A014300 -id:A014300 - cp's OEIS frontend

A172064 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=7.

1, 8, 46, 230, 1068, 4744, 20476, 86662, 361711, 1494384, 6126818, 24972326, 101320712, 409609664, 1651162688, 6640469816, 26655382802, 106830738224, 427612715516, 1709790470780, 6830461107736, 27266848437608

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 7th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

Apparently the number of peaks in all Dyck paths of semilength n+7 that are 5 steps higher than the preceding peak. - David Scambler, Apr 22 2013

			a(4) = C(15,4) - C(14,3) + C(13,2) - C(12,1) + C(11,0) = 7*13*15 - 14*13*2 + 78 - 12 + 1 = 1068.

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

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172065 (k=8), A172066 (k=9), A172067 (k=10).

k:=7; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
# 2nd program
for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^7, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=7; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=7; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j, n-j), with k=7.
a(n) ~ 2^(2*n+8)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+7)*(3*n+11)*a(n) -(21*n^3+212*n^2+719*n+840)*a(n-1) -2*(2*n+5)*(n+3)*(3*n+14)*a(n-2)=0. - R. J. Mathar, Feb 19 2016

A172065 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=8.

Original entry on oeis.org

1, 9, 56, 297, 1444, 6656, 29618, 128603, 548591, 2309467, 9624964, 39799813, 163556776, 668796712, 2723729944, 11055878188, 44753742226, 180746332690, 728571706240, 2932018571370, 11783070278816, 47297147250204

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 8th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

			a(4) = C(16,4) - C(15,3) + C(14,2) - C(13,1) + C(12,0) = 20*91 - 35*13 + 91 - 13 + 1 = 1820 - 455 + 79 = 1444.

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

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172066 (k=9), A172067 (k=10)

k:=8; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

a:= n-> add((-1)^(p)*binomial(2*n-p+8, n-p), p=0..n):
seq(a(n), n=0..40);
# 2nd program
a:= n-> coeff(series((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))
        /(2*z))^8, z, n+20), z, n):
seq(a(n), n=0..40);

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^8, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=8; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=8; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j *binomial(2*n+k-j, n-j), with k=8.
a(n) ~ 2^(2*n+9)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+8)*(3*n+13)*a(n) -(21*n^3 + 247*n^2 + 980*n + 1344)*a(n-1) - 2*(n+3)*(3*n+16)*(2*n+7)*a(n-2) = 0. - R. J. Mathar, Feb 29 2016

A172066 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=9.

Original entry on oeis.org

1, 10, 67, 376, 1912, 9142, 41941, 186880, 815083, 3498146, 14827487, 62236064, 259187048, 1072567256, 4415408372, 18098359424, 73915594466, 300958990724, 1222228100590, 4952609171080, 20030298812596, 80876902778482

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 9th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

			a(4) = C(17,4) - C(16,3) + C(15,2) - C(14,1) + C(13,0) = 17*4*5*7 - 16*5*7 + 105 - 14 + 1 = 5*7*(68-16) + 92 = 1912.

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

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172067 (k=10).

k:=9; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
# 2nd program
for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^9, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=9; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

k=9; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j,n-j), with k=9.
a(n) ~ 2^(2*n+10)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+9)*(n+5)*a(n) -(7*n^3+94*n^2+427*n+672)*a(n-1) -2*(2*n+7)*(n+6)*(n+4)*a(n-2)=0. - R. J. Mathar, Feb 19 2016

A172067 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=10.

Original entry on oeis.org

1, 11, 79, 468, 2486, 12323, 58277, 266492, 1188679, 5202523, 22436251, 95630272, 403770544, 1691678428, 7042481236, 29161852240, 120212658034, 493656394350, 2020590599710, 8247228533780, 33579755528278, 136434358356201

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 10th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

			a(4) = C(18,4) - C(17,3) + C(16,2) - C(15,1) + C(14,0) = 60*51 - 680 + 120 - 15 + 1 = 2486.

