A027307 -id:A027307 - cp's OEIS frontend

A333715 a(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n.

1, 2, 24, 326, 4672, 69002, 1038984, 15856206, 244396544, 3795731282, 59307908024, 931222155030, 14680871849152, 232236016459098, 3684420837693480, 58600075142247326, 934064636705476608, 14917333936933664674, 238641621366613695576, 3823510794994321546214, 61344017874989324388672

Offset: 0

Peter Bala, Apr 03 2020

a(n) is also equal to [x^n] G(x)^n, where G(x) is the o.g.f. of A027307.
Compare with A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n and A103885(n) = [x^(2*n)] ( (1 + x)/(1 - x) )^n = [x^n] S(x)^n, where S(x) is the o.g.f. of the large Schröder numbers A006318.
If we define an operator I acting on power series f(x) = 1 + f_1*x + f_2*x^2 + ... by I(f(x)) = 1/x * Revert( x/f(x) ) then S(x) = I( (1 + x)/(1 - x) ) and G(x) = (I o I)( (1 + x)/(1 - x )).
It can be shown that a(n) satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that a(n) satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples of these congruences are given below.
The same congruences may hold more generally for the sequences a(r,s,n) := [x^(r*n)]( (1 + x)/(1 - x) )^(s*n), r a positive integer and s an integer. This is the case a(3,1,n).
For fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form P(6*m,n)*b(n+1) + P(6*m,-n)*b(n-1) = Q(3*m,n^2)*b(n), where the polynomials P(6*m,n) and Q(3*m,n^2) have degree 6*m. Conjecturally, the 6*m zeros of the polynomial Q(3*m,n^2) belong to the interval [-1, 1] and 6*m - 4 of these zeros appear to be approximated by the rational numbers +- k/(4*m), where 1 <= k <= 4*m - 3, k not a multiple of 4.

			Examples of congruences:
a(11) - a(1) = 931222155030 - 2 = (2^2)*(11^3)*163*1073069 == ( mod 11^3 )
a(3*7) - a(3) = 985413034951400888962602 - 326 = (2^2)*(7^4)*263* 390130947874776863 == 0 ( mod 7^3 )
a(5^2) - a(5) = 66292579025690123511768694002 - 69002 = (2^3)*(5^6)*39461* 13439614612035199009 == 0 ( mod 5^6 )

Seiichi Manyama, Table of n, a(n) for n = 0..824

Cf. A002003, A027307, A103885, A144097.

seq(add(binomial(n,k)*binomial(3*n+k-1,n-1), k = 0..n), n = 0..20);

Table[Binomial[3*n-1, n-1] * Hypergeometric2F1[-n, 3*n, 2*n+1, -1], {n, 0, 20}] (* Vaclav Kotesovec, Apr 04 2020 *)

a(n) = Sum_{k = 0..n} C(n,k)*C(3*n+k-1,n-1).
a(n) = (1/3) * Sum_{k = 0..n} C(3*n,n-k)*C(3*n+k-1,k) for n >= 1.
a(n) = (1/3) * [x^n] ( (1 + x)/(1 - x) )^(3*n) for n >= 1.
a(n) = Sum_{k = 1..n} (2^k)*C(n,k)*C(3*n-1,k-1) for n >= 1.
P-recursive:
P(6,n)*a(n+1) + P(6,-n)*a(n-1) = Q(3,n^2)*a(n), where P(6,n) = (2*n-1)*(3*n+1)*(3*n+2)*(3*n+3)*(35*n^2 - 35*n + 6) and the polynomial Q(3,n) = 4*(7805*n^3 - 7132*n^2 + 1559*n - 72).
Congruences: a(p) == 2 ( mod p^3 ) for prime p >= 3.
a(n) ~ (223 + 70*sqrt(10))^n / (2^(3/4) * 5^(1/4) * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Apr 04 2020
exp( Sum_{n>=1} a(n)*x^n/n ) = B(x) where B(x) = 1 + x*(B(x)^3 + B(x)^4) is the g.f. of A144097. - Paul D. Hanna, May 31 2023

A363006 a(n) = 1/((d-1)n + 1)Sum_{i=0..n} binomial((d - 1)n+1, n-i) binomial((d-1)*n+i, i), with d = 6.

