A084773 -id:A084773 - cp's OEIS frontend

A152254 Twice A084773.

2, 12, 104, 1008, 10272, 107712, 1150592, 12451584, 136053248, 1497664512, 16583583744, 184511361024, 2061074178048, 23100352413696, 259648659554304, 2925683135152128, 33037383972814848, 373774017998094336, 4235888981636022272, 48076611822491271168

Offset: 0

Eric Bach (bach(AT)cs.wisc.edu), Oct 15 2009

nonn

Andras Kornai, Mathematical Linguistics (Springer-Verlag 2008), p. 36. (Gives a sequence which matches the first five terms.)

G. C. Greubel, Table of n, a(n) for n = 0..930

Cf. A000079, A001850, A084773.

[2^(n+1)*Evaluate(LegendrePolynomial(n), 3): n in [0..40]]; // G. C. Greubel, May 22 2023

Table[2^(n+1)*LegendreP[n, 3], {n,0,40}] (* G. C. Greubel, May 22 2023 *)

[2^(n+1)*gen_legendre_P(n,0,3) for n in range(41)] # G. C. Greubel, May 22 2023

Offset changed by Petros Hadjicostas, Feb 09 2021

A001850 Central Delannoy numbers: a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k).

Original entry on oeis.org

1, 3, 13, 63, 321, 1683, 8989, 48639, 265729, 1462563, 8097453, 45046719, 251595969, 1409933619, 7923848253, 44642381823, 252055236609, 1425834724419, 8079317057869, 45849429914943, 260543813797441, 1482376214227923, 8443414161166173, 48141245001931263

Offset: 0

N. J. A. Sloane

Number of paths from (0,0) to (n,n) in an n X n grid using only steps north, northeast and east (i.e., steps (1,0), (1,1), and (0,1)).
Also the number of ways of aligning two sequences (e.g., of nucleotides or amino acids) of length n, with at most 2*n gaps (-) inserted, so that while unnecessary gappings: - -a a- - are forbidden, both b- and -b are allowed. (If only other of the latter is allowed, then the sequence A000984 gives the number of alignments.) There is an easy bijection from grid walks given by Dickau to such set of alignments (e.g., the straight diagonal corresponds to the perfect alignment with no gaps). - Antti Karttunen, Oct 10 2001
Also main diagonal of array A008288 defined by m(i,1) = m(1,j) = 1, m(i,j) = m(i-1,j-1) + m(i-1,j) + m(i,j-1). - Benoit Cloitre, May 03 2002
So, as a special case of Dmitry Zaitsev's Dec 10 2015 comment on A008288, a(n) is the number of points in Z^n that are L1 (Manhattan) distance <= n from any given point. These terms occur in the crystal ball sequences: a(n) here is the n-th term in the sequence for the n-dimensional cubic lattice. See A008288 for a list of crystal ball sequences (rows or columns of A008288). - Shel Kaphan, Dec 26 2022
a(n) is the number of n-matchings of a comb-like graph with 2*n teeth. Example: a(2) = 13 because the graph consisting of a horizontal path ABCD and the teeth Aa, Bb, Cc, Dd has 13 2-matchings: any of the six possible pairs of teeth and {Aa, BC}, {Aa, CD}, {Bb, CD}, {Cc, AB}, {Dd, AB}, {Dd, BC}, {AB, CD}. - Emeric Deutsch, Jul 02 2002
Number of ordered trees with 2*n+1 edges, having root of odd degree, nonroot nodes of outdegree at most 2 and branches of odd length. - Emeric Deutsch, Aug 02 2002
The sum of the first n coefficients of ((1 - x) / (1 - 2*x))^n is a(n-1). - Michael Somos, Sep 28 2003
Row sums of A063007 and A105870. - Paul Barry, Apr 23 2005
The Hankel transform (see A001906 for definition) of this sequence is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005
Also number of paths from (0,0) to (n,0) using only steps U = (1,1), H = (1,0) and D =(1,-1), U can have 2 colors and H can have 3 colors. - N-E. Fahssi, Jan 27 2008
Equals row sums of triangle A152250 and INVERT transform of A109980: (1, 2, 8, 36, 172, 852, ...). - Gary W. Adamson, Nov 30 2008
Number of overpartitions in the n X n box (treat a walk of the type in the first comment as an overpartition, by interpreting a NE step as N, E with the part thus created being overlined). - William J. Keith, May 19 2017
Diagonal of rational functions 1/(1 - x - y - x*y), 1/(1 - x - y*z - x*y*z). - Gheorghe Coserea, Jul 03 2018
Dimensions of endomorphism algebras End(R^{(n)}) in the Delannoy category attached to the oligomorphic group of order preserving self-bijections of the real line. - Noah Snyder, Mar 22 2023
a(n) is the number of ways to tile a strip of length n with white squares, black squares, and red dominos, where we must have an equal number of white and black squares. - Greg Dresden and Leo Zhang, Jul 11 2025

			G.f. = 1 + 3*x + 13*x^2 + 63*x^3 + 321*x^4 + 1683*x^5 + 8989*x^6 + ...