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

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), this sequence (k=10).

k:=10; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 27 2019

a:= n-> add((-1)^(p)*binomial(2*n-p+10, n-p), p=0..n):
seq(a(n), n=0..40);
# 2nd program
a:= n-> coeff(series((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))
        /(2*z))^10, z, n+20), z, n):
seq(a(n), n=0..40);

With[{k=10}, CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^k, {x, 0, 30}], x]] (* G. C. Greubel, Feb 27 2019 *)

k=10; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 27 2019

k=10; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 27 2019

a(n) = Sum_{j=0..n} (-1)^j*binomial(2*n+k-j,n-j), with k=10.

Conjecture: 2*n*(n+10)*(3*n+17)*a(n) - (21*n^3 + 317*n^2 + 1622*n + 2880)*a(n-1) - 2*(3*n+20)*(n+4)*(2*n+9)*a(n-2) = 0. - R. J. Mathar, Feb 21 2016

A333564 a(n) = [x^n] ( c(x)/c(-x) )^n, where c(x) = (1 - sqrt( 1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

2, 8, 56, 384, 2752, 20096, 148864, 1114112, 8403968, 63787008, 486584320, 3727196160, 28649455616, 220869853184, 1707123245056, 13223868760064, 102636144295936, 797982357192704, 6213784327684096, 48452953790480384, 378291752487878656, 2956824500391378944

Offset: 1

Peter Bala, Apr 07 2020

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. 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 >= 3 and positive integers n and k. Some examples are given below.

More generally, for integer r and positive integer s, we conjecture that the sequence a(r,s;n) := 2^(r*n) * Sum_{k = 0..s*n-1} (-1)^(s*n-1+k)*( binomial(n+k,k) )^r satisfies the same congruences. This is the case r = s = 1.

			Examples of congruences:
a(11) - a(1) = 486584320 - 2 = 2*(11^3)*182789 == 0 ( mod 11^3 ).
a(2*11) - a(2) = 2956824500391378944 - 8 = (2^3)*(3^2)*(11^3)*
107*288357478039 == 0 ( mod 11^3 ).
a(5^2) - a(5) = 1420245922851693002752 - 2752 = (2^6)*(3^3)*(5^6)* 7*19*1123*352183001 == 0 ( mod 5^6 ).

Cf. A000108, A014300, A119259, A333565, A349648.

seq( (2^n)*add( (-1)^(n-1+k)*binomial(n+k,k), k = 0..n-1), n = 1..25);
# alternative program
a := proc (n) option remember; `if`(n = 1, 2, `if`(n = 2, 8, `if`(n = 3, 56, ((3*n+4)*a(n-1)+(36*n-76)*a(n-2)+(32*n-80)*a(n-3))/n)))
end proc:
seq(a(n), n = 1..25);

a[n_] := -2^(n) Binomial[2n, n-1] Hypergeometric2F1[1, 2n +1, n + 2, 2];
Table[Simplify[a[n]], {n, 1, 22}] (* Peter Luschny, Apr 13 2020 *)
c[x_] := (1-Sqrt[1-4x])/(2x); ser[n_] := Series[(c[x]/c[-x])^n, {x, 0, 22}];
Table[SeriesCoefficient[ser[n], n], {n, 1, 22}] (* Peter Luschny, Apr 14 2020 *)

from itertools import count, islice
def A333564_gen(): # generator of terms
    yield (a:=2)
    c = 1
    for n in count(1):
        yield a<>1
A333564_list = list(islice(A333564_gen(),20)) # Chai Wah Wu, Apr 26 2023