Original entry on oeis.org

1, 2, 22, 342, 6202, 122762, 2571326, 56031470, 1257199154, 28849835538, 673953255142, 15973925161030, 383186776643946, 9285457458463770, 226959074854361742, 5588974707042304222, 138529985051020001634, 3453373395317346136610, 86526667346028323084726, 2177844556015530807952438

Offset: 0

Michael De Vlieger, May 16 2023

nonn

See Yang-Jiang paper, related to large Schröder numbers, which correspond to the formula in the Name, instead with d=2.

Seiichi Manyama, Table of n, a(n) for n = 0..500
Lin Yang, Yu-Yuan Zhang, and Sheng-Liang Yang, The halves of Delannoy matrix and Chung-Feller properties of the m-Schröder paths, Linear Alg. Appl. (2024).
Sheng-liang Yang and Mei-yang Jiang, Pattern avoiding problems on the hybrid d-trees, J. Lanzhou Univ. Tech., (China, 2023) Vol. 49, No. 2, 144-150. See p. 145. (in Mandarin.)

Cf. A006318 (d=2), A027307 (d=3), A144097 (d=4), A260332 (d=5).

Cf. A217364, A363305, A364195, A364196.

With[{d = 6}, Table[(1/((d - 1) n + 1)) Sum[Binomial[(d - 1) n + 1, n - i] Binomial[(d - 1) n + i, i], {i, 0, n}], {n, 0, 12}] ]

a(n) = my(d=6); sum(i=0, n, binomial((d - 1)*n+1, n-i) * binomial((d-1)*n+i, i))/((d-1)*n + 1); \\ Michel Marcus, May 16 2023

G.f. satisfies A(x) = 1 + x * A(x)^5 * (1 + A(x)). - Seiichi Manyama, May 29 2023
From Seiichi Manyama, Aug 09 2023: (Start)
a(n) = (1/n) * Sum_{k=0..n-1} (-1)^k * 2^(n-k) * binomial(n,k) * binomial(6*n-k,n-1-k) for n > 0.
a(n) = (1/n) * Sum_{k=1..n} 2^k * binomial(n,k) * binomial(5*n,k-1) for n > 0.  (End)

A364394 G.f. satisfies A(x) = 1 + x/A(x)*(1 + 1/A(x)).

Original entry on oeis.org

1, 2, -6, 34, -238, 1858, -15510, 135490, -1223134, 11320066, -106830502, 1024144482, -9945711566, 97634828354, -967298498358, 9659274283650, -97119829841854, 982391779220482, -9990160542904134, 102074758837531810, -1047391288012377774, 10788532748880319298

Offset: 0

Seiichi Manyama, Jul 22 2023

sign

Cf. A112478, A364396, A364398.

Cf. A027307, A087197, A108424.

A364394 := proc(n)
    if n = 0 then
        1;
    else
    (-1)^(n-1)*add( binomial(n,k) * binomial(2*n+k-2,n-1),k=0..n)/n ;
    end if;
end proc:
seq(A364394(n),n=0..80); # R. J. Mathar, Jul 25 2023

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

G.f.: A(x) = 1/B(-x) where B(x) is the g.f. of A027307.
a(n) = (-1)^(n-1) * (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(2*n+k-2,n-1) = (-1)^(n-1) * A108424(n) for n > 0.
D-finite with recurrence n*(2*n-1)*a(n) +3*(6*n^2-10*n+3)*a(n-1) +(-46*n^2+227*n-279)*a(n-2) +2*(n-3)*(2*n-7)*a(n-3)=0. - R. J. Mathar, Jul 25 2023
a(n) ~ c*(-1)^(n-1)*4^n*2F1([-n, 2*n-1], [n], -1)*n^(-3/2), with c = 1/(4*sqrt(Pi)) = A087197/4. - Stefano Spezia, Oct 21 2023

A084075 Length of list created by n substitutions k -> Range( -abs(k+1), abs(k-1), 2) starting with {1}.