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Cf. A008288, bisection of A026003, A027618, A047665, A052141, A084773, A152250, A109980, A000129, A078057, A241023, A243949.
Main diagonal of A064861.
Column k=2 of A262809 and A263159.

seq(add(multinomial(n+k,n-k,k,k),k=0..n),n=0..20); # Zerinvary Lajos, Oct 18 2006
seq(orthopoly[P](n,3), n=0..100); # Robert Israel, Nov 03 2015

f[n_] := Sum[ Binomial[n, k] Binomial[n + k, k], {k, 0, n}]; Array[f, 21, 0] (* Or *)
a[0] = 1; a[1] = 3; a[n_] := a[n] = (3(2 n - 1)a[n - 1] - (n - 1)a[n - 2])/n; Array[a, 21, 0] (* Or *)
CoefficientList[ Series[1/Sqrt[1 - 6x + x^2], {x, 0, 20}], x] (* Robert G. Wilson v *)
Table[LegendreP[n, 3], {n, 0, 22}] (* Jean-François Alcover, Jul 16 2012, from first formula *)
a[n_] := Hypergeometric2F1[-n, n+1, 1, -1]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Feb 26 2013 *)
a[ n_] := With[ {m = If[n < 0, -1 - n, n]}, SeriesCoefficient[ (1 - 6 x + x^2)^(-1/2), {x, 0, m}]]; (* Michael Somos, Jun 10 2015 *)

a(n):=coeff(expand((1+3*x+2*x^2)^n),x,n);
makelist(a(n),n,0,12); /* Emanuele Munarini, Mar 02 2011 */

{a(n) = if( n<0, n = -1 - n); polcoeff( 1 / sqrt(1 - 6*x + x^2 + x * O(x^n)), n)}; /* Michael Somos, Sep 23 2006 */

{a(n) = if( n<0, n = -1 - n); subst( pollegendre(n), x, 3)}; /* Michael Somos, Sep 23 2006 */

{a(n) = if( n<0, n = -1 - n); n++; subst( Pol(((1 - x) / (1 - 2*x) + O(x^n))^n), x, 1);} /* Michael Somos, Sep 23 2006 */

a(n)=if(n<0, 0, polcoeff((1+3*x+2*x^2)^n, n)) \\ Paul Barry, Aug 22 2007

/* same as in A092566 but use */
steps=[[1,0], [0,1], [1,1]]; /* Joerg Arndt, Jun 30 2011 */

a(n)=sum(k=0,n,binomial(n,k)*binomial(n+k,k)); \\ Joerg Arndt, May 11 2013

my(x='x+O('x^30)); Vec(1/sqrt(1 - 6*x + x^2)) \\ Altug Alkan, Oct 17 2015

# from Nick Hobson.
def f(a, b):
    if a == 0 or b == 0:
        return 1
    return f(a, b - 1) + f(a - 1, b) + f(a - 1, b - 1)
[f(n, n) for n in range(7)]

from gmpy2 import divexact
A001850 = [1, 3]
for n in range(2,10**3):
    A001850.append(divexact(A001850[-1]*(6*n-3)-(n-1)*A001850[-2],n))
# Chai Wah Wu, Sep 01 2014

a = lambda n: hypergeometric([-n, -n], [1], 2)
[simplify(a(n)) for n in range(23)] # Peter Luschny, Nov 19 2014

