A092392 -id:A092392 - cp's OEIS frontend

A293469 a(n) = Sum_{k=0..n} (2k-1)!!binomial(2*n-k, n).

1, 3, 12, 57, 330, 2436, 23226, 277389, 3966534, 65517210, 1220999208, 25279328958, 575024187192, 14247595540542, 381846383109030, 11004598454925405, 339324532631899110, 11146022446431209490, 388535338484934710040, 14324570939127320452350, 556887682690152668745660

Offset: 0

Ilya Gutkovskiy, Oct 09 2017

nonn

Robert Israel, Table of n, a(n) for n = 0..403
Index entries for sequences related to factorial numbers

Cf. A001147, A076795, A084262, A092392, A270447, A293468.

seq(add(doublefactorial(2*k-1)*binomial(2*n-k,n),k=0..n),n=0..40); # Robert Israel, Oct 09 2017

Table[Sum[(2 k - 1)!! Binomial[2 n - k, n], {k, 0, n}], {n, 0, 20}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) 1/(1 + ContinuedFractionK[-k x, 1, {k, 1, n}]), {x, 0, n}], {n, 0, 20}]
Table[SeriesCoefficient[(1/(1 - x)^(n + 1)) Sum[(2 k - 1)!! x^k, {k, 0, n}], {x, 0, n}], {n, 0, 20}]

a(n) = [x^n] 1/((1 - x)^(n+1)*(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - 6*x/(1 - ...)))))))), a continued fraction.
a(n) = Gamma(n+1/2)*hypergeom([1/2, 1, -n], [-2*n], 2)*4^n/(n!*sqrt(Pi)). - Robert Israel, Oct 09 2017
a(n) ~ 2^(n + 1/2) * n^n / exp(n - 1/2). - Vaclav Kotesovec, Oct 18 2017

A329096 Row sums of A329057.

Original entry on oeis.org

1, 2, 8, 47, 374, 3852, 49398, 762785, 13805702, 286796072, 6727496456, 175903776622, 5073226515772, 160000741383368, 5478160073933490, 202366832844684645, 8022796547785815878, 339769654607776375824, 15309183806159727889536, 731253261602981693567090, 36909816019024064633444820

Offset: 0

Stefano Spezia, Nov 04 2019

nonn

Cf. A007318, A329057, A092392, A000108.

Table[(Binomial[2n,n]Hypergeometric2F1[1,-n,-2n,1+n])/(1+n),{n,0,20}]

{a(n) = sum(k=0,n, binomial(2*n-k, n) * (n+1)^(k-1) )}
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Sep 12 2024

{a(n) = my(C = (1 - sqrt(1-4*x +x^2*O(x^n)))/2);
(1/(n+1)) * polcoef( C'/(1 - (n+1)*C), n)}
for(n=0,25,print1(a(n),", ")) \\ Paul D. Hanna, Sep 12 2024

a(n) = binomial(2*n, n)*2F1([1, -n], [-2*n], 1 + n)/(1 + n), where 2F1 is the hypergeometric function.
a(n) ~ exp(2) * n^(n-1). - Vaclav Kotesovec, Nov 04 2019
From Paul D. Hanna, Sep 12 2024: (Start)
a(n) = Sum_{k=0..n} binomial(2*n-k, n) * (n+1)^(k-1).
a(n) = (1/(n+1)) * [x^n] C(x)'/(1 - (n+1)*C(x)) for n >= 0 where C(x) = (1 - sqrt(1-4*x))/2 is the g.f. of the Catalan numbers (A000108). (End)

A113214 Riordan array (1+2x,x(1+x)).

Original entry on oeis.org

1, 2, 1, 0, 3, 1, 0, 2, 4, 1, 0, 0, 5, 5, 1, 0, 0, 2, 9, 6, 1, 0, 0, 0, 7, 14, 7, 1, 0, 0, 0, 2, 16, 20, 8, 1, 0, 0, 0, 0, 9, 30, 27, 9, 1, 0, 0, 0, 0, 2, 25, 50, 35, 10, 1, 0, 0, 0, 0, 0, 11, 55, 77, 44, 11, 1, 0, 0, 0, 0, 0, 2, 36, 105, 112, 54, 12, 1, 0, 0, 0, 0, 0, 0, 13, 91, 182, 156, 65, 13, 1

Offset: 0

Paul Barry, Oct 18 2005

Row sums are Lucas numbers A000204. Diagonal sums are A007307(n+1). Inverse is (-1)^(n-k)A092392(n,k). Product with Pascal triangle (1/(1-x),x/(1-x)) is A111125.