Original entry on oeis.org

1, 2, 5, 12, 33, 86, 249, 680, 2033, 5722, 17485, 50260, 156033, 455534, 1431281, 4228752, 13412193, 40003058, 127840085, 384232156, 1235575201, 3737280582, 12080678505, 36736735672, 119276490193, 364372758986, 1187542872989

Offset: 0

Wouter Meeussen, May 11 2003

nonn

			{1}, {-2,0}, {-1,1,3,-1,1}, {0,2,-2,0,-4,-2,0,2,0,2,-2,0}

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

Cf. A084076, A084077, A084078.

Cf. A027307, A215067, A034015 (even bisection).

I:=[1,2,5,12]; [n le 4 select I[n] else (6*(35*n^2-55*n-76)*Self(n-1) + (275*n^4-770*n^3-203*n^2+1736*n-912)*Self(n-2) -6*(5*n^2+5*n-28)*Self(n-3) + (n-4)*(n-2)*(25*n^2+5*n-48)*Self(n-4))/(n*(n+2)*(25*n^2-45*n-28)): n in [1..41]]; // G. C. Greubel, Nov 24 2022

Rest@CoefficientList[InverseSeries[Series[ (-1-6n-8n^2+(1+2n)^2 Sqrt[1+4n])/( 2(n+4n^2+4n^3)), {n, 0, 40}]], n]
Length/@Flatten/@NestList[ #/.k_Integer:>Range[-Abs[k+1], Abs[k-1], 2] &, {1}, 8]

# replace iterates lists as described in Example.
def replace(L):
    return [i for k in L for i in range(-abs(k + 1), 1 + abs(k - 1), 2)]
def a(n):
  L = [1]
  for k in range(n): L=replace(L)
  return len(L)
print([a(n) for n in range(12)]) # F. Chapoton, Nov 15 2024

@CachedFunction
def a(n): # a = A084075
    if n < 4: return (1, 2, 5, 12)[n]
    else: return (6*(35*n^2 +15*n -96)*a(n-1) +(275*n^4+330*n^3-863*n^2+120*n+126)*a(n-2) -6*(5*n^2+15*n-18)*a(n-3) +(n-3)*(n-1)*(25*n^2+55*n-18)*a(n-4))/((n+1)*(n+3)*(25*n^2+5*n-48))
[a(n) for n in range(41)] # G. C. Greubel, Nov 24 2022

G.f. is the series reversion of (-1 -6*x -8*x^2 + (1+2*x)^2 * sqrt(1+4*x))/(2*(x +4*x^2 +4*x^3)).
a(2*n) = A027307(n)/2, n >= 1.
a(n) = ( 6*(35*n^2 +15*n -96)*a(n-1) + (275*n^4 +330*n^3 -863*n^2 +120*n +126)*a(n-2) - 6*(5*n^2 +15*n -18)*a(n-3) + (n-3)*(n-1)*(25*n^2 +55*n -18)*a(n-4) )/((n+1)*(n+3)*(25*n^2 +5*n -48)), n >= 4. - G. C. Greubel, Nov 24 2022

A108424 Number of paths from (0,0) to (3n,0) that stay in the first quadrant, consist of steps u=(2,1), U=(1,2), or d=(1,-1) and do not touch the x-axis, except at the endpoints.

Original entry on oeis.org

2, 6, 34, 238, 1858, 15510, 135490, 1223134, 11320066, 106830502, 1024144482, 9945711566, 97634828354, 967298498358, 9659274283650, 97119829841854, 982391779220482, 9990160542904134, 102074758837531810, 1047391288012377774, 10788532748880319298

Offset: 1

Emeric Deutsch, Jun 03 2005

nonn

These are the large nu-Schröder numbers with nu=NE(NEE)^(n-1). - Matias von Bell, Jun 02 2021

			a(2) = 6 because we have uudd, uUddd, Ududd, UdUddd, Uuddd and UUdddd.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.
M. von Bell and M. Yip, Schröder combinatorics and nu-associahedra, arXiv:2006.09804 [math.CO], 2020.

