4), A123967 (k = -5).

With[{n = 8}, DeleteCases[#, 0] & /@ CoefficientList[Series[1/(1 - 5 x + x^2 - y x), {x, 0, n}, {y, 0, n}], {x, y}]] // Flatten (* Michael De Vlieger, Apr 25 2018 *)

row(n) = Vecrev(polchebyshev(n, 2, (x+5)/2)); \\ Michel Marcus, Apr 26 2018

Recurrence : T(n,k) = 5*T(n-1,k) + T(n-1,k-1) - T(n-2,k).
G.f.: 1/(1-5*x+x^2-y*x).
Diagonal sums are 5^n = A000351(n).
Row sums are A001109(n+1).
T(n,0) = A004254(n+1), T(n+1,n) = 5n+5 = A008587(n+1).

A053112 Expansion of (-1 + 1/(1-9x)^9)/(81x); related to A053108.

Original entry on oeis.org

1, 45, 1485, 40095, 938223, 19702683, 379980315, 6839645670, 116273976390, 1883638417518, 29282015399598, 439230230993970, 6385731819835410, 90312492880529370, 1246312401751305306, 16825217423642621631

Offset: 0

Wolfdieter Lang

G. C. Greubel, Table of n, a(n) for n = 0..400
Index entries for linear recurrences with constant coefficients, signature (81, -2916, 61236, -826686, 7440174, -44641044, 172186884, -387420489, 387420489).

Cf. A053108, A053110 - A053113.

Without signs: A078812. With zeros: A049310. Cf. A008310 (T(n, x)), A008312 (U(n, x)).

[9^(n-1)*Binomial(n+9, 8): n in [0..30]]; // G. C. Greubel, Aug 16 2018

CoefficientList[Series[(-1+1/(1-9*x)^9)/(81*x),{x,0,30}],x] (* or *) LinearRecurrence[{81,-2916,61236,-826686,7440174,-44641044,172186884,-387420489,387420489}, {1,45,1485,40095,938223,19702683, 379980315, 6839645670,116273976390},20] (* Harvey P. Dale, Apr 27 2013 *)
Table[9^(n - 1)*Binomial[n + 9, 8], {n, 0, 30}] (* G. C. Greubel, Aug 16 2018 *)

vector(30,n,n--; 9^(n-1)*binomial(n+9, 8)) \\ G. C. Greubel, Aug 16 2018

G.f.: (-1 + 1/(1-9*x)^9)/(81*x).
a(n) = 9^(n-1)*binomial(n+9, 8).
a(0)=1, a(1)=45, a(2)=1485, a(3)=40095, a(4)=938223, a(5)=19702683, a(6)=379980315, a(7)=6839645670, a(8)=116273976390, a(n)=81*a(n-1)- 2916*a(n-2)+ 61236*a(n-3)- 826686*a(n-4)+ 7440174*a(n-5)- 44641044*a(n-6)+ 172186884*a(n-7)- 387420489*a(n-8)+ 387420489*a(n-9). - Harvey P. Dale, Apr 27 2013

A179320 E.g.f. satisfies: A(x) = A( x/(1-x)^2 ) * (1-x)/(1+x) with A(0)=0.

Original entry on oeis.org

0, 2, -2, 6, -28, 160, -936, 4536, -20448, 627264, -19699200, 43908480, 17788273920, -211715112960, -41219197125120, 1301670191808000, 160057006683033600, -10037518414724505600, -1007362871616478003200