a(n) = P_n(3), where P_n is n-th Legendre polynomial.
G.f.: 1 / sqrt(1 - 6*x + x^2).
a(n) = a(n-1) + 2*A002002(n) = Sum_{j} A063007(n, j). - Henry Bottomley, Jul 02 2001
Dominant term in asymptotic expansion is binomial(2*n, n)/2^(1/4)*((sqrt(2) + 1)/2)^(2*n + 1)*(1 + c_1/n + c_2/n^2 + ...). - Michael David Hirschhorn
a(n) = Sum_{i=0..n} (A000079(i)*A008459(n, i)) = Sum_{i=0..n} (2^i * C(n, i)^2). - Antti Karttunen, Oct 10 2001
a(n) = Sum_{k=0..n} C(n+k, n-k)*C(2*k, k). - Benoit Cloitre, Feb 13 2003
a(n) = Sum_{k=0..n} C(n, k)^2 * 2^k. - Michael Somos, Oct 08 2003
a(n - 1) = coefficient of x^n in A120588(x)^n if n>=0. - Michael Somos, Apr 11 2012
G.f. of a(n-1) = 1 / (1 - x / (1 - 2*x / (1 - 2*x / (1 - x / (1 - 2*x / (1 - x / ...)))))). - Michael Somos, May 11 2012
INVERT transform is A109980. BINOMIAL transform is A080609. BINOMIAL transform of A006139. PSUM transform is A089165. PSUMSIGN transform is A026933. First backward difference is A110170. - Michael Somos, May 11 2012
E.g.f.: exp(3*x)*BesselI(0, 2*sqrt(2)*x). - Vladeta Jovovic, Mar 21 2004
a(n) = Sum_{k=0..n} C(2*n-k, n)*C(n, k). - Paul Barry, Apr 23 2005
a(n) = Sum_{k>=n} binomial(k, n)^2/2^(k+1). - Vladeta Jovovic, Aug 25 2006
a(n) = a(-1 - n) for all n in Z. - Michael Somos, Sep 23 2006
D-finite with recurrence: a(-1) = a(0) = 1; n*a(n) = 3*(2*n-1)*a(n-1) - (n-1)*a(n-2). Eq (4) in T. D. Noe's article in JIS 9 (2006) #06.2.7.
Define general Delannoy numbers by (i,j > 0): d(i,0) = d(0,j) = 1 =: d(0,0) and d(i,j) = d(i-1,j-1) + d(i-2,j-1) + d(i-1,j). Then a(k) = Sum_{j >= 0} d(k,j)^2 + d(k-1,j)^2 = A026933(n)+A026933(n-1). This is a special case of the following formula for general Delannoy numbers: d(k,j) = Sum_{i >= 0, p=0..n} d(p, i) * d(n-p, j-i) + d(p-1, i) * d(n-p-1, j-i-1). - Peter E John, Oct 19 2006
Coefficient of x^n in (1 + 3*x + 2*x^2)^n. - N-E. Fahssi, Jan 11 2008
a(n) = A008288(A046092(n)). - Philippe Deléham, Apr 08 2009
G.f.: 1/(1 - x - 2*x/(1 - x - x/(1 - x - x/(1 - x - x/(1 - ... (continued fraction). - Paul Barry, May 28 2009
G.f.: d/dx log(1/(1 - x*A001003(x))). - Vladimir Kruchinin, Apr 19 2011
G.f.: 1/(2*Q(0) + x - 1) where Q(k) = 1 + k*(1-x) - x - x*(k + 1)*(k + 2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n) = Sum_{k=0..n} C(n,k) * C(n+k,k). - Joerg Arndt, May 11 2013
G.f.: G(0), where G(k) = 1 + x*(6 - x)*(4*k + 1)/(4*k + 2 - 2*x*(6-x)*(2*k + 1)*(4*k + 3)/(x*(6 - x)*(4*k + 3) + 4*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k - 1)/(x*(6 - x)*(2*k - 1) + 2*(k + 1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 16 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(6 - x)*(2*k + 1)/(x*(6 - x)*(2*k + 1) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jul 17 2013
a(n)^2 = Sum_{k=0..n} 2^k * C(2*k, k)^2 * C(n+k, n-k) = A243949(n). - Paul D. Hanna, Aug 17 2014
a(n) = hypergeom([-n, -n], [1], 2). - Peter Luschny, Nov 19 2014
a(n) = Sum_{k=0..n/2} C(n-k,k) * 3^(n-2*k) * 2^k * C(n,k). - Vladimir Kruchinin, Jun 29 2015
a(n) = A049600(n, n-1).
a(n) = Sum_{0 <= j, k <= n} (-1)^(n+j)*C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). Cf. A126086 and A274668. - Peter Bala, Jan 15 2020
a(n) ~ c * (3 + 2*sqrt(2))^n / sqrt(n), where c = 1/sqrt(4*Pi*(3*sqrt(2)-4)) = 0.572681... (Banderier and Schwer, 2005). - Amiram Eldar, Jun 07 2020
a(n+1) = 3*a(n) + 2*Sum_{l=1..n} A006318(l)*a(n-l). [Eq. (1.16) in Qi-Shi-Guo (2016)]
a(n) ~ (1 + sqrt(2))^(2*n+1) / (2^(5/4) * sqrt(Pi*n)). - Vaclav Kotesovec, Jan 09 2023
a(n-1) + a(n) = A241023(n) for n >= 1. - Peter Bala, Sep 18 2024
a(n) = Sum_{k=0..n} C(n+k, 2*k) * C(2*k, k). - Greg Dresden and Leo Zhang, Jul 11 2025

New name and reference Sep 15 1995

Formula and more references from Don Knuth, May 15 1996

A052141 Number of paths from (0,0) to (n,n) that always move closer to (n,n) (and do not pass (n,n) and backtrack).