			Triangle begins
  1;
  2,  1;
  0,  3,  1;
  0,  2,  4,  1;
  0,  0,  5,  5,  1;
  0,  0,  2,  9,  6,  1;
  0,  0,  0,  7, 14,  7,  1;
  0,  0,  0,  2, 16, 20,  8,  1;
Row 4: (1 + x*c(-x))^5 = 1 + 5*x + 5*x^2 + O(x^5). - _Peter Bala_, Sep 10 2021

R. Tauraso, Edge cover time for regular graphs, JIS 11 (2008) 08.4.4.

Cf. A000108, A000204, A007307, A092392, A111125.

T(n, k) = C(k, n-k) + 2*C(k, n-k-1).
T(n, k) = Sum_{j = 0..n} (-1)^(n-j)*C(n, j)*C(j+k, 2*k)*(2*j+1)/(2*k+1).
From Peter Bala, Sep 10 2021: (Start)
T(n,k) = T(n-1,k-1) + T(n-2,k-1) with boundary conditions T(n,n) = 1, T(1,0) = 2 and T(n,k) = 0 for k < 0 or k > n.
The entries in row n, read in reverse order, are the coefficients in the n-th degree Taylor polynomial of (1 + x*c(-x))^(n+1) at x = 0, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. (End)

A117362 Riordan array (1-2x,x(1-x)).

Original entry on oeis.org

1, -2, 1, 0, -3, 1, 0, 2, -4, 1, 0, 0, 5, -5, 1, 0, 0, -2, 9, -6, 1, 0, 0, 0, -7, 14, -7, 1, 0, 0, 0, 2, -16, 20, -8, 1, 0, 0, 0, 0, 9, -30, 27, -9, 1, 0, 0, 0, 0, -2, 25, -50, 35, -10, 1, 0, 0, 0, 0, 0, -11, 55, -77, 44, -11, 1

Offset: 0

Paul Barry, Mar 10 2006

A signed version of A113214. Inverse of A092392. Row sums are A100051(n+1) with g.f. (1-2x)/(1-x+x^2). Diagonal sums are A117363.

			Triangle begins
1,
-2, 1,
0, -3, 1,
0, 2, -4, 1,
0, 0, 5, -5, 1,
0, 0, -2, 9, -6, 1,
0, 0, 0, -7, 14, -7, 1,
0, 0, 0, 2, -16, 20, -8, 1

Number triangle T(n,k)=(-1)^(n-k)(C(k,n-k)+2*C(k, n-k-1))

A171824 Triangle T(n,k)= binomial(n + k,n) + binomial(2*n-k,n) read by rows.

Original entry on oeis.org

2, 3, 3, 7, 6, 7, 21, 14, 14, 21, 71, 40, 30, 40, 71, 253, 132, 77, 77, 132, 253, 925, 469, 238, 168, 238, 469, 925, 3433, 1724, 828, 450, 450, 828, 1724, 3433, 12871, 6444, 3048, 1452, 990, 1452, 3048, 6444, 12871, 48621, 24320, 11495, 5225, 2717, 2717, 5225, 11495, 24320, 48621

Offset: 0

Roger L. Bagula, Dec 19 2009

			Triangle begins as:
       2;
       3,     3;
       7,     6,     7;
      21,    14,    14,    21;
      71,    40,    30,    40,   71;
     253,   132,    77,    77,  132,  253;
     925,   469,   238,   168,  238,  469, 925;
    3433,  1724,   828,   450,  450,  828, 1724,  3433;
   12871,  6444,  3048,  1452,  990, 1452, 3048,  6444, 12871;
   48621, 24320, 11495,  5225, 2717, 2717, 5225, 11495, 24320, 48621;
  184757, 92389, 43824, 19734, 9009, 6006, 9009, 19734, 43824, 92389, 184757;

G. C. Greubel, Table of n, a(n) for n = 0..1325

Row sums are A000984(n+1).

Cf. A001700, A007318, A054142, A085478.