Cf. A027307, A032349, A344553.

Cf. A006318 (d = 2, signed version at d = 0), A027307 (d = 3), A144097 (d = 4), A260332 (d = 5, conjecturally), A363006 (d = 6).

A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: G:=z*A+z*A^2: Gser:=series(G,z=0,28): seq(coeff(Gser,z^n),n=1..25);
a:=proc(n) if n=1 then 2 else (n*2^n*binomial(2*n,n)/((2*n-1)*(n+1)))*sum(binomial(n-1,j)^2/2^j/binomial(n+j+1,j),j=0..n-1) fi end: seq(a(n),n=1..19);
# Alternative:
a := n -> 2*binomial(3*n - 2, 2*n - 1)*hypergeom([2 - 2*n, 1 - n], [2 - 3*n], -1)/n:
seq(simplify(a(n)), n = 1..21); # Peter Luschny, Jun 14 2021

Table[(n*2^n*Binomial[2*n,n]/((2n-1)*(n+1))) * Sum[(Binomial[n-1,j])^2/ (2^j * Binomial[n+j+1,j]), {j,0,n-1}], {n,1,20}] (* Vaclav Kotesovec, Oct 17 2012 *)

a(n) = A027307(n-1) + A032349(n).
G.f.: z*A+z*A^2, where A=1+z*A^2+z*A^3 or, equivalently, A=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3.
a(n) = (n*2^n*C(2*n, n)/((2n-1)(n+1))) * Sum_{j=0..n-1} (C(n-1, j))^2 / (2^j*C(n+j+1,j)).
Recurrence: n*(2*n-1)*a(n) = 3*(6*n^2-10*n+3)*a(n-1) + (46*n^2-227*n+279)*a(n-2) + 2*(n-3)*(2*n-7)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ sqrt(30*sqrt(5) - 50)*((11 + 5*sqrt(5))/2)^n/(20*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
a(n) = Sum_{i=0..n} (2*n+i-2)!/((n-i)!*(n+i-1)!*i!), n>0. - Vladimir Kruchinin, Feb 16 2013
From Matias von Bell, Jun 02 2021: (Start)
a(n) = 2*Sum_{i>=0} (1/n)*binomial(2*n-2,i)*binomial(3*n-2-i,2*n-1).
a(n) = 2*A344553(n). (End)
a(n) = 2*binomial(3*n - 2, 2*n - 1)*hypergeom([2 - 2*n, 1 - n], [2 - 3*n], -1) / n. - Peter Luschny, Jun 14 2021
From Peter Bala, Jun 17 2023: (Start)
a(n) = (-1)^(n+1) * (1/((d-1)*n + 1))*Sum_{i = 0..n} binomial((d - 1)*n+1, n-i) * binomial((d-1)*n+i, i), with d = -1.
P-recursive: n*(2*n - 1)*(5*n - 8)*a(n) = (110*n^3 - 396*n^2 + 445*n - 150)*a(n-1) + (n - 2)*(2*n - 5)*(5*n - 3)*a(n-2) with a(1) = 2 and a(2) = 6.
The g.f. A(x) = 2*x + 6*x^2 + 34*x^3 + .... Then 1/(1 - A(x)) = 1 + 2*x + 10*x^2 + 66*x^3 + .. is the g.f. of A027307.
(1/x) * the series reversion of x*(1 - A(x)) = 1 + 2*x + 14*x^2 + 134*x^3 + ... is the g.f. of A144097.
(1/x) * the series reversion of x/(1 - A(x)) = 1 - 2*x - 2*x^2 - 6*x^3 - 22*x^4 - 90*x^5 - ... =  1 - x - x*S(x), where S(x) is the g.f. of A006318. (End)

A108426 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have k peaks of the form Ud.

