A288470 -id:A288470 - cp's OEIS frontend

A218045 Number of truth tables of bracketed formulas (case 3).

0, 0, 1, 2, 9, 46, 262, 1588, 10053, 65686, 439658, 2999116, 20774154, 145726348, 1033125004, 7390626280, 53281906861, 386732675046, 2823690230850, 20725376703324, 152833785130398, 1131770853856100, 8412813651862868

Offset: 0

N. J. A. Sloane, Oct 23 2012

nonn

Equals the self-convolution of A186997 (up to offset). - Paul D. Hanna, Jul 03 2023

			G.f. A(x) = x^2 + 2*x^3 + 9*x^4 + 46*x^5 + 262*x^6 + 1588*x^7 + 10053*x^8 + 65686*x^9 + 439658*x^10 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Volkan Yildiz, General combinatorical structure of truth tables of bracketed formulas connected by implication, arXiv:1205.5595 [math.CO], 2012.

Cf. A186997, A218182, A288470.

CoefficientList[Series[(2+2*Sqrt[1-8*x]-(1+Sqrt[1-8*x])*Sqrt[2+2*Sqrt[1-8*x]+8*x])/8, {x, 0, 20}], x] (* Vaclav Kotesovec, Nov 19 2014 after Yildiz *)
Flatten[{0,0,Table[Sum[(Sum[Binomial[k,2*k+i+2-n]*Binomial[k+i-1,i],{i,0,n-k-1}]*Binomial[2*n-2,k])/(n-1),{k,0,n-1}],{n,2,20}]}] (* Vaclav Kotesovec, Nov 19 2014 after Vladimir Kruchinin *)

a(n):=sum((sum(binomial(k,2*k+i-n)*binomial(k+i-1,i),i,0,n-k+1))*binomial(2*n+2,k),k,0,n+1)/(n+1); /* Vladimir Kruchinin, Nov 19 2014  */

x='x+O('x^50); concat([0,0], Vec((2+2*sqrt(1-8*x)-(1+sqrt(1-8*x))*sqrt(2 + 2*sqrt(1-8*x)+8*x))/8)) \\ G. C. Greubel, Apr 01 2017

Yildiz gives a g.f.: (2+2*sqrt(1-8*x)-(1+sqrt(1-8*x))*sqrt(2+2*sqrt(1-8*x)+8*x))/8.
a(n+1) = (Sum_{k = 0..n} (Sum_{i=0..n-k} (binomial(k, 2*k+i+1-n)*binomial(k+i-1, i)))*binomial(2*n,k))/n. - Vladimir Kruchinin, Nov 19 2014
G.f. G(x) = A(x)/x satisfies G(x) = x*((G(x)*(G(x)+1))/(1-G(x))+1)^2. - Vladimir Kruchinin, Nov 19 2014
a(n) ~ (2*sqrt(3)-3) * 2^(3*n-3) / (3 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Nov 19 2014
From Paul D. Hanna, Jul 03 2023: (Start)
G.f. A(x) = Series_Reversion( x*(1 + sqrt(1 - 4*x - 4*x^2)) / 2 )^2.
G.f. A(x) = exp( Sum_{n>=1} A288470(n) * x^n/n ), where A288470(n) = Sum_{k=0..n} binomial(n,k) * binomial(2*n,2*k). (End)

A155495 Triangle read by rows: t(n,m) = binomial(2n,2m) * binomial(n,m).

Original entry on oeis.org

1, 1, 1, 1, 12, 1, 1, 45, 45, 1, 1, 112, 420, 112, 1, 1, 225, 2100, 2100, 225, 1, 1, 396, 7425, 18480, 7425, 396, 1, 1, 637, 21021, 105105, 105105, 21021, 637, 1, 1, 960, 50960, 448448, 900900, 448448, 50960, 960, 1, 1, 1377, 110160, 1559376, 5513508, 5513508, 1559376, 110160, 1377, 1

Offset: 0

Roger L. Bagula, Jan 23 2009

T(n,k) equals (-1)^k times the coefficient of x^k in 3F2(-n,-n,-n+1/2;1,1/2;x); see Mathematica code below. - John M. Campbell, Oct 23 2011

			Table starts:
  1;
  1,    1;
  1,   12,      1;
  1,   45,     45,       1;
  1,  112,    420,     112,        1;
  1,  225,   2100,    2100,      225,        1;
  1,  396,   7425,   18480,     7425,      396,        1;
  1,  637,  21021,  105105,   105105,    21021,      637,       1;
  1,  960,  50960,  448448,   900900,   448448,    50960,     960,      1;
  1, 1377, 110160, 1559376,  5513508,  5513508,  1559376,  110160,   1377,    1;
  1, 1900, 218025, 4651200, 26453700, 46558512, 26453700, 4651200, 218025, 1900, 1;

Robert Israel, Table of n, a(n) for n = 0..10010(rows 0 to 140, flattened)

Cf. A155497, A155516, A288470 (row sums).