Offset: 0

Paul D. Hanna, Jul 11 2010

sign

			E.g.f.: A(x) = 2*x - 2*x^2/2! + 6*x^3/3! - 28*x^4/4! + 160*x^5/5! - 936*x^6/6! + 4536*x^7/7! - 20448*x^8/8! + 627264*x^9/9! - 19699200*x^10/10! + 43908480*x^11/11! + 17788273920*x^12/12! -+ ...
A(x/(1-x)^2) = 2*x + 6*x^2/2! + 18*x^3/3! + 68*x^4/4! + 360*x^5/5! + 2184*x^6/6! + 13272*x^7/7! + 122016*x^8/8! + 1541376*x^9/9! + 1987200*x^10/10! - 150923520*x^11/11! + 16504093440*x^12/12! + ...
where A(x/(1-x)^2) = (1+x)/(1-x)*A(x).
...
Related expansions begin:
. A = 2*x - 2*x^2/2! + 6*x^3/3! - 28*x^4/4! + 160*x^5/5! + ...
. A*Dx(A)/2! = 8*x^2/2! - 30*x^3/3! + 180*x^4/4! - 1400*x^5/5! + ...
. A*Dx(A*Dx(A))/3! = 48*x^3/3! - 416*x^4/4! + 4280*x^5/5! + ...
. A*Dx(A*Dx(A*Dx(A)))/4! = 384*x^4/4! - 6160*x^5/5! + 98400*x^6/6! -+ ...
. A*Dx(A*Dx(A*Dx(A*Dx(A))))/5! = 3840*x^5/5! - 100224*x^6/6! +- ...
where Catalan(-x)^2 = 1 - A + A*Dx(A)/2! - A*Dx(A*Dx(A))/3! +- ... = 1 - 2*x + 5*x^2 - 14*x^3 + 42*x^4 + ... + A000108(n)*(-x)^n + ...

Paul D. Hanna, Table of n, a(n) for n = 0..200
J. Cigler, Some results and conjectures about a class of q-polynomials with simple moments, 2014.

Cf. A027614, A078812, A179321, A261885.

Variant: A179199.

T[n_, k_]:= T[n, k]= If[k==0, 1, If[k==1, 2*(n-1), T[n-2, k-2] + Binomial[2n-k-1, 2n-2k-1] ]]; (* T = A123521 *)
b[n_]:= b[n]= If[n==1, 1, (-1/(2*(n-1)))*Sum[b[n-j+1]*T[n,j], {j, 2, 2*Floor[n/2]}]];
A179320[n_] := 2*n!*b[n];
Table[A179320[n], {n, 0, 40}] (* G. C. Greubel, Sep 01 2022 *)

/* E.g.f. satisfies: A(x) = (1-x)/(1+x)*A(x/(1-x)^2): */
{a(n)=local(A=2*x, B); for(m=2, n, B=(1-x)/(1+x+O(x^(n+3)))*subst(A,x,x/(1-x+O(x^(n+3)))^2); A=A-polcoeff(B, m+1)*x^m/(m-1)/2); n!*polcoeff(A, n)}

/* 1/(1-x)^2 = 1 + A + A*Dx(A)/2! + A*Dx(A*Dx(A))/3! +...: */
{a(n)=local(A=0+sum(m=1,n-1,a(m)*x^m/m!),D=1,R=0);R=-1/(1-x+x*O(x^n))^2+1+sum(m=1,n,(D=A*deriv(x*D+x*O(x^n)))/m!);-n!*polcoeff(R,n)}

/* As column 0 of the matrix log of triangle A078812: */
{a(n)=local(A078812=matrix(n+1, n+1, r, c, if(r>=c, binomial(r+c-1,r-c))), LOG, ID=A078812^0); LOG=sum(m=1, n+1, -(ID-A078812)^m/m); n!*LOG[n+1, 1]}

@CachedFunction
def T(n,k): # T = A123521
    if (k==0): return 1
    elif (k==1): return 2*(n-1)
    else: return T(n-2, k-2) + binomial(2*n-k-1, 2*n-2*k-1)
@CachedFunction
def b(n):
    if (n==1): return 1
    else: return (-1/(2*(n-1)))*sum(T(n,j)*b(n-j+1) for j in (2..2*floor(n/2)))
def A179320(n): return 0 if (n==0) else 2*factorial(n)*b(n)
[A179320(n) for n in (0..40)] # G. C. Greubel, Sep 01 2022