Original entry on oeis.org

1, 1, 1, 3, 5, 2, 12, 28, 21, 5, 55, 165, 180, 84, 14, 273, 1001, 1430, 990, 330, 42, 1428, 6188, 10920, 10010, 5005, 1287, 132, 7752, 38760, 81396, 92820, 61880, 24024, 5005, 429, 43263, 245157, 596904, 813960, 678300, 352716, 111384, 19448, 1430, 246675

Offset: 0

Emeric Deutsch, Jun 03 2005

Row sums yield A027307.
T(n,n) = A000108(n) (the Catalan numbers).
T(n,0) = A001764(n) = binomial(3n,n)/(2n+1).
Number of Ud peaks in all paths from (0,0) to (3n,0) is given by A108427.

			Example T(2,1) = 5 because we have udUdd, uUddd, Uddud, Ududd and UUdddd.
Triangle begins:
1;
1,1;
3,5,2;
12,28,21,5;
...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A000108, A001764, A108425, A108427.

T:=(n,k)->binomial(n,k)*binomial(3*n-k,n-1)/n: print(1); for n from 1 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

Table[If[n == 0, 1, (1/n)*Binomial[n, k]*Binomial[3 n - k, n - 1]], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Nov 29 2017 *)

for(n=0,10, for(k=0,n, print1(if(n==0, 1, (1/n)*binomial(n,k) *binomial(3*n-k,n-1)), ", "))) \\ G. C. Greubel, Nov 29 2017

T(n,k) = (1/n)*binomial(n,k)*binomial(3*n-k,n-1).

G.f.: G = G(t,z) satisfies G=1+z(t+G)G^2.

A336534 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(kn+j+1,n)/(kn+j+1).

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 1, 2, 6, 2, 1, 2, 10, 22, 2, 1, 2, 14, 66, 90, 2, 1, 2, 18, 134, 498, 394, 2, 1, 2, 22, 226, 1482, 4066, 1806, 2, 1, 2, 26, 342, 3298, 17818, 34970, 8558, 2, 1, 2, 30, 482, 6202, 52450, 226214, 312066, 41586, 2, 1, 2, 34, 646, 10450, 122762, 881970, 2984206, 2862562, 206098, 2

Offset: 0

Seiichi Manyama, Jul 25 2020

			Square array begins:
  1,   1,    1,     1,     1,      1, ...
  2,   2,    2,     2,     2,      2, ...
  2,   6,   10,    14,    18,     22, ...
  2,  22,   66,   134,   226,    342, ...
  2,  90,  498,  1482,  3298,   6202, ...
  2, 394, 4066, 17818, 52450, 122762, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0-3 give A040000, A006318, A027307, A144097.
If Michael D. Weiner's conjecture on A260332 is correct, column 4 is A260332 for n > 0.
Main diagonal gives A336537.
Cf. A336521, A336574. A336575.

T[n_, k_] := Sum[Binomial[n, j] * Binomial[k*n+j+1, n]/(k*n+j+1), {j, 0, n}]; Table[T[k, n-k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 01 2021 *)

T(n, k) = sum(j=0, n, binomial(k*n+1, j)*binomial((k+1)*n-j, n-j))/(k*n+1);

G.f. A_k(x) of column k satisfies A_k(x) = 1 + x * A_k(x)^k * (1 + A_k(x)).
T(n,k) = (1/n) * Sum_{j=1..n} 2^j * binomial(n,j) * binomial(k*n,j-1) for n > 0.
T(n,k) = (1/(k*n+1)) * Sum_{j=0..n} binomial(k*n+1,j) * binomial((k+1)*n-j,n-j).
T(n,k) = binomial(1+k*n, n)*hypergeom([-n, 1+k*n], [2+(k-1)*n], -1)/(1 + k*n) for k > 0. - Stefano Spezia, Aug 09 2025

A363304 Expansion of g.f. A(x) satisfying A(x) = 1 + x*(A(x)^4 + A(x)^7).