T:= func< n,k | Binomial(n+k,n) + Binomial(2*n-k,n) >;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 29 2021

T[n_, k_] = Binomial[n+k, k] + Binomial[2*n-k, n-k];
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten

def T(n, k): return binomial(n+k,n) + binomial(2*n-k,n)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 29 2021

T(n,k) = A046899(n,k) + A092392(n,k).

Sum_{k=0..n} T(n,k) = binomial(2*n+2, n+1) = 2*A001700(n) = A000984(n+1). - G. C. Greubel, Apr 29 2021

Formula and row sums reference added by the Assoc. Editors of the OEIS, Feb 24 2010

A191528 Triangle read by rows: T(n,k) is the number of left factors of Dyck paths of length n that have k returns to the axis.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 2, 1, 6, 3, 1, 10, 6, 3, 1, 20, 10, 4, 1, 35, 20, 10, 4, 1, 70, 35, 15, 5, 1, 126, 70, 35, 15, 5, 1, 252, 126, 56, 21, 6, 1, 462, 252, 126, 56, 21, 6, 1, 924, 462, 210, 84, 28, 7, 1, 1716, 924, 462, 210, 84, 28, 7, 1, 3432, 1716, 792, 330, 120, 36, 8, 1, 6435, 3432, 1716, 792, 330, 120, 36, 8, 1

Offset: 0

Emeric Deutsch, Jun 06 2011

Row n contains 1+floor(n/2) entries.
Sum of entries in row n is binomial(n, floor(n/2)) =A001405(n).
T(n,0) = A001405(n-1).
Rows 0, 2, 4, ... form triangle A100100.
Rows 1, 3, 5, ... form triangle A092392.
Sum_{k>=0} k*T(n,k) = A037955(n).
From Roger Ford, Oct 16 2020: (Start)
This is an empirical observation. T(n,k) = the number of different semi-meander arch depth models with n+2 top arches and k+1 arches at depth 0.  T(3,1) = the number of different semi-meander arch depth models with 5 top arches and 2 arches at depth 0.
Example: The depth of a semi-meander arch is the number of covering arches directly above the arch.  The arch depth model is the number of arches at each depth starting at 0 for a specific semi-meander.  The following is the arch depth models for semi-meanders with 5 top arches.
            /\                                   /\
           //\\                                 /  \
          ///\\\             depth             //\  \             depth
         ////\\\\  /\       (0)(1)(2)(3)      ///\\/\\ /\        (0)(1)(2)
depth    0123      0  model= 2  1  1  1       012  1   0   model= 2  2  1
         /\
        //\\    /\         depth         /\   /\            depth
       ///\\\  //\\       (0)(1)(2)     //\\ //\\ /\       (0)(1)
depth  012     01   model= 2  2  1      01   01   0  model= 3  2
         /\
        /  \               depth
       //\/\\ /\ /\       (0)(1)
depth  01 1   0  0  model= 3  2
There are 5 more semi-meanders with 5 top arches.  They are reflections of the above semi-meanders over a center vertical line and they yield the same arch depth models as the semi-meanders above.
T(3,1) = 2    different models=  2 2 1 and 2 1 1 1;
T(3,2) = 1     model=  3 2    (End).

			T(6,2)=3 because we have U(D)U(D)UU, U(D)UUD(D), and UUD(D)U(D), where U=(1,1) and D=(1,-1) (the return steps to the axis are shown between parentheses).
Triangle starts:
   1:
   1;
   1, 1;
   2, 1;
   3, 2, 1;
   6, 3, 1;
  10, 6, 3, 1;

Indranil Ghosh, Rows 0..100, flattened

Cf. A001405, A037955, A092392, A100100.

T := proc (n, k) if k <= floor((1/2)*n) then binomial(n-k-1, ceil((1/2)*n)-1) else 0 end if end proc: for n from 0 to 16 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form

Flatten[Table[Binomial[n-k-1,Ceiling[(n/2)-1]],{n,0,16},{k,0,Floor[n/2]}]] (* Indranil Ghosh, Mar 05 2017 *)

tabf(nn) = if(n==0, print1(1,", "), {for (n=1, nn, for(k=0, floor(n/2), print1(binomial(n-k-1, ceil((n/2)-1)),", ");); print();); });
tabf(16); \\ Indranil Ghosh, Mar 05 2017

T(n,k) = binomial(n-k-1, ceiling(n/2)-1) if 0 <= k <= floor(n/2).