E.g.f. A = A(x) satisfies:
(1) 1/(1-x)^2 = 1 + A + A*Dx(A)/2! + A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! + ...
(2) Catalan(-x)^2 = 1 - A + A*Dx(A)/2! - A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! -+ ...
(3) (1-x)^2/(1-3*x+x^2)^2 = 1 + 2*A + 2^2*A*Dx(A)/2! + 2^3*A*Dx(A*Dx(A))/3! + 2^4*A*Dx(A*Dx(A*Dx(A)))/4! + ...
where Dx(F) = d/dx(x*F).
INVERSION FORMULA:
More generally, if A(x) = A(G(x))*G(x)/(x*G'(x)) with G(0)=0, G'(0)=1,
then G(x) can be obtained from A=A(x) by the series:
G(x)/x = 1 + A + A*Dx(A)/2! + A*Dx(A*Dx(A))/3! + A*Dx(A*Dx(A*Dx(A)))/4! + ... where Dx(F) = d/dx(x*F).
ITERATION FORMULA:
Let G_{n}(x) denote the n-th iteration of G(x) = x/(1-x)^2, and A=A(x), then:
G_{n}(x)/x = 1 + n*A + n^2*A*Dx(A)/2! + n^3*A*Dx(A*Dx(A))/3! + n^4*A*Dx(A*Dx(A*Dx(A)))/4! + ...
MATRIX LOG OF RIORDAN ARRAY (G(x)/x, G(x)) where G(x) = x/(1-x)^2:
E.g.f. A(x) forms column 0 of A179321, the matrix log of triangle A078812, where A078812(n,k) = C(n+k+1,n-k); the g.f. of column k in A078812 is [x/(1-x)^2]^(k+1)/x.
A179321(n,k) = (k+1)*a(n-k)/(n-k)! for n>0, k>=0, where A179321 = matrix log of triangle A078812.
...
a(n) = (-1)^(n-1)*2*A027614(n), where A027614 is related to Clebsch-Gordan formulas.
a(n) = 2*n!*b(n), with a(0) = 0, where b(n) = (-1/(2*(n-1))) * Sum_{j=2..2*floor(n/2)} A123521(n, j)*b(n-j+1), and b(1) = 1. - G. C. Greubel, Sep 01 2022

A176992 Triangle T(n,m) = binomial(2n-k+1, n+1) read by rows, 0 <= k <= n.

Original entry on oeis.org

1, 3, 1, 10, 4, 1, 35, 15, 5, 1, 126, 56, 21, 6, 1, 462, 210, 84, 28, 7, 1, 1716, 792, 330, 120, 36, 8, 1, 6435, 3003, 1287, 495, 165, 45, 9, 1, 24310, 11440, 5005, 2002, 715, 220, 55, 10, 1, 92378, 43758, 19448, 8008, 3003, 1001, 286, 66, 11, 1, 352716, 167960, 75582, 31824, 12376, 4368, 1365, 364, 78, 12, 1

Offset: 0

Roger L. Bagula, Dec 08 2010

Row sums are A001791.
Obtained from A059481 by removal of the last two terms in each row, followed by row reversal.
Riordan array (c(x)/sqrt(1 - 4*x), x*c(x)) where c(x) is the g.f. of A000108. - Philippe Deléham, Jul 12 2015

			Triangle begins:
       1;
       3,      1;
      10,      4,     1;
      35,     15,     5,     1;
     126,     56,    21,     6,    1;
     462,    210,    84,    28,    7,     1;
    1716,    792,   330,   120,   36,     8,    1;
    6435,   3003,  1287,   495,   165,   45,    9,   1;
   24310,  11440,  5005,  2002,   715,  220,   55,  10,  1;
   92378,  43758, 19448,  8008,  3003, 1001,  286,  66, 11,  1;
  352716, 167960, 75582, 31824, 12376, 4368, 1365, 364, 78, 12, 1;

Cf. A092392, A001791, A078812.
Cf. Similar triangle: A033184, A054445.
Cf. A178300 (reversal).

