A118384 -id:A118384 - cp's OEIS frontend

A190726 Central coefficients of Riordan matrix A118384.

1, 6, 62, 720, 8806, 110916, 1423796, 18520788, 243289670, 3220011684, 42872967012, 573608356272, 7705343534716, 103857425975400, 1403902871946000, 19024773303675420, 258372666772083270, 3515644245559211172, 47918193512409831380

Offset: 0

Emanuele Munarini, May 17 2011

nonn

This sequence gives the integer part of an integral approximation to log(2), thus bears strong similarity to A123178. Quality of rational approximants appears entirely sufficient to prove irrationality. - Bradley Klee, Jun 29 2018

			From _Bradley Klee_, Jul 16 2018: (Start)
I_2 = Integral_{t=0..1} ((1-t)^4*t^4)/(4*(1+t)^3)*dt = 62*log(2) - 1719/40 < 10^(-3).
I_3 = Integral_{t=0..1} - ((1-t)^6*t^6)/(8*(1+t)^4)*dt = 720*log(2) - 143731/288 < 10^(-5). (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..88
Wadim Zudilin, An essay on irrationality measures of pi and other logarithms, arXiv:math/0404523 [math.NT], 2004.

Cf. A118384, A123178.
Log(2) approximation rationals: A316911, A316912.
Cf. A123178.

Table[Sum[Binomial[2n,k]Binomial[2n,n-k]2^k,{k,0,n}],{n,0,100}]
RecurrenceTable[{2*(n-1)*(2*n-3)*(2*n-1)*(33*n-8)*a[n-2]+ 9*(2*n-1)*(693*n^3-1554*n^2+989*n-160)*a[n-1] -3*n*(3*n-2)*(3*n-1)*(33*n-41)*a[n]==0, a[0]==1,a[1]==6},a,{n,0,10}] (* Bradley Klee, Jun 29 2018 *)

makelist(sum(binomial(2*n,k)*binomial(2*n,n-k)*2^k,k,0,n),n,0,12);

a(n)=sum(k=0,n,binomial(2*n,k)*binomial(2*n,n-k)<Charles R Greathouse IV, Jun 29 2011

a(n) = T(2*n,n), where T(n,k) = A118384(n,k).
a(n) = Sum_{k=0..n} binomial(2*n, k)*binomial(2*n, n-k)*2^k.
a(n) = Sum_{k=0..n} binomial(2*n, k)*binomial(k, n-k)*2^(n-k)*3^(2*k-n).
From Bradley Klee, Jun 29 2018: (Start)
a(n)*log(2) - A316911(n)/A316912(n) = I_n = Integral_{t=0..1}(-1/2)^n/(1+t)*((1-t)^2*t^2/(1+t))^n*dt.
Lim_{n->oo} I_n = 0, therefore:
Lim_{n->oo} A316911(n)/A316912(n)/a(n) = log(2).
G.f. G(x) and derivatives G^(n)(x) = d^n/dx^n G(x) satisfy a Picard-Fuchs type differential equation, 0 = Sum_{m=0..5,n=0..3} M_{m,n} x^m G^(n)(x), with integer matrix: M = {{324,-54,0,0}, {-36,10842,-486,0}, {84,8352,14931,-243}, {0,756,19026,3024}, {0,0,672,5364}, {0,0,0,112}}.
2*(n-1)*(2*n-3)*(2*n-1)*(33*n-8)*a(n-2)+ 9*(2*n-1)*(693*n^3-1554*n^2+989*n-160)*a(n-1) -3*n*(3*n-2)*(3*n-1)*(33*n-41)*a(n)=0.
(End)

A190724 Row sums of Riordan matrix A118384.

Original entry on oeis.org

1, 4, 20, 106, 576, 3174, 17648, 98746, 555104, 3131854, 17720880, 100507554, 571179792, 3251459670, 18535914480, 105803208906, 604598535360, 3458315246238, 19799128470896, 113441876080306, 650450158678256, 3731985710892454, 21425304596140080

Offset: 0

Emanuele Munarini, May 17 2011

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A001850, A118384.