Original entry on oeis.org

1, 2, 22, 350, 6538, 133658, 2895214, 65294502, 1516963346, 36056007602, 872615973766, 21430572885422, 532737957899290, 13379121740808266, 338941379999841758, 8651415618928816886, 222278432539991439906, 5743974149517874477922, 149192980850883703986166

Offset: 0

Paul D. Hanna, May 29 2023

nonn

			G.f.: A(x) = 1 + 2*x + 22*x^2 + 350*x^3 + 6538*x^4 + 133658*x^5 + 2895214*x^6 + 65294502*x^7 + 1516963346*x^8 + 36056007602*x^9 + ...
where A(x) = 1 + x*(A(x)^4 + A(x)^7).
RELATED SERIES.
A(x)^4 = 1 + 8*x + 112*x^2 + 1960*x^3 + 38528*x^4 + 813064*x^5 + 17998512*x^6 + 412364968*x^7 + ...
A(x)^7 = 1 + 14*x + 238*x^2 + 4578*x^3 + 95130*x^4 + 2082150*x^5 + 47295990*x^6 + 1104598378*x^7 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A027307, A363311, A363111.

{a(n) = sum(k=0, n, binomial(n, k)*binomial(4*n+3*k+1, n)/(4*n+3*k+1) )}
for(n=0, 20, print1(a(n), ", "))

G.f.: A(x) = Sum_{n>=0} a(n)*x^n may be defined by the following.
(1) A(x) = 1 + x*(A(x)^4 + A(x)^7).
(2) a(n) = Sum_{k=0..n} binomial(n, k)*binomial(4*n+3*k+1, n)/(4*n+3*k+1) for n >= 0.

A071949 Triangle read by rows of numbers of paths in a lattice satisfying certain conditions.

Original entry on oeis.org

1, 1, 2, 1, 4, 10, 1, 6, 24, 66, 1, 8, 42, 172, 498, 1, 10, 64, 326, 1360, 4066, 1, 12, 90, 536, 2706, 11444, 34970, 1, 14, 120, 810, 4672, 23526, 100520, 312066, 1, 16, 154, 1156, 7410, 42024, 211546, 911068, 2862562, 1, 18, 192, 1582, 11088, 69002, 387456, 1951494, 8457504, 26824386

Offset: 0

N. J. A. Sloane, Jun 15 2002

			Triangle begins:
  1;
  1, 2;
  1, 4, 10;
  1, 6, 24,  66;
  1, 8, 42, 172, 498;
  ...

D. Merlini, D. G. Rogers, R. Sprugnoli and M. C. Verri, On some alternative characterizations of Riordan arrays, Canad J. Math., 49 (1997), 301-320.

T(n, n)=A027307(n).

T := proc(n,k) if k>0 and k<=n then (n-k+1)*sum(2^(j+1)*binomial(k,j+1)*binomial(n+k,j),j=0..k-1)/k elif k=0 then 1 else 0 fi end: seq(seq(T(n,k),k=0..n),n=0..10);

T[_, 0] = 1;
T[n_, n_] := T[n, n] = T[n, n-1] + T[n+1, n-1];
T[n_, k_] /; 0 <= k < n := T[n, k] = T[n, k-1] + T[n+1, k-1] + T[n-1, k];
T[, ] = 0;
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 15 2019 *)

T(n, k) = (n-k+1)*(Sum_{j=0..k-1} (2^(j+1)*binomial(k, j+1)*binomial(n+k, j)))/k for 0n.

T(n,0) = 1, T(n,n) = T(n,n-1) + T(n+1,n-1), otherwise T(n,k) = T(n,k-1) + T(n+1,k-1) + T(n-1,k). [Gerald McGarvey, Oct 09 2008]

Edited by Emeric Deutsch, Mar 04 2004

A084078 Length of list created by n substitutions k -> Range[-abs(k+1), abs(k-1), 2] starting with {0}.