/* As triangle */ [[Binomial(2*n-k+1,n+1): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 12 2015

A176992 := proc(n,k) binomial(1+2*n-k,n+1) ; end proc: # R. J. Mathar, Dec 09 2010

p[t_, j_] = ((-1)^(j + 1)/2)*Sum[Binomial[k - j - 1, j + 1]*t^k, {k, 0, Infinity}];
Flatten[Table[CoefficientList[ExpandAll[p[t, j]], t], {j, 0, 10}]]

n-th row of the triangle = top row of M^n, where M is the following infinite square production matrix:
  3, 1, 0, 0, 0, ...
  1, 1, 1, 0, 0, ...
  1, 1, 1, 1, 0, ...
  1, 1, 1, 1, 1, ...
  ... - Philippe Deléham, Jul 12 2015

A215042 a(n) = F(8n)/L(2n) with n >= 0, F = A000045 (Fibonacci numbers) and L = A000032 (Lucas numbers).

Original entry on oeis.org

0, 7, 141, 2576, 46347, 831985, 14930208, 267913919, 4807525989, 86267568688, 1548008749155, 27777890017577, 498454011832896, 8944394323670071, 160500643816049277, 2880067194369984080, 51680708854856144763, 927372692193073296289

Offset: 0

Wolfdieter Lang, Aug 31 2012

This provides the second example for the Riordan transition matrix R mentioned in a comment to A078812 (here the column called there n=1 is relevant).

Index entries for linear recurrences with constant coefficients, signature (21,-56,21,-1).

Cf. A001906 (for F(4*n)/L(2*n) = F(2*n)), 24*A215043 (for F(12*n)/L(2*n)).

Cf. A000032, A000045.

[Fibonacci(8*n)/Lucas(2*n): n in [0..17]]; // Bruno Berselli, Aug 31 2012

Table[Fibonacci[8 n]/LucasL[2 n], {n, 0, 17}] (* Bruno Berselli, Aug 31 2012 *)
LinearRecurrence[{21,-56,21,-1},{0,7,141,2576},20] (* Harvey P. Dale, Jul 18 2021 *)

a(n) = 2*F(2*n) + 1*5*F(2*n)^3, n >= 0 (for the coefficients 2, 1,  see the second row of the Riordan matrix R = A078812 (with offset [0,0])).
a(n) = F(6*n) - F(2*n), n >= 0, (from the preceding line and a 5*F(2*n)^3 formula given in a comment on the signed triangle A111418, with l->2*n, n->1; see also 5*A215039).
O.g.f.: x*(7-6*x+7*x^2)/((1-3*x+x^2)*(1-18*x+x^2)). The partial fraction decomposition and recurrences lead to the preceding formula.

A215043 a(n) = F(12n)/(24L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).

Original entry on oeis.org

0, 2, 276, 34561, 4261992, 524393210, 64499742738, 7933009283134, 975696814205904, 120002796170968643, 14759368609635548580, 1815282342961539780022, 223264968937188026209956, 27459775899111901985784506

Offset: 0

Wolfdieter Lang, Aug 31 2012

24*a(n) is the third example for the Riordan transition matrix introduced in a comment on A078812 (with offset [0,0]). Take there l -> n, n -> 2. See the second formula below.

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Index entries for linear recurrences with constant coefficients, signature (144,-2640,6930,-2640,144,-1).

Cf. A215042 (for F(8*n)/L(2*n)).