CoefficientList[Series[(1-7x+Sqrt[1-6x+x^2])/((2-12x)Sqrt[1-6x+x^2]),{x,0,100}],x]

x='x+O('x^50); Vec((1-7*x+sqrt(1-6*x+x^2))/((2-12*x)*sqrt(1-6*x+x^2))) \\ G. C. Greubel, Mar 26 2017

a(n) = (6^n+d(n)-sum(6^(k-1)*d(n-k),k=1..n))/2, where d(n) = central Delannoy number (A001850).
G.f.: (1-7*x+sqrt(1-6*x+x^2))/((2-12*x)*sqrt(1-6*x+x^2)).
Recurrence: (n^2+9*n+20)*a(n+5)-8*(3*n^2+23*n+44)*a(n+4)+2*(108*n^2+683*n+1089)*a(n+3)-2*(435*n^2+2159*n+2716)*a(n+2)+(1367*n^2+4917*n+4366)*a(n+1)-210*(n^2+3*n+2)*a(n)=0.
Conjecture: n*(2*n+3)*a(n) +2*(-12*n^2-15*n+22)*a(n-1) +(74*n^2+73*n-274)*a(n-2) -6*(2*n+5)*(n-2)*a(n-3)=0. - R. J. Mathar, Jul 24 2012
a(n) ~ (2+sqrt(2))/(2*sqrt(3*sqrt(2)-4)) * (3+2*sqrt(2))^n/sqrt(Pi*n). - Vaclav Kotesovec, Oct 20 2012

A190725 Diagonal sums of Riordan matrix A118384.

Original entry on oeis.org

1, 3, 14, 69, 355, 1872, 10037, 54459, 298138, 1643565, 9111191, 50739120, 283635481, 1590648819, 8945090870, 50423423685, 284831065723, 1611918320688, 9137141645645, 51869777201595, 294843392318146, 1677980087882013, 9559901907126959

Offset: 0

Emanuele Munarini, May 17 2011

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A000045, A001850, A118384.

CoefficientList[Series[(3+3x-Sqrt[1-6x+x^2])/(2(1+3x+x^2)Sqrt[1-6x+x^2]),{x,0,100}],x]

x='x+O('x^50); Vec((3+3*x-sqrt(1-6*x+x^2))/(2*(1+3*x+x^2)*sqrt(1-6*x+x^2))) \\ G. C. Greubel, Mar 26 2017

a(n) = (3*sum((-1)^k*F(2k+1)*d(n-k),k=0..n)-(-1)^n*F(2n+2))/2, where d(n) = central Delannoy number (A001850) and F(n) = Fibonacci number (A000045).
G.f.: (3+3*x-sqrt(1-6*x+x^2))/(2*(1+3*x+x^2)*sqrt(1-6*x+x^2)).
Recurrence: (n^2+11*n+30)*a(n+6)-(3*n^2+29*n+70)*a(n+5)-(17*n^2+177*n+458)*a(n+4)-34*(n+4)*a(n+3)+(17*n^2+95*n+130)*a(n+2)+(3*n^2+19*n+30)*a(n+1)-(n^2+5*n+6)*a(n)=0.
conjecture: n*(2*n-3)*a(n) +(-6*n^2+15*n-8)*a(n-1) +2*(-16*n^2+32*n-11)*a(n-2) +(-6*n^2+9*n-2)*a(n-3) +(2*n-1)*(n-2)*a(n-4) =0. - R. J. Mathar, Jul 24 2012
a(n) ~ sqrt(24+17*sqrt(2))*(3+2*sqrt(2))^n/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 24 2012

A200536 Triangle, read by rows of 2n+1 terms, where row n lists the coefficients in (1+3x+2*x^2)^n.