a(n) = 2^n * Sum_{k = 0..n-1} (-1)^(n-1+k)*binomial(n+k,k) = 2^n * A014300(n).
a(n) = (-1)^(n+1) + Sum_{k = 0..n-1} n^2/((n-k)*(2*n-k))*C(n-k,k)*C(3*n-2*k-1,n-k).
Congruences: a(p) == 2 ( mod p^3 ) for all prime p >= 3, follows from previous formula.
a(n) = Sum_{k = 1..n} (-1)^(n+k)*(3*k-1)*2^(k-1)*A000108(k-1).
a(n) = (1/2)*(A119259(n) - (-1)^n).
a(n) ~ 8^n / (3*sqrt(Pi*n)).
P-recursive with recurrence n*(3*n - 4)*a(n) = (21*n^2 - 40*n + 12)*a(n-1) + 4*(3*n - 1)*(2*n - 3)*a(n-2) with a(1) = 2 and a(2) = 8. Cf. A333565.
Alternative form: (a(n) + a(n-1))/(a(n) - a(n-2)) = P(n)/Q(n), where P(n) = 4*(3*n - 1)*(2*n - 3) and Q(n) = (21*n^2 - 40*n + 12).
Also, n*a(n) = (3*n + 4)*a(n-1) + 4*(9*n - 19)*a(n-2) + 16*(2*n - 5)*a(n-3) with a(1) = 2, a(2) = 8 and a(3) = 56.
O.g.f.: A(x) = 4*x/(1 - 8*x + (1 + 4*x)*sqrt(1 - 8*x)), which satisfies the differential equation (x + 1)*(4*x + 1)*(8*x - 1)*A'(x) + (16*x^2 - 4*x + 7)*A(x) + 2*(1 - 2*x) = 0.

A167772 Riordan array (c(x)/(1+xc(x)), xc(x)), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 6, 8, 6, 3, 1, 18, 24, 18, 10, 4, 1, 57, 75, 57, 33, 15, 5, 1, 186, 243, 186, 111, 54, 21, 6, 1, 622, 808, 622, 379, 193, 82, 28, 7, 1, 2120, 2742, 2120, 1312, 690, 311, 118, 36, 8, 1, 7338, 9458, 7338, 4596, 2476, 1164, 474, 163, 45, 9, 1

Offset: 0

Philippe Deléham, Nov 11 2009, corrected Nov 12 2009

			Triangle begins:
     1;
     0,    1;
     1,    1,    1;
     2,    3,    2,    1;
     6,    8,    6,    3,   1;
    18,   24,   18,   10,   4,   1;
    57,   75,   57,   33,  15,   5,   1;
   186,  243,  186,  111,  54,  21,   6,  1;
   622,  808,  622,  379, 193,  82,  28,  7,  1;
  2120, 2742, 2120, 1312, 690, 311, 118, 36,  8,  1;
Production matrix begins:
  0, 1;
  1, 1, 1;
  1, 1, 1, 1;
  1, 1, 1, 1, 1;
  1, 1, 1, 1, 1, 1;
  1, 1, 1, 1, 1, 1, 1;
  1, 1, 1, 1, 1, 1, 1, 1;
  1, 1, 1, 1, 1, 1, 1, 1, 1;
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
  ... - _Philippe Deléham_, Mar 03 2013

Reinhard Zumkeller, Rows n=0..125 of triangle, flattened

Cf. A000957, A000958 (row sums), A001558, A001559, A104629, A167769.

Diagonals: A000012, A000217, A001477, A166830.

import Data.List (genericIndex)
a167772 n k = genericIndex (a167772_row n) k
a167772_row n = genericIndex a167772_tabl n
a167772_tabl = [1] : [0, 1] :
               map (\xs@(:x:) -> x : xs) (tail a065602_tabl)
-- Reinhard Zumkeller, May 15 2014

A065602[n_, k_]:= A065602[n,k]= Sum[(k-1+2*j)*Binomial[2*(n-j)-k-1, n-1]/(2*(n-j) -k-1), {j, 0, (n-k)/2}];
T[n_, k_]:= If[k==0, A065602[n+1,3] + Boole[n==0], A065602[n+1, k+1]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 26 2022 *)

def A065602(n,k): return sum( (k+2*j-1)*binomial(2*n-2*j-k-1, n-1)/(2*n-2*j-k-1) for j in (0..(n-k)//2) )
def A167772(n,k):
    if (k==0): return A065602(n+1,3) + bool(n==0)
    else: return A065602(n+1,k+1)
flatten([[A167772(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 26 2022