[Fibonacci(12*n)/(24*Lucas(2*n)): n in [0..15]]; // Vincenzo Librandi, Sep 02 2012

Table[Fibonacci[12*n]/(24*LucasL[2*n]), {n,0,15}] (* G. C. Greubel, Jun 30 2019 *)

lucas(n) = fibonacci(n+1) + fibonacci(n-1);
vector(15, n, n--; fibonacci(12*n)/(24*lucas(2*n))) \\ G. C. Greubel, Jun 30 2019

[fibonacci(12*n)/(24*lucas_number2(2*n,1,-1)) for n in (0..15)] # G. C. Greubel, Jun 30 2019

a(n) = F(12*n)/(24*L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).
a(n) = 3*F(2*n) + 20*F(2*n)^3 + 25*F(2*n)^5, n >= 0 (see the comment above).
O.g.f.: x*(2 - 12*x + 97*x^2 - 12*x^3 + 2*x^4)/((1 - 3*x + x^2)*(1 - 18*x + x^2)*(1 - 123*x + x^2)). From the o.g.f.s for the sequences appearing in the preceding formula, see A001906, A215039 and A215044.
a(n) = (L(8*n) + 1)*F(2*n)/24. - Ehren Metcalfe, Jun 04 2019

A305309 Array read by rows: a(n, k) = A048996(n, k) * A118851(n, k), n >= 1, k = 1..A000041(n).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 6, 4, 6, 1, 5, 8, 12, 9, 12, 8, 1, 6, 10, 16, 9, 12, 36, 8, 12, 24, 10, 1, 7, 12, 20, 24, 15, 48, 27, 36, 16, 72, 32, 15, 40, 12, 1, 8, 14, 24, 30, 16, 18, 60, 72, 48, 54, 20, 96, 54, 144, 16, 20, 120, 80, 18, 60, 14, 1, 9, 16, 28, 36, 40, 21, 72, 90, 48, 60, 144, 27, 24, 120, 144, 192, 216, 96, 25, 160, 90, 360, 80, 24, 180, 160, 21, 84, 16, 1, 10, 18, 32, 42, 48, 25, 24, 84, 108, 120, 72, 180, 96, 108, 28, 144, 180, 96, 240, 576, 108, 128, 216, 30, 200, 240, 480, 540, 480, 32, 30, 240, 135, 720, 240, 28, 252, 280, 24, 112, 18, 1

Offset: 1

Wolfdieter Lang, May 31 2018

The Data section here is longer than usual.  Do not shorten it! - N. J. A. Sloane, Jan 10 2019
The length of row n is A000041(n), the number of partitions of n.
Partitions follow the Abramowitz-Stegun (A-St) order (see the link).
The row sums give A001906(n) = Fibonacci(2*n).
The triangle T(n, m) obtained by summing in row n the entries of the columns k with identical part number m is A078812(n, m) = binomial(n+m-1, 2*m-1) (with offsets n >= 1, m = 1..n). The array of the number of parts m = m(n,k) = A036043(n, k) in A-St order.
This array is the elementwise product of the array A048996, the composition numbers, and A118851, the products of the parts of partitions, both arrays are in A-St order.
Therefore a(n, k) is the sum of the number of products of the block lengths of all the A048996(n, k) set partition of [n] := {1,2, ..., n} with m = m(n, k) blocks consisting of consecutive numbers corresponding to the k-th partition of n in A-St order. Because the block structure depends only on the exponents (signature) of the underlying partition this leads to the product of the two array entries. Equivalently, one can consider compositions. Then a(n, k) gives the sum of the products of the parts of all A048996(n, k) compositions originating from the k-th partition of n.
This array is the result of an attempt to understand the comment of Kevin Long, May 11 2018, on A001906.
This array is similar to A085643 but some pairs of numbers like  (27, 36), (72,48), (54,144), ... are there swapped.

			For the rows n = 1..10, and comments on compositions and set partitions with blocks of consecutive numbers, see the link.
Example: n = 5, k = 4: the partition is (1^2, 3^1) = [1,1,3] with m = m(n,k) = 3. The A048996(5, 4) = 3 compositions are 1 + 1 + 3, 1 + 3 + 1 and 3 + 1 + 1. The corresponding three consecutive 3-block partitions of [5] := {1, 2, ..., 5} are {1}, {2}, {3,4,5} and {1}, {2,3,4}, {5} and {1,2,3}, {4}, {5}, Therefore, a(5, 4) = 1*1*3 + 1*3*1 + 3*1*1 = 3*3 = 9. For the compositions one has the same sum from the products of the parts.