Original entry on oeis.org

1, 1, 3, 2, 1, 6, 13, 12, 4, 1, 9, 33, 63, 66, 36, 8, 1, 12, 62, 180, 321, 360, 248, 96, 16, 1, 15, 100, 390, 985, 1683, 1970, 1560, 800, 240, 32, 1, 18, 147, 720, 2355, 5418, 8989, 10836, 9420, 5760, 2352, 576, 64, 1, 21, 203, 1197, 4809, 13923, 29953, 48639, 59906, 55692, 38472, 19152, 6496, 1344, 128

Offset: 0

Paul D. Hanna, Nov 18 2011

			The triangle begins:
1;
1, 3, 2;
1, 6, 13, 12, 4;
1, 9, 33, 63, 66, 36, 8;
1, 12, 62, 180, 321, 360, 248, 96, 16;
1, 15, 100, 390, 985, 1683, 1970, 1560, 800, 240, 32;
1, 18, 147, 720, 2355, 5418, 8989, 10836, 9420, 5760, 2352, 576, 64;
1, 21, 203, 1197, 4809, 13923, 29953, 48639, 59906, 55692, 38472, 19152, 6496, 1344, 128;
1, 24, 268, 1848, 8806, 30744, 81340, 166344, 265729, 332688, 325360, 245952, 140896, 59136, 17152, 3072, 256; ...
where row n equals the coefficients in (1+x)^n*(1+2*x)^n.

Paul D. Hanna, Table of n, a(n) for n = 0..1295; Rows n = 0..35, flattened.
We-jin Woan, The Lagrange Inversion formula and divisibility properties, JIS 10 (2007) 07.7.8, Example 3.

Cf. A001850 (central Delannoy numbers), A006318, A190726; related triangle: A118384.

Cf. A200537.

{T(n,k)=polcoeff((1+3*x+2*x^2+x*O(x^k))^n,k)}
{for(n=0,10,for(k=0,2*n,print1(T(n,k),","));print(""))}

Central terms in rows form the central Delannoy numbers: T(n,n) = A001850(n).
T(2*n,n) = A190726(n).
T(n,n+1) = n*A006318(n), where A006318 form the large Schroeder numbers.

A110171 Triangle read by rows: T(n,k) (0 <= k <= n) is the number of Delannoy paths of length n that start with exactly k (0,1) steps (or, equivalently, with exactly k (1,0) steps).

Original entry on oeis.org

1, 2, 1, 8, 4, 1, 38, 18, 6, 1, 192, 88, 32, 8, 1, 1002, 450, 170, 50, 10, 1, 5336, 2364, 912, 292, 72, 12, 1, 28814, 12642, 4942, 1666, 462, 98, 14, 1, 157184, 68464, 27008, 9424, 2816, 688, 128, 16, 1, 864146, 374274, 148626, 53154, 16722, 4482, 978, 162, 18, 1

Offset: 0

Emeric Deutsch, Jul 14 2005

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).
Column k for k >= 1 has g.f. z^k*R^(k-1)*g*(1+z*R), where R = 1 + zR + zR^2 = (1 - z - sqrt(1-6z+z^2))/(2z) is the g.f. of the large Schroeder numbers (A006318) and g = 1/sqrt(1-6z+z^2) is the g.f. of the central Delannoy numbers (A001850).
Sum_{k=0..n} k*T(n,k) = A050151(n) (the partial sums of the central Delannoy numbers) = (1/2)*n*R(n), where R(n) = A006318(n) is the n-th large Schroeder number.
From Paul Barry, May 07 2009: (Start)
Riordan array ((1+x+sqrt(1-6x+x^2))/(2*sqrt(1-6x+x^2)), (1-x-sqrt(1-6x+x^2))/2).
Inverse of Riordan array ((1-2x-x^2)/(1-x^2), x(1-x)/(1+x)). (End)