[Binomial(n, k)*Binomial(2*n, 2*k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 29 2021

seq(seq(binomial(2*n,2*m)*binomial(n,m), m=0..n),n=0..10); # Robert Israel, Jun 12 2017

T[n_, k_]:= Binomial[2*n,2*k]*Binomial[n,k]; Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten
Abs[Flatten[Table[CoefficientList[HypergeometricPFQ[{-n,-n,-n+1/2}, {1,1/2}, x], x], {n, 1, 20}]]] (* or *)
T[n_,k_]:= (-1)^k*Coefficient[HypergeometricPFQ[{-n,-n,-n+1/2}, {1,1/2}, x], x^k] (* John M. Campbell, Oct 23 2011 *)

flatten([[binomial(n, k)*binomial(2*n, 2*k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 29 2021

T(n, k) = binomial(n, k)*binomial(2*n, 2*k).

Sum_{k=0..n} T(n, k) = A288470(n).

A240688 Expansion of -(xsqrt(-4x^2-4x+1)-2x^2-3x) / ((x+1)sqrt(-4x^2-4x+1)+ 4x^3+8x^2+3*x-1).

Original entry on oeis.org

1, 1, 5, 19, 81, 351, 1553, 6959, 31489, 143551, 658305, 3033471, 14034177, 65147135, 303285505, 1415422719, 6620053505, 31021657087, 145613977601, 684537354239, 3222408929281, 15187861143551, 71663163121665

Offset: 0

Vladimir Kruchinin, Apr 10 2014

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)

Cf. A052709.

CoefficientList[Series[-(x Sqrt[-4 x^2 - 4 x + 1] - 2 x^2 - 3 x) / ((x + 1) Sqrt[-4 x^2 - 4 x + 1] + 4 x^3 + 8 x^2 + 3 x - 1), {x, 0, 25}], x] (* Vaclav Kotesovec, Apr 12 2014 *)

a(n):=sum((sum(binomial(k,n-k-i)*binomial(k+i-1,i),i,0,n-k))*binomial(n,k),k,0,n);

x='x+O('x^50); Vec(-(x*sqrt(-4*x^2-4*x+1)-2*x^2-3*x) / ((x+1)*sqrt(-4*x^2-4*x+1)+ 4*x^3+8*x^2+3*x-1)) \\ G. C. Greubel, Apr 05 2017

a(n) = Sum_{k=0..n} Sum_{i=0..(n-k)} binomial(k,n-k-i)*binomial(k+i-1,i)*binomial(n,k).
A(x) = x*D'(x)/D(x) where D(x)=(1-sqrt(1-4*x-4*x^2))/(2*(1+x)) is g.f. of A052709.
a(n) ~ 2^(n-1/4) * (1+sqrt(2))^(n-1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Apr 12 2014
a(n) = Sum_{i=0..n/2} binomial(n,i)*binomial(2*n-2*i-1,n-2*i). - Vladimir Kruchinin, Mar 10 2015
Conjecture: n*(n-1)*a(n) -(3*n-2)*(n-1)*a(n-1) +2*(-4*n^2+7*n-1)*a(n-2) -4*n*(n-2)*a(n-3)=0. - R. J. Mathar, Jun 14 2016
From Peter Bala, Feb 13 2022: (Start)
The o.g.f. A(x) satisfies the differential equation (8*x^4 + 20*x^3 + 14*x^2 + x - 1)*A(x)' + (8*x^3 + 12*x^2 + 6*x + 3)*A(x) - 2 = 0 with A(0) = 1.
n*a(n) = (n+2)*a(n-1) + (14*n-22)*a(n-2) + (20*n-48)*a(n-3) + (8*n-24)*a(n-4).
Mathar's conjectural third-order recurrence above can be verified using Maple's gfun:-rectodiffeq command.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)
From Peter Bala, Aug 30 2025: (Start)
a(n) = Sum_{0 <= i, j <= n/2} binomial(n, j)*binomial(n+i-1, i)*binomial(n, 2*i+2*j) (verified to satisfy Mathar's third-order recurrence using the MulZeil procedure in Doron Zeilberger's MultiZeilberger Maple package).
Equivalently, a(n) = [x^n] ( (1 + x + x^2 + x^3)/(1 - x^2) )^n. Hence the Gauss congruences hold as stated above. Cf. A288470.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 5 and positive integers n and k. (End)

A329816 Triangular array, read by rows: T(n,k) = [(xy)^k] (-1 + (1 + x + 1/x)(1 + y + 1/y))^n for -n <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 2, 8, 2, 1, 1, 6, 27, 24, 27, 6, 1, 1, 12, 70, 132, 216, 132, 70, 12, 1, 1, 20, 155, 480, 1070, 1200, 1070, 480, 155, 20, 1, 1, 30, 306, 1370, 4035, 6900, 8840, 6900, 4035, 1370, 306, 30, 1, 1, 42, 553, 3332, 12621, 29750, 51065, 58800, 51065, 29750, 12621, 3332, 553, 42, 1

Offset: 0

Seiichi Manyama, Nov 21 2019

Also the coefficient of (x/y)^k in the expansion of (-1 + (1 + x + 1/x)*(1 + y + 1/y))^n for -n <= k <= n.