Original entry on oeis.org

1, 3, 26, 252, 2568, 26928, 287648, 3112896, 34013312, 374416128, 4145895936, 46127840256, 515268544512, 5775088103424, 64912164888576, 731420783788032, 8259345993203712, 93443504499523584, 1058972245409005568, 12019152955622817792, 136599995048232747008

Offset: 0

N. J. A. Sloane, Jan 23 2000

From Michel Marcus and Petros Hadjicostas, Jul 16 2020: (Start)
a(n) is the number of subdivisions of a 2 x n grid as defined in Robeva and Sun (2020). We have a(n) = A059576(n-1, n-1) for n >= 1 privided the latter is viewed as a square array (rather than a triangle).
In general, A059576(m-1, n-1) is the number of subdivisions of a 2-row grid with m points at the top row and n points at the bottom. (End)
The title condition is unclear: the path (0,0) -> (0,n) -> (n,n-1) -> (n,n) arguably meets the title condition but is not allowed, because steps with negative slope are proscribed. Steps must move east (slope 0) or have finite positive slope or move north (infinite slope). On the other hand, for lattice paths subject only to the condition that each successive point on the path is closer to the terminal point than its predecessor, see the question "Why are the numbers counting "ever-closer" lattice paths so round?" on the mathoverflow website. - David Callan, Nov 21 2021

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 6.3.9.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Elina Robeva and Melinda Sun, Bimonotone Subdivisions of Point Configurations in the Plane, arXiv:2007.00877 [math.CO], 2020. See A(2,n) column in Table 3 (p. 10).

Main diagonal of A059576.
Column k=2 of A316674.
Cf. A084773, A316673.

[n eq 0 select 1 else 2^(n-1)*Evaluate(LegendrePolynomial(n), 3) : n in [0..40]]; // G. C. Greubel, May 21 2023

a[0]=1; a[n_]:= Hypergeometric2F1[-n, n+1, 1, -1]*2^(n-1); Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Feb 23 2012, after Jon Stadler *)
Table[2^(n-1)*LegendreP[n,3] +Boole[n==0]/2, {n,0,40}] (* G. C. Greubel, May 21 2023 *)
CoefficientList[Series[(1+1/Sqrt[1-12x+4x^2])/2,{x,0,30}],x] (* Harvey P. Dale, Mar 10 2024 *)

def A052141(n): return 2^(n-1)*gen_legendre_P(n,0,3) + int(n==0)/2
[A052141(n) for n in range(41)] # G. C. Greubel, May 21 2023

G.f.: (1/2)*( 1 + 1/sqrt(1 - 12*x + 4*x^2) ).
a(n) = 2^(n-1) * A001850(n). - Jon Stadler (jstadler(AT)capital.edu), Apr 30 2003
D-finite with recurrence: n*a(n) = 6*(2*n-1)*a(n-1) - 4*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 08 2012
a(n) ~ sqrt(8+6*sqrt(2))*(6+4*sqrt(2))^n/(8*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 08 2012

A098270 a(n) = 2^n*P_n(5), 2^n times the Legendre polynomial of order n at 5.

Original entry on oeis.org

1, 10, 148, 2440, 42256, 752800, 13660480, 251113600, 4660568320, 87140108800, 1638884021248, 30970912737280, 587599919386624, 11185644310405120, 213540626285805568, 4086692369433395200, 78378887309200261120

Offset: 0

Paul Barry, Sep 01 2004

Central coefficients of (1 + 10*x + 24*x^2)^n. 2^n*LegendreP(n,k) yields the central coefficients of (1 + 2*k*x + (k^2-1)*x^2)^n, with g.f. 1/sqrt(1 - 4*k*x + 4*x^2).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Eric Weisstein's World of Mathematics, Legendre Polynomial.

Sequences of the form 2^n*LegendreP(n, 2*m+1): A000079 (m=0), A084773 (m=1), this sequence (m=2).

Cf. A006442, A059473, A069835.

[2^n*Evaluate(LegendrePolynomial(n), 5): n in [0..40]]; // G. C. Greubel, May 21 2023

Table[SeriesCoefficient[1/Sqrt[1-20*x+4*x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
Table[2^n*LegendreP[n, 5], {n,0,40}] (* G. C. Greubel, May 21 2023 *)

a(n)=pollegendre(n,5)<Charles R Greathouse IV, Oct 25 2011

def A098270(n): return 2^n*gen_legendre_P(n, 0, 5)
[A098270(n) for n in (0..16)] # Peter Luschny, Oct 14 2012