Milton Abramowitz and Irene A. Stegun, editors, Multinomials and Partitions, Handbook of Mathematical Functions, December 1972, pp. 831-2.
Wolfdieter Lang, Rows n = 1..10, and more.

Cf. A000041, A001906, A036043, A048996, A078812, A085643, A118851.

a(n, k) = A048996(n, k) * A118851(n, k), n >= 1, k = 1..A000041(n).

A185331 Riordan array ((1-x+x^2)/(1+x^2), x/(1+x^2)).

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 1, -1, -1, 1, 0, 2, -2, -1, 1, -1, 1, 3, -3, -1, 1, 0, -3, 3, 4, -4, -1, 1, 1, -1, -6, 6, 5, -5, -1, 1, 0, 4, -4, -10, 10, 6, -6, -1, 1, -1, 1, 10, -10, -15, 15, 7, -7, -1, 1, 0, -5, 5, 20, -20, -21, 21, 8, -8, -1, 1

Offset: 0

Philippe Deléham, Feb 08 2012

Triangle T(n,k), read by rows, given by (-1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

			Triangle begins:
   1;
  -1,  1;
   0, -1,   1;
   1, -1,  -1,   1;
   0,  2,  -2,  -1,   1;
  -1,  1,   3,  -3,  -1,   1;
   0, -3,   3,   4,  -4,  -1,   1;
   1, -1,  -6,   6,   5,  -5,  -1,  1;
   0,  4,  -4, -10,  10,   6,  -6, -1,  1;
  -1,  1,  10, -10, -15,  15,   7, -7, -1,  1;
   0, -5,   5,  20, -20, -21,  21,  8, -8, -1,  1;
   1, -1, -15,  15,  35, -35, -28, 28,  9, -9, -1, 1;

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. A206474 (unsigned version).

Cf. A007318, A053122, A078812, A085478, A128908, A129818,

CoefficientList[Series[CoefficientList[Series[(1 - x + x^2)/(1 - y*x + x^2), {x, 0, 10}], x], {y, 0, 10}], y] // Flatten (* G. C. Greubel, Jun 27 2017 *)

T(n,k) = T(n-1,k-1) - T(n-2,k), T(0,0) = 1, T(0,1) = -1, T(0,2) = 0.
G.f.: (1-x+x^2)/(1-y*x+x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = (-1)^n*A184334(n), A163805(n), A000007(n), A028310(n), A025169(n-1), A005320(n) (n>0) for x = -1, 0, 1, 2, 3, 4 respectively.
T(n,n) = 1, T(n+1,n) = -1, T(n+2,n) = -n, T(n+3,n) = n+1, T(n+4,n) = n(n+1)/2 = A000217(n).
T(2n,2k) = (-1)^(n-k) * A128908(n,k), T(2n+1,2k+1) = -T(2n+1,2k) = A129818(n,k), T(2n+2,2k+1) = (-1)*A053122(n,k). - Philippe Deléham, Feb 09 2012

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A207824 Triangle of coefficients of Chebyshev's S(n,x+5) polynomials (exponents of x in increasing order).

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A053112 Expansion of (-1 + 1/(1-9*x)^9)/(81*x); related to A053108.

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A179320 E.g.f. satisfies: A(x) = A( x/(1-x)^2 ) * (1-x)/(1+x) with A(0)=0.

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A176992 Triangle T(n,m) = binomial(2n-k+1, n+1) read by rows, 0 <= k <= n.

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A106195 Riordan array (1/(1-2x), x(1-x)/(1-2*x)).

A053112 Expansion of (-1 + 1/(1-9x)^9)/(81x); related to A053108.

A215042 a(n) = F(8n)/L(2n) with n >= 0, F = A000045 (Fibonacci numbers) and L = A000032 (Lucas numbers).

A215043 a(n) = F(12n)/(24L(2*n)), n >= 0, with F = A000045 (Fibonacci) and L = A000032 (Lucas).