T(n,k) is the number of n step walks a chess king can take from (0,0) to (k,k). For example, for n=3 starting from (0,0) there is 1 walk to (3,3), 6 walks to (2,2), 27 walks to (1,1), 24 walks to (0,0), 27 walks to (-1,-1), 6 walks to (-2,-2) and 1 walk to (-3,-3). - Martin Clever, May 27 2023

			-1 + (1 + x + 1/x)*(1 + y + 1/y) = x*y + 1/(x*y) + x/y + y/x + x + 1/x + y + 1/y. So T(1,-1) = 1, T(1,0) = 0, T(1,1) = 1.
Triangle begins:
                          1;
                    1,    0,    1;
              1,    2,    8,    2,   1;
         1,   6,   27,   24,   27,   6,   1;
    1,  12,  70,  132,  216,  132,  70,  12,  1;
1, 20, 155, 480, 1070, 1200, 1070, 480, 155, 20, 1;

Seiichi Manyama, Rows n = 0..50, flattened

T(n,0) gives A094061.
Row sums give A288470.
Cf. A260492, A329819, A329820.

{T(n, k) = polcoef(polcoef((-1+(1+x+1/x)*(1+y+1/y))^n, k), k)}

T(n,k) = T(n,-k).

A370245 Coefficient of x^n in the expansion of ( 1/(1-x)^2 * (1+x^2)^3 )^n.

Original entry on oeis.org

1, 2, 16, 110, 828, 6352, 49696, 393668, 3148316, 25362992, 205519616, 1673272702, 13677016932, 112165564656, 922490228032, 7605558361960, 62839438825244, 520180768020464, 4313251202569216, 35818392770702104, 297846498752214128, 2479748570715505472

Offset: 0

Seiichi Manyama, Feb 13 2024

nonn

Cf. A288470, A360242.

Cf. A369263.

a(n, s=2, t=3, u=2) = sum(k=0, n\s, binomial(t*n, k)*binomial((u+1)*n-s*k-1, n-s*k));

a(n) = Sum_{k=0..floor(n/2)} binomial(3*n,k) * binomial(3*n-2*k-1,n-2*k).

The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^2)^3 ). See A369263.

A370242 Coefficient of x^n in the expansion of ( 1/(1-x)^2 * (1+x^2) )^n.

Original entry on oeis.org

1, 2, 12, 74, 480, 3202, 21756, 149746, 1040640, 7285538, 51307212, 363057114, 2579270304, 18385404546, 131429288828, 941857237474, 6764184258560, 48671099313730, 350799656912652, 2532218940625642, 18303373070813280, 132462237913391362, 959699439413581692

Offset: 0

Seiichi Manyama, Feb 13 2024

nonn

Cf. A288470, A370245.

Cf. A240688, A369208.

a(n, s=2, t=1, u=2) = sum(k=0, n\s, binomial(t*n, k)*binomial((u+1)*n-s*k-1, n-s*k));

a(n) = Sum_{k=0..floor(n/2)} binomial(n,k) * binomial(3*n-2*k-1,n-2*k).

The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^2) ). See A369208.

cp's OEIS Frontend

A218045 Number of truth tables of bracketed formulas (case 3).

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A155495 Triangle read by rows: t(n,m) = binomial(2*n,2*m) * binomial(n,m).

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A240688 Expansion of -(x*sqrt(-4*x^2-4*x+1)-2*x^2-3*x) / ((x+1)*sqrt(-4*x^2-4*x+1)+ 4*x^3+8*x^2+3*x-1).

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A329816 Triangular array, read by rows: T(n,k) = [(x*y)^k] (-1 + (1 + x + 1/x)*(1 + y + 1/y))^n for -n <= k <= n.

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A370245 Coefficient of x^n in the expansion of ( 1/(1-x)^2 * (1+x^2)^3 )^n.

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A370242 Coefficient of x^n in the expansion of ( 1/(1-x)^2 * (1+x^2) )^n.

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A155495 Triangle read by rows: t(n,m) = binomial(2n,2m) * binomial(n,m).

A240688 Expansion of -(xsqrt(-4x^2-4x+1)-2x^2-3x) / ((x+1)sqrt(-4x^2-4x+1)+ 4x^3+8x^2+3*x-1).

A329816 Triangular array, read by rows: T(n,k) = [(xy)^k] (-1 + (1 + x + 1/x)(1 + y + 1/y))^n for -n <= k <= n.