G.f.: 1/sqrt(1-20*x+4*x^2).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n,k)*binomial(2*(n-k), n)*5^(n-2*k).
D-finite with recurrence: n*a(n) +10*(1-2*n)*a(n-1) +4*(n-1)*a(n-2) = 0. - R. J. Mathar, Sep 26 2012
a(n) ~ sqrt(72+30*sqrt(6))*(10+4*sqrt(6))^n/(12*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = A059473(n,n). - Alois P. Heinz, Oct 05 2017
From Peter Bala, Nov 28 2021: (Start)
a(n) = (1/3)*Sum_{k >= n} binomial(k,n)^2*(2/3)^k.
a(n) = (4^n)*Sum_{k = 0..n} binomial(n,k)^2*(3/2)^k.
a(n) = (1/3)*(2/3)^n*hypergeometric2F1([n+1, n+1], [1], 2/3).
a(n) = (4^n)*hypergeometric2F1([-n, -n], [1], 3/2)
a(n) = [x^n] ((2*x - 2)*(3 - 2*x))^n.
a(n) = (2^n)*A006442(n). (End)

A059474 Triangle read by rows: T(n,k) is coefficient of z^nw^k in 1/(1 - 2z - 2w + 2z*w) read by rows in order 00, 10, 01, 20, 11, 02, ...

Original entry on oeis.org

1, 2, 2, 4, 6, 4, 8, 16, 16, 8, 16, 40, 52, 40, 16, 32, 96, 152, 152, 96, 32, 64, 224, 416, 504, 416, 224, 64, 128, 512, 1088, 1536, 1536, 1088, 512, 128, 256, 1152, 2752, 4416, 5136, 4416, 2752, 1152, 256, 512, 2560, 6784, 12160, 16032, 16032, 12160, 6784, 2560, 512

Offset: 0

N. J. A. Sloane, Feb 03 2001; revised Jun 12 2005

Pascal-like triangle: start with 1 at top; every subsequent entry is the sum of everything above you, plus 1.

			Triangle begins as:
   n\k [0]  [1]  [2]  [3]  [4]  [5]  [6] ...
  [0]   1;
  [1]   2,   2;
  [2]   4,   6,   4;
  [3]   8,  16,  16,   8;
  [4]  16,  40,  52,  40,  16;
  [5]  32,  96, 152, 152,  96,  32;
  [6]  64, 224, 416, 504, 416, 224,  64;
       ...

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000079, A001792, A007070, A057711, A077957, A084773.

See A059576 for a similar triangle.

A059474:= func< n,k | (&+[(-1)^j*2^(n-j)*Binomial(n-k,j)*Binomial(n-j,n-k): j in [0..n-k]]) >;
[A059474(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 21 2023

read transforms; SERIES2(1/(1-2*z-2*w+2*z*w),x,y,12): SERIES2TOLIST(%,x,y,12);
# Alternative
T := (n, k) -> 2^n*binomial(n, k)*hypergeom([-k, -n + k], [-n], 1/2):
for n from 0 to 10 do seq(simplify(T(n, k)), k = 0 .. n) end do; # Peter Luschny, Nov 26 2021

Table[(-1)^k*2^n*JacobiP[k, -n-1,0,0], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 04 2017; May 21 2023 *)

def A059474(n,k): return 2^n*binomial(n, k)*simplify(hypergeometric([-k, k-n], [-n], 1/2))
flatten([[A059474(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, May 21 2023

G.f.: 1/(1 - 2*z - 2*w + 2*z*w).
T(n, k) = Sum_{j=0..n} (-1)^j*2^(n + k - j)*C(n, j)*C(n + k - j, n).
T(n, 0) = T(n, n) = A000079(n).
T(2*n, n) = A084773(n).
T(n, k) = 2^n*binomial(n, k)*hypergeom([-k, k - n], [-n], 1/2). - Peter Luschny, Nov 26 2021
From G. C. Greubel, May 21 2023: (Start)
T(n, n-k) = T(n, k).
Sum_{k=0..n} T(n, k) = A007070(n).
Sum_{k=0..n} (-1)^k * T(n, k) = A077957(n).
T(n, 1) = A057711(n+1) = 2*A001792(n) - [n=0].
T(n, 2) = 4*A049611(n-1). (End)

A098269 a(n) = 2^n*P_n(4), 2^n times the Legendre polynomial of order n at 4.

Original entry on oeis.org

1, 8, 94, 1232, 16966, 240368, 3468844, 50712992, 748553926, 11131168688, 166498969924, 2502416381792, 37759888297756, 571681667171168, 8679980422677784, 132116085646644032, 2015249400937940806

Offset: 0

Paul Barry, Sep 01 2004

Central coefficients of (1+8x+15x^2)^n. 2^n*LegendreP(n,k) yields the central coefficients of (1+2kx+(k^2-1)x^2)^n, with g.f. 1/sqrt(1-4kx+4x^2).
16th binomial transform of 2^n*LegendreP(n,-4) = (-1)^n*A098269(n). - Paul Barry, Sep 03 2004
Diagonal of rational functions 1/(1 + x + 3*y + x*z - 2*x*y*z), 1/(1 - x + y + 3*x*z - 2*x*y*z), 1/(1 - x - x*y - 3*y*z - 2*x*y*z). - Gheorghe Coserea, Jul 03 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

Cf. A069835, A084773.