			T(2,1)=4 because we have NED, NENE, NEEN and NDE.
Triangle starts:
    1;
    2,  1;
    8,  4,  1;
   38, 18,  6,  1;
  192, 88, 32,  8,  1;
From _Paul Barry_, May 07 2009: (Start)
Production matrix is
   2, 1,
   4, 2, 1,
   6, 2, 2, 1,
   8, 2, 2, 2, 1,
  10, 2, 2, 2, 2, 1,
  12, 2, 2, 2, 2, 2, 1,
  14, 2, 2, 2, 2, 2, 2, 1,
  16, 2, 2, 2, 2, 2, 2, 2, 1,
  18, 2, 2, 2, 2, 2, 2, 2, 2, 1 (End)

G. C. Greubel, Rows n = 0..100 of triangle, flattened
P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
Robert A. Sulanke, Objects Counted by the Central Delannoy Numbers, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.
W.-j. Woan, The Lagrange Inversion Formula and Divisibility Properties, JIS 10 (2007) 07.7.8, example 5.

Cf. A001850, A002003, A006318, A050146, A050151, A113139, A118384.

Q:=sqrt(1-6*z+z^2): G:=(1+z+Q)/Q/(2-t+t*z+t*Q): Gser:=simplify(series(G,z=0,13)): P[0]:=1: for n from 1 to 10 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 10 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields sequence in triangular form

T[n_, n_] = 1;
T[n_, k_] := Sum[Binomial[n, i] Binomial[2n-k-i-1, n-k-i], {i, 0, n}];
Table[T[n, k], {n, 0, 9}, {k, 0, n}]//Flatten (* Jean-François Alcover, Jun 13 2019 *)

A110171 = lambda n,k : binomial(n, k)*hypergeometric([k-n, n], [k+1], -1)
for n in (0..9): [round(A110171(n,k).n(100)) for k in (0..n)] # Peter Luschny, Sep 17 2014

T(n,0) = A002003(n) for n >= 1.
T(n,1) = A050146(n) for n >= 1.
Row sums are the central Delannoy numbers (A001850).
G.f.: (1+z+Q)/(Q(2-t+tz+tQ)), where Q=sqrt(1-6z+z^2).
T(n,k) = x^(n-k)*((1+x)/(1-x))^n. - Paul Barry, May 07 2009
T(n,k) = C(n, k)*hypergeometric([k-n, n], [k+1], -1). - Peter Luschny, Sep 17 2014
From Peter Bala, Jun 29 2015: (Start)
T(n,k) = Sum_{i = 0..n} binomial(n,i)*binomial(2*n-k-i-1,n-k-i).
Matrix product A118384 * A007318^(-1)
Riordan array has the form ( x*h'(x)/h(x), h(x) ) with h(x) = ( 1 - x - sqrt(1 - 6*x + x^2) )/2 and so belongs to the hitting time subgroup H of the Riordan group (see Peart and Woan). (End)
T(n,k) = P(n-k, k, -1, 3), where P(n, alpha, beta, x) is the n-th Jacobi polynomial with parameters alpha and beta. Cf. A113139. - Peter Bala, Feb 16 2020

A376467 Triangular array read by rows: A063007 * A007318.

Original entry on oeis.org

1, 3, 2, 13, 18, 6, 63, 132, 90, 20, 321, 900, 930, 420, 70, 1683, 5910, 8190, 5600, 1890, 252, 8989, 37926, 65940, 60480, 30870, 8316, 924, 48639, 239624, 501228, 577080, 395010, 160776, 36036, 3432, 265729, 1497096, 3660300, 5072760, 4358970, 2378376, 804804, 154440, 12870, 1462563, 9274410, 25951860, 42060480, 43513470, 29801772, 13513500, 3912480, 656370, 48620

Offset: 0

Peter Bala, Sep 30 2024

Note that the n-th row generating polynomial of A063007 is equal to P(n,2*x + 1), where P(n,x) denotes the n-th Legendre polynomial.