Original entry on oeis.org

1, 2, 4, 10, 24, 66, 172, 498, 1360, 4066, 11444, 34970, 100520, 312066, 911068, 2862562, 8457504, 26824386, 80006116, 255680170, 768464312, 2471150402, 7474561164, 24161357010, 73473471344, 238552980386, 728745517972

Offset: 0

Wouter Meeussen, May 11 2003

nonn

			{0}, {-1,1}, {0,2,-2,0}, {-1,1,-3,-1,1,-1,1,3,-1,1}

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

Cf. A027307, A084075.

I:=[1,2,4,10]; [n le 4 select I[n] else (6*(35*n^2-125*n+14)*Self(n-1) + (275*n^4 -1870*n^3 +3757*n^2 -1268*n -1806)*Self(n-2) -6*(5*n^2-5*n-28)*Self(n-3) + (n-5)*(n-3)*(25*n^2-45*n-28)*Self(n-4))/((n-1)*(n+1)*(25*n^2-95*n+42)): n in [1..41]]; // G. C. Greubel, Nov 24 2022

Join[{1}, 2*Rest@CoefficientList[InverseSeries[Series[(-1 -6*n -8*n^2 + (1+ 2*n)^2*Sqrt[1+4*n])/(2*(n +4*n^2 +4*n^3)), {n, 0, 40}]], n]]
Length/@ Flatten/@ NestList[# /. k_Integer :> Range[-Abs[k+1], Abs[k-1], 2] &, {0}, 12]

def replace(L): return [i for k in L for i in range(-abs(k + 1), 1 + abs(k - 1), 2)]
def aList(upto, L=[0]): return [1] + [len((L := replace(L))) for _ in range(upto)]
print(aList(12))  # Peter Luschny, Nov 16 2024

@CachedFunction
def a(n): # a = A084078
    if (n<4): return (1,2,4,10)[n]
    else: return (6*(35*n^2 -55*n -76)*a(n-1) +(275*n^4-770*n^3-203*n^2+1736*n-912)*a(n-2) -6*(5*n^2+5*n-28)*a(n-3) +(n-4)*(n-2)*(25*n^2+5*n-48)*a(n-4))/(n*(n+2)*(25*n^2-45*n-28))
[a(n) for n in range(41)] # G. C. Greubel, Nov 24 2022

a(2*n-1) = A027307(n), n >= 1.
a(n) = 2*A084075(n-1), n >= 1.
a(n) = ( 6*(35*n^2 -55*n -76)*a(n-1) + (275*n^4 -770*n^3 -203*n^2 +1736*n -912)*a(n-2) - 6*(5*n^2 +5*n -28)*a(n-3) + (n-4)*(n-2)*(25*n^2+5*n-48)*a(n-4) )/(n*(n+2)*(25*n^2 -45*n -28)), for n >= 4. - G. C. Greubel, Nov 24 2022
a(2*n) = A032349(n+1), n >= 0. - Alexander Burstein, Nov 19 2023

cp's OEIS Frontend

A333715 a(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n.

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A363006 a(n) = 1/((d-1)*n + 1)*Sum_{i=0..n} binomial((d - 1)*n+1, n-i) * binomial((d-1)*n+i, i), with d = 6.

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A364394 G.f. satisfies A(x) = 1 + x/A(x)*(1 + 1/A(x)).

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A084075 Length of list created by n substitutions k -> Range( -abs(k+1), abs(k-1), 2) starting with {1}.

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A108424 Number of paths from (0,0) to (3n,0) that stay in the first quadrant, consist of steps u=(2,1), U=(1,2), or d=(1,-1) and do not touch the x-axis, except at the endpoints.

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A108426 Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1),U=(1,2), or d=(1,-1) and have k peaks of the form Ud.

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A336534 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(k*n+j+1,n)/(k*n+j+1).

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A363006 a(n) = 1/((d-1)n + 1)Sum_{i=0..n} binomial((d - 1)n+1, n-i) binomial((d-1)*n+i, i), with d = 6.

A336534 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,k) = Sum_{j=0..n} binomial(n,j) * binomial(kn+j+1,n)/(kn+j+1).