Table[SeriesCoefficient[1/Sqrt[1-16*x+4*x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
a[n_] := 3^n*HypergeometricPFQ[{-n, -n}, {1}, 5/3]; Flatten[Table[a[n], {n,0,16}]] (* Detlef Meya, May 21 2024 *)

a(n)=pollegendre(n,4)<Charles R Greathouse IV, Oct 24 2011

{a(n)=sum(k=0, n, binomial(n, k)^2*3^k*5^(n-k))} \\ Paul D. Hanna, Sep 29 2012

G.f.: 1/sqrt(1-16x+4x^2).
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*binomial(n, k)*binomial(2(n-k), n)*4^(n-2k).
E.g.f.: exp(8*x)*BesselI(0, 2*sqrt(15)*x), cf. A084770. - Vladeta Jovovic, Sep 01 2004
a(n) = Sum_{k=0..n} binomial(n,k)^2 * 3^k * 5^(n-k). - Paul D. Hanna, Sep 29 2012
D-finite with recurrence: n*a(n) = 8*(2*n-1)*a(n-1) - 4*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ sqrt(450+120*sqrt(15))*(8+2*sqrt(15))^n/(30*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = 3^n*hypergeom([-n, -n], [1], 5/3) = 5^n*hypergeom([-n, -n], [1], 3/5). - Detlef Meya, May 21 2024

A110124 A scaled Legendre triangle.

Original entry on oeis.org

1, 0, 1, -2, 2, 1, 0, 4, 4, 1, 6, 8, 22, 6, 1, 0, 16, 136, 52, 8, 1, -20, 32, 886, 504, 94, 10, 1, 0, 64, 5944, 5136, 1232, 148, 12, 1, 70, 128, 40636, 53856, 16966, 2440, 214, 14, 1, 0, 256, 281488, 575296, 240368, 42256, 4248, 292, 16, 1, -252, 512, 1968934, 6225792, 3468844, 752800, 88566, 6776, 382, 18, 1

Offset: 0

Paul Barry, Jul 13 2005

Row sums are A110125. Diagonal sums are A110126. Columns include A000079, A069835, A084773, and A098269.

			Rows begin
1;
0,1;
-2,2,1;
0,4,4,1;
6,8,22,6,1;
0,16,136,62,8,1;
-20,32,886,504,94,10,1;

T(n, k)=pollegendre(n-k, k)<<(n-k) \\ Charles R Greathouse IV, Mar 18 2017

Number triangle T(n, k)=2^(n-k)*LegendreP(n-k, k); T(n, k)=sum{j=0..floor((n-k)/2), (-1)^j*C(n-k, j)C(2n-2k-2j, n-k)k^(n-k-2j)}.

A098329 Expansion of 1/(1-2x-31x^2)^(1/2).

Original entry on oeis.org

1, 1, 17, 49, 481, 2081, 16241, 85457, 600769, 3489601, 23391697, 143000177, 938797729, 5897385313, 38397492017, 244866166289, 1590355308929, 10231490804353, 66456634775441, 429898281869489, 2795449543782241, 18150017431150241, 118194927388259057, 769438418283309649

Offset: 0

Paul Barry, Sep 03 2004

Central coefficient of (1+x+8x^2)^n. 7th binomial transform of 2^n*LegendreP(n,-3) (signed version of A084773).

Also number of paths from (0,0) to (n,0) using steps U=(1,1), H=(1,0) and D=(1,-1), the U steps can have 8 colors. - N-E. Fahssi, Mar 31 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A084603.

Table[SeriesCoefficient[1/Sqrt[1-2*x-31*x^2],{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
CoefficientList[Series[1/Sqrt[1-2x-31x^2],{x,0,30}],x] (* Harvey P. Dale, May 14 2017 *)
a[n_] := Hypergeometric2F1[1/2 - n/2, -n/2, 1, 32];
Table[a[n], {n, 0, 23}] (* Peter Luschny, Mar 18 2018 *)

x='x+O('x^66); Vec(1/(1-2*x-31*x^2)^(1/2)) \\ Joerg Arndt, May 11 2013

a(n) = sum{k=0..floor(n/2), binomial(n-k, k)*binomial(n, k)*8^k}.
E.g.f.: exp(x)*BesselI(0, 4*sqrt(2)*x)
Recurrence: n*a(n) = (2*n-1)*a(n-1) + 31*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ sqrt(8+sqrt(2))*(1+4*sqrt(2))^n/(4*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = hypergeom([1/2 - n/2, -n/2], [1], 32). - Peter Luschny, Mar 18 2018