Sum_{k=0..n} T(n, k) = A000958(n+1).
From Philippe Deléham, Nov 12 2009: (Start)
Sum_{k=0..n} T(n,k)*2^k = A014300(n).
Sum_{k=0..n} T(n,k)*2^(n-k) = A064306(n). (End)
For n > 0: T(n,0) = A065602(n+1,3), T(n,k) = A065602(n+1,k+1), k = 1..n. - Reinhard Zumkeller, May 15 2014

A109244 A tree-node counting triangle.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 13, 7, 3, 1, 46, 24, 11, 4, 1, 166, 86, 40, 16, 5, 1, 610, 314, 148, 62, 22, 6, 1, 2269, 1163, 553, 239, 91, 29, 7, 1, 8518, 4352, 2083, 920, 367, 128, 37, 8, 1, 32206, 16414, 7896, 3544, 1461, 541, 174, 46, 9, 1, 122464, 62292, 30086, 13672, 5776, 2232

Offset: 0

Paul Barry, Jun 23 2005

Columns include A026641,A014300,A014301. Inverse matrix is A109246. Row sums are A014300. Diagonal sums are A109245.

			Rows begin:
  1;
  1,1;
  4,2,1;
  13,7,3,1;
  46,24,11,4,1;
  166,86,40,16,5,1;

G. C. Greubel, Rows n=0..100 of triangle, flattened

Flat(List([0..12], n-> List([0..n], k-> Sum([0..n], j-> (-1)^(n-j)*Binomial(n+j-k, j-k) )))); # G. C. Greubel, Feb 19 2019

[[(&+[(-1)^(n-j)*Binomial(n+j-k, j-k): j in [0..n]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 19 2019

Table[Sum[(-1)^(n-j)*Binomial[n+j-k, j-k], {j,0,n}], {n,0,12}, {k,0,n}] //Flatten  (* G. C. Greubel, Feb 19 2019 *)

{T(n,k) = sum(j=0,n, (-1)^(n-j)*binomial(n+j-k, j-k))};
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 19 2019

[[sum((-1)^(n-j)*binomial(n+j-k, j-k) for j in (0..n)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 19 2019

Number triangle T(n, k) = Sum_{i=0..n} (-1)^(n-i)*binomial(n+i-k, i-k).
Riordan array (1/(1-x*c(x)-2*x^2*c(x)^2), x*c(x)) where c(x)=g.f. of A000108.
The production matrix M (discarding the zeros) is:
  1, 1;
  3, 1, 1;
  3, 1, 1, 1;
  3, 1, 1, 1, 1;
... such that the n-th row of the triangle is the top row of M^n. - Gary W. Adamson, Feb 16 2012

A128730 Number of UDL's in all skew Dyck paths of semilength n.

Original entry on oeis.org

0, 0, 1, 4, 16, 68, 301, 1366, 6301, 29400, 138355, 655424, 3121438, 14930540, 71675839, 345148892, 1666432816, 8064278288, 39103576699, 189949958332, 924163714216, 4502711570988, 21966152501239, 107284324830302

Offset: 0

Emeric Deutsch, Mar 31 2007

nonn

A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of a path is defined to be the number of steps in it.

			a(3) = 4 because we have UDUUDL, UUUDLD, UUDUDL and UUUDLL (the other six skew Dyck paths of semilength 3 are the five Dyck paths and UUUDDL).

G. C. Greubel, Table of n, a(n) for n = 0..1000
E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203

Cf. A128728, A014300.