G.f.: G(t,z) = 1/((1-z*c)*(1-t*z^2*c)), where c = (1-sqrt(1-4*z^2))/(2*z^2) is the Catalan function with argument z^2.

A270489 Sum_{k=0..n} ((binomial(3k,k)binomial(2n-k,n))/(2k+1)).

Original entry on oeis.org

1, 3, 12, 54, 265, 1401, 7903, 47088, 293319, 1892440, 12548041, 84988566, 585314652, 4085026386, 28820064810, 205156454376, 1471492171068, 10622954509803, 77122189800121, 562684397212060, 4123449352097229, 30336562360256955

Offset: 0

Vladimir Kruchinin, Mar 18 2016

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A001764, A092392.

Table[Sum[Binomial[3*k,k]*Binomial[2*n-k,n]/(2*k+1), {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Mar 18 2016 *)

taylor((sqrt(2)*sin(asin((3^(3/2)*sqrt(1-sqrt(1-4*x)))/2^(3/2))/3)*
sqrt(1-sqrt(1-4*x)))/(sqrt(3)*(1-x/(1-(1-sqrt(1-4*x))/2)^2))/x,x,0,20);

for(n=0,25, print1(sum(k=0,n, binomial(3*k,k)*binomial(2*n-k,n)/(2*k+1)), ", ")) \\ G. C. Greubel, Jun 05 2017

G.f.: P(C(x))/(1-x/(1-C(x)))^2/x, where C(x)=(1-sqrt(1-4*x))/2, P(x)/x is g.f. of A001764.
Recurrence: 46*(n-2)*(n-1)*n*(2*n + 1)*(133*n - 239)*a(n) = 5*(n-2)*(n-1)*(38969*n^3 - 108996*n^2 + 78733*n - 18774)*a(n-1) - 4*(n-2)*(242858*n^4 - 1286417*n^3 + 2496793*n^2 - 2103937*n + 643755)*a(n-2) + 36*(3*n - 7)*(3*n - 5)*(6*n - 13)*(6*n - 11)*(133*n - 106)*a(n-3). - Vaclav Kotesovec, Mar 18 2016
a(n) ~ 3^(6*n + 7/2) / (19^(3/2) * sqrt(Pi) * 2^(2*n+2) * 23^(n - 1/2) * n^(3/2)). - Vaclav Kotesovec, Mar 18 2016

A270490 a(n) = Sum_{i=0..(n+1)/2} binomial(2i+1,i)binomial(2n-2i,n)/(2*i+1).

Original entry on oeis.org

1, 2, 7, 24, 87, 320, 1195, 4504, 17102, 65304, 250501, 964480, 3724996, 14424504, 55983091, 217702880, 848042197, 3308490496, 12924954514, 50553798696, 197948515868, 775853655760, 3043672637457, 11950142769664, 46954356540812

Offset: 0

Vladimir Kruchinin, Mar 18 2016

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A092392.

Table[Sum[Binomial[2*i+1, i]*Binomial[2*n-2*i, n]/(2*i+1), {i, 0, (n+1)/2}], {n,0,20}] (* Vaclav Kotesovec, Mar 18 2016 *)

a(n):=sum(binomial(2*i+1,i)*binomial(2*n-2*i,n)/(2*i+1),i,0,(n+1)/2);

for(n=0,25, print1(sum(k=0,n, binomial(2*k+1,k)*binomial(2*n-2*k,n)/(2*k+1)), ", ")) \\ G. C. Greubel, Jun 05 2017

G.f.: 1/x*C(C(x)^2)/(C(x)*(1-x/(1-C(x))^2)), where C(x)=(1-sqrt(1-4*x))/2.
a(n) ~ 2^(2*n+1)/sqrt(Pi*n) * (1 - Gamma(3/4)/(sqrt(Pi)*n^(1/4)) + 7*sqrt(2*Pi) / (16*n^(3/4)*Gamma(3/4))). - Vaclav Kotesovec, Mar 18 2016
Conjecture: 3*n*(n-2)*(n+2)*a(n) -4*(n+1)*(8*n^2-23*n+12)*a(n-1) +16*n *(3*n-4)*(2*n-5)*a(n-2) +8*(2*n-3)*(4*n-7)*a(n-3) -64*(2*n-5)*(n-3)*(2*n-3)*a(n-4)=0. - R. J. Mathar, Jun 07 2016

A270561 Binomial transform(2) of Motzkin numbers.