The matrix product A063007 * A007318^(-1) is equal to a signed version of A063007 and A007318^(-1) * A063007 = A115951.

			Triangle begins
 n\k|     0       1       2       3       4       5      6     7
 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |     1
  1 |     3       2
  2 |    13      18       6
  3 |    63     132      90      20
  4 |   321     900     930     420      70
  5 |  1683    5910    8190    5600    1890     252
  6 |  8989   37926   65940   60480   30870    8316    924
  7 | 48639  239624  501228  577080  395010  160776  36036  3432
  ...

A000984 (main diagonal), A002457( (1/3)*first subdiagonal ), A001850 (Column 0), A002695 ( (1/2)*Column 1 ), A277660 ( (1/3)*Column 2 ), A006442 (row sums).

Cf. A063007, A007318, A115951, A118384.

A376467 := proc(n, k); add(binomial(n, j)*binomial(n+j, j)*binomial(j, k), j = k..n) end:
seq(print(seq(A376467(n, k) , k = 0..n)), n = 0..10);

T(n, k) = Sum_{j = k..n} binomial(n, j)*binomial(n+j, j)*binomial(j, k).
(n - k)*T(n, k) = 3*(2*n - 1)*T(n-1, k) - (n + k - 1)*T(n-2, k).
T(n, k) = (1/k!) * (d/dx)^k (P(n, 2*x+1)) evaluated at x = 1, where P(n,x) denotes the n-th Legendre polynomial.
G.f. for triangle: 1/sqrt(1 - 6*t + t^2 - 4*t*x) = 1 + (3 + 2*x)*t + (13 + 18*x + 6*x^2)*t^2 + ....
G.f. for column k: binomial(2*k, k) * x^k/(1 - 6*x + x^2)^(k+1/2).
T(n, k) is divisible by binomial(2*k, k) and the array ( T(n, k)/binomial(2*k, k) )n,k >= 0 is the Riordan array (1/sqrt(1 - 6*x + x^2), x/(1 - 6*x + x^2)).
T(n, k) is divisible by binomial(n+k, k) and the array ( T(n, k)/binomial(n+k, k) )n,k >= 0 is the Riordan array A118384.
T(n, n) = binomial(2*n, n); T(n, n-1) = 3*(2*n-1)!/(n-1)!^2 = 3 * A002457(n-1) for n >= 1.
The n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*(1 + x)^k = P(n, 2*x+3) = hypergeom([-n, n+1], [1], -1-x).
Recurrence: n*R(n, x) = (2*x + 3)*(2*n - 1)*R(n-1, x) - (n - 1)*R(n-2, x) with R(0, x) = 1.
If we set R(-1,x) = 1, we can run the recurrence backwards to give R(-n, x) = Sum_{k = 0..n} binomial(-n, k)*binomial(-n+k, k)*(1 + x)^k = R(n-1, x).
R(n, x) = (-1)^n * R(n, -x-3).
R(n, x) = 1/n! * (d/dx)^n( ((1 + x)*(2 + x))^n ).
R(1, x) = 3 + 2*x divides R(2*n+1, x) in the polynomial ring Z[x].

cp's OEIS Frontend

A190726 Central coefficients of Riordan matrix A118384.

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A190724 Row sums of Riordan matrix A118384.

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A190725 Diagonal sums of Riordan matrix A118384.

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A200536 Triangle, read by rows of 2*n+1 terms, where row n lists the coefficients in (1+3*x+2*x^2)^n.

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A110171 Triangle read by rows: T(n,k) (0 <= k <= n) is the number of Delannoy paths of length n that start with exactly k (0,1) steps (or, equivalently, with exactly k (1,0) steps).

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A376467 Triangular array read by rows: A063007 * A007318.

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A200536 Triangle, read by rows of 2n+1 terms, where row n lists the coefficients in (1+3x+2*x^2)^n.