A100631 Triangle read by rows: T(n,k) = 2*(T(n-1,k-1) - T(n-2,k-1) + T(n-1,k)) for 0 < k < n, T(n,0) = T(n,n) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 8, 12, 8, 1, 1, 16, 32, 32, 16, 1, 1, 32, 80, 104, 80, 32, 1, 1, 64, 192, 304, 304, 192, 64, 1, 1, 128, 448, 832, 1008, 832, 448, 128, 1, 1, 256, 1024, 2176, 3072, 3072, 2176, 1024, 256, 1, 1, 512, 2304, 5504, 8832, 10272, 8832, 5504, 2304, 512, 1

Offset: 0

Reinhard Zumkeller, Dec 03 2004

From Petros Hadjicostas, Feb 09 2021: (Start)
The rectangular version (R(n,k): n,k >= 0) of this symmetric triangular array (T(n,k): 0 <= k <= n) is given by R(n,k) = T(n+k,k) for n,k >= 0. Conversely, T(n,k) = R(n-k, k) for 0 <= k <= n.
Note that [o.g.f of R](x,y) = [o.g.f. of T](x, y/x) and [o.g.f of T](x,y) = [o.g.f of R](x,x*y). (End)
From Petros Hadjicostas, Feb 10 2021: (Start)
All the conjectures below are true because one has to prove only one of them, and the rest follow from the proved one.
As Peter Luschny pointed out, one has to show only that the function S(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1) satisfies the recurrence S(n,k) = 2*(S(n,k-1) - S(n-1,k-1) + S(n-1,k)) for n, k > 0 and the initial conditions S(n,0) = S(0,n) = 1 for n >= 0.
This is quite easy to achieve because S(n,k) = 2^n*Sum_{s=0}^{k-1} binomial(k-1,s)*binomial(n+s-1,s) for n >= 0 and k >= 1. The proof of the recurrence relies on the identity binomial(m,n) = binomial(m-1, n) + binomial(m-1,n-1).
Note that without the 2^n in the formula R(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1), we essentially get array A049600.
In addition, note that without the 2^(n-k-1) in the formula T(n,k+1) = 2^(n-k-1)*hypergeom([-k, n-k+1], [1], -1), we essentially get A208341 (without the first column and the main diagonal of T). (End)

			From _Petros Hadjicostas_, Feb 09 2021: (Start)
Triangle T(n,k) (with rows n >= 0 and columns 0 <= k <= n) begins:
  1,
  1,   1,
  1,   2,    1,
  1,   4,    4,    1,
  1,   8,   12,    8,    1,
  1,  16,   32,   32,   16,    1,
  1,  32,   80,  104,   80,   32,    1,
  1,  64,  192,  304,  304,  192,   64,    1,
  1, 128,  448,  832, 1008,  832,  448,  128,   1,
  1, 256, 1024, 2176, 3072, 3072, 2176, 1024, 256, 1,
  ...
Rectangular array R(n,k) (with rows n >= 0 and columns k >= 0) begins:
  1,   1,    1,    1,     1,     1,      1,       1, ...
  1,   2,    4,    8,    16,    32,     64,     128, ...
  1,   4,   12,   32,    80,   192,    448,    1024, ...
  1,   8,   32,  104,   304,   832,   2176,    5504, ...
  1,  16,   80,  304,  1008,  3072,   8832,   24320, ...
  1,  32,  192,  832,  3072, 10272,  32064,   95104, ...
  1,  64,  448, 2176,  8832, 32064, 107712,  341504, ...
  1, 128, 1024, 5504, 24320, 95104, 341504, 1150592, ...
  ... (End)

Index entries for triangles and arrays related to Pascal's triangle

Cf. A000079, A001787, A049600, A084773, A087161, A100312, A152254, A208341, A341344.

T[, 0] = T[n, n_] = 1;
T[n_, k_] /; 0, ] = 0;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 03 2021 *)