Original entry on oeis.org

1, 3, 11, 42, 164, 649, 2592, 10423, 42140, 171133, 697641, 2853587, 11707542, 48166629, 198677283, 821495226, 3404577572, 14140959469, 58859315929, 245493952745, 1025954717376, 4295887639272, 18021572480109, 75740267331717

Offset: 0

Vladimir Kruchinin, Mar 19 2016

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A001006, A092392.

Table[Sum[Sum[Binomial[i, 2 k] Binomial[2 k, k]/(k + 1), {k, 0, i}] Binomial[2 n - i, n - i], {i, 0, n}], {n, 0, 23}] (* or *)
nn = 23; m = CoefficientList[Series[(1 - x - (1 - 2 x - 3 x^2)^(1/2))/(2 x^2), {x, 0, nn}], x]; Table[Sum[Binomial[2 n - k, n] m[[k + 1]], {k, 0, n}], {n, 0, nn}] (* Michael De Vlieger, Mar 19 2016, latter after Jean-François Alcover at A001006 *)

A(x):=(1-sqrt(1-4*x))/2;
M(x) := ( 1 - x - (1-2*x-3*x^2)^(1/2) ) / (2*x^2);
makelist(coeff(taylor(M(A(x))*A(x)/(2*x-A(x)),x,0,10),x,n),n,0,10);

a(n):=sum((sum((binomial(i,2*k)*binomial(2*k,k))/(k+1),k,0,i))*binomial(2*n-i,n-i),i,0,n);

a(n) = sum(i=0, n, sum(k=0, i, binomial(i, 2*k) * binomial(2*k, k) / (k+1)) * binomial(2*n-i, n-i)); \\ Indranil Ghosh, Mar 04 2017

G.f.: M(A(x))*A(x)/(2*x-A(x)), where M(x) is g.f. of Motzkin numbers (A001006) and A(x)/x is the g.f. of Catalan numbers (A000108).
a(n) = Sum_{i=0..n}((Sum_{k=0..i}((binomial(i,2*k)*binomial(2*k,k))/(k+1)))* binomial(2*n-i,n-i)).
a(n) = Sum_{k=0,n} (T(n,k)*m(k)), where m(k) is Motzkin numbers (A001006), T(n,k) = binomial(2*n-k,n) (triangle A092392).
a(n) ~ 3^(2*n + 5/2) / (sqrt(Pi) * n^(3/2) * 2^(n + 1/2)). - Vaclav Kotesovec, Mar 19 2016
a(n) = [x^n] (1 - x - sqrt(1 - 2*x - 3*x^2))/(2*x^2*(1 - x)^(n+1)). - Ilya Gutkovskiy, Oct 30 2017

cp's OEIS Frontend

A293469 a(n) = Sum_{k=0..n} (2*k-1)!!*binomial(2*n-k, n).

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A329096 Row sums of A329057.

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A113214 Riordan array (1+2x,x(1+x)).

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A117362 Riordan array (1-2x,x(1-x)).

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A171824 Triangle T(n,k)= binomial(n + k,n) + binomial(2*n-k,n) read by rows.

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Magma

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A191528 Triangle read by rows: T(n,k) is the number of left factors of Dyck paths of length n that have k returns to the axis.

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A270489 Sum_{k=0..n} ((binomial(3*k,k)*binomial(2*n-k,n))/(2*k+1)).

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A270490 a(n) = Sum_{i=0..(n+1)/2} binomial(2*i+1,i)*binomial(2*n-2*i,n)/(2*i+1).

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A293469 a(n) = Sum_{k=0..n} (2k-1)!!binomial(2*n-k, n).

A270489 Sum_{k=0..n} ((binomial(3k,k)binomial(2n-k,n))/(2k+1)).

A270490 a(n) = Sum_{i=0..(n+1)/2} binomial(2i+1,i)binomial(2n-2i,n)/(2*i+1).