From Petros Hadjicostas, Feb 09 2021: (Start)
Formulas for the triangular array (T(n,k): 0 <= k <= n):
T(n,k) = T(n,n-k) for 0 <= k <= n.
Sum_{k=0..n} T(n,k) = A087161(n+1).
T(n,1) = T(n,n-1) = 2^(n-1) = A000079(n-1) for n >= 1.
T(n,2) = T(n,n-2) = (n-1)*2^(n-2) = A001787(n-1) for n >= 2.
T(n,3) = T(n,n-3) = (n^2-n-4)*2^(n-4) = A100312(n-3) for n >= 3.
T(n,floor(n/2)) = T(n,ceiling(n/2)) = A341344(n).
Bivariate o.g.f.: Sum_{n,k >= 0} T(n,k)*x^n*y^k = (3*x^2*y - 2*x*y - 2*x + 1)/((1 - x)*(-x*y + 1)*(2*x^2*y - 2*x*y - 2*x + 1)).
Conjecture based on Peter Luschny's formulas in other sequences: T(n,k) = 2^(n-k)*hypergeom([-k + 1, n-k], [1], -1) = 2^k*hypergeom([-(n-k) + 1, k], [1], -1).
Formulas for the rectangular array (R(n,k): n,k >= 0):
R(n,k) = 2*(R(n,k-1) - R(n-1,k-1) + R(n-1,k)) for n,k > 0 with R(n,0) = R(0,n) = 1 for n >= 0.
R(n,k) = R(k, n) for n,k >= 0.
R(1,n) = R(n,1) = 2^n = A000079(n).
R(2,n) = R(n,2) = (n+1)*2^n = A001787(n+1).
R(3,n) = R(n,3) = (n^2+5*n+2)*2^(n-1) = A100312(n).
R(n,n) = A152254(n-1) = 2*A084773(n-1) for n >= 1.
Bivariate o.g.f.: Sum_{n,k >= 0} R(n,k)*x^n*y^k = (3*x*y - 2*x - 2*y - 1)/((1 - x)*(1 - y)*(2*x*y - 2*x - 2*y - 1)).
Conjecture based on Peter Luschny's formulas in other sequences: R(n,k) = 2^n*hypergeom([-k + 1, n], [1], -1) = 2^k*hypergeom([-n + 1, k], [1], -1). (End)
From Petros Hadjicostas, Feb 10 2021: (Start)
The above conjecture is true (see the comments).
R(n,k) = 2^k*Sum_{s=0}^{n-1} binomial(n-1,s)*binomial(k+s-1,s) = 2^n*Sum_{s=0}^{k-1} binomial(k-1,s)*binomial(n+s-1,s) for n, k >= 1.
To get two binomial formulas for T(n,k), use the equation T(n,k) = R(n-k, k) for 1 <= k <= n and the above formulas for R(n,k). (End)

Offset changed by Petros Hadjicostas, Feb 09 2021

A341344 a(n) = A100631(n, floor(n/2)).

Original entry on oeis.org

1, 1, 2, 4, 12, 32, 104, 304, 1008, 3072, 10272, 32064, 107712, 341504, 1150592, 3688192, 12451584, 40239104, 136053248, 442442752, 1497664512, 4894728192, 16583583744, 54419632128, 184511361024, 607524225024, 2061074178048, 6805625192448, 23100352413696, 76462341095424, 259648659554304

Offset: 0

Petros Hadjicostas, Feb 09 2021

nonn

"Middle" diagonal of Reinhard Zumkeller's symmetric (Pascal-like) triangular array A100631.

Cf. A084773, A100631, A152254.

a(n) = {my(m=matrix(n+1, n+1)); for (i=1, n+1, for (j=1, n+1, if ((j==1) || (j==i), m[i, j] = 1, if (j<=n, m[i,j] = 2*(if (i>1, m[i-1,j-1] + m[i-1,j], 0) - if (i>2, m[i-2,j-1], 0) ))););); m[n+1, (n+2)\2];} \\ Michel Marcus, Feb 10 2021

a(n) = A100631(n, floor(n/2)) = A100631(n, ceiling(n/2)).
a(2*n) = A152254(n-1) = 2*A084773(n-1) for n >= 1.
a(n) = 2^ceiling(n/2)*hypergeom([-floor(n/2) + 1, ceiling(n/2)], [1], -1); see the comments for A100631. - Petros Hadjicostas, Feb 10 2021

cp's OEIS Frontend

A152254 Twice A084773.

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A001850 Central Delannoy numbers: a(n) = Sum_{k=0..n} C(n,k)*C(n+k,k).

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A052141 Number of paths from (0,0) to (n,n) that always move closer to (n,n) (and do not pass (n,n) and backtrack).

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A098270 a(n) = 2^n*P_n(5), 2^n times the Legendre polynomial of order n at 5.

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A059474 Triangle read by rows: T(n,k) is coefficient of z^n*w^k in 1/(1 - 2*z - 2*w + 2*z*w) read by rows in order 00, 10, 01, 20, 11, 02, ...

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A098269 a(n) = 2^n*P_n(4), 2^n times the Legendre polynomial of order n at 4.

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A059474 Triangle read by rows: T(n,k) is coefficient of z^nw^k in 1/(1 - 2z - 2w + 2z*w) read by rows in order 00, 10, 01, 20, 11, 02, ...