G:=2*z^2/(1-6*z+5*z^2+(1+z)*sqrt(1-6*z+5*z^2)): Gser:=series(G,z=0,30): seq(coeff(Gser,z,n),n=0..26);

CoefficientList[Series[2*x^2/(1-6*x+5*x^2+(1+x)*Sqrt[1-6*x+5*x^2]), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 20 2014 *)

z='z+O('z^50); concat([0,0], Vec(2*z^2/(1-6*z+5*z^2+(1+z)*sqrt(1-6*z+5*z^2)))) \\ G. C. Greubel, Mar 19 2017

a(n) = Sum_{k>=0} k*A128728(n,k).
G.f.: 2*z^2/(1-6*z+5*z^2+(1+z)*sqrt(1-6*z+5*z^2)).
a(n) ~ 5^(n-1/2)/(6*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 20 2014
D-finite with recurrence: +2*(n-1)*(3*n-8)*a(n) +(-39*n^2+161*n-148)*a(n-1) +(48*n^2-215*n+220)*a(n-2) -5*(3*n-5)*(n-3)*a(n-3)=0. - R. J. Mathar, Jun 17 2016
For n >= 2, a(n) = Sum_{k=1..n-1} binomial(n,k)*A014300(k). - Jianing Song, Apr 20 2019

A306409 a(n) = -Sum_{0<=i

Original entry on oeis.org

0, 1, 3, 10, 34, 120, 434, 1597, 5949, 22363, 84655, 322245, 1232205, 4729453, 18210279, 70307546, 272087770, 1055139408, 4099200524, 15951053566, 62159391150, 242542955378, 947504851414, 3705431067156, 14505084243860, 56831711106496, 222853334131080

Offset: 0

Seiichi Manyama, Apr 05 2019

nonn

			n | a(n) | A307354 | A006134 | A120305
--+------+---------+---------+---------
0 |    0 |       1 |       1 |       1
1 |    1 |       2 |       3 |       1
2 |    3 |       6 |       9 |       3
3 |   10 |      19 |      29 |       9
4 |   34 |      65 |      99 |      31
5 |  120 |     231 |     351 |     111

Seiichi Manyama, Table of n, a(n) for n = 0..1664

Partial sums of A014300. - Seiichi Manyama, Jan 30 2023

Cf. A000108, A000957, A006134, A057552, A120305, A307354.

Table[-Sum[Sum[(-1)^(i+j) * (i+j)!/(i!*j!), {i, 0, j-1}], {j, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Apr 05 2019 *)

a(n) = -sum(i=0, n, sum(j=i+1, n, (-1)^(i+j)*(i+j)!/(i!*j!)));

my(N=30, x='x+O('x^N)); concat(0, Vec((1-sqrt(1-4*x))/(sqrt(1-4*x)*(1-x)*(3-sqrt(1-4*x))))) \\ Seiichi Manyama, Jan 30 2023

a(n) = A006134(n) - A307354(n).
a(n) = (A006134(n) - A120305(n))/2.
a(n) ~ 4^(n+1) / (9*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 05 2019
G.f.: ( 1/(sqrt(1-4*x) * (1-x)) ) * ( x *c(x)/(1 + x *c(x)) ), where c(x) is the g.f. of A000108. - Seiichi Manyama, Jan 30 2023

A119307 Triangle read by rows: T(n, k) = Sum_{j=0..n} C(j, k)*C(j, n - k).

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 11, 11, 1, 1, 19, 46, 19, 1, 1, 29, 127, 127, 29, 1, 1, 41, 281, 517, 281, 41, 1, 1, 55, 541, 1579, 1579, 541, 55, 1, 1, 71, 946, 4001, 6376, 4001, 946, 71, 1, 1, 89, 1541, 8889, 20626, 20626, 8889, 1541, 89, 1, 1, 109, 2377, 17907, 56904, 82994

Offset: 0

Paul Barry, May 13 2006

			Triangle begins
  1,
  1, 1,
  1, 5, 1,
  1, 11, 11, 1,
  1, 19, 46, 19, 1,
  1, 29, 127, 127, 29, 1,
  1, 41, 281, 517, 281, 41, 1
  ...

Indranil Ghosh, Rows 0..100, flattened

Second column is A028387.
Row sums are A014300.
Central coefficients T(2*n, n) are A112029.

T := (n, k) -> if n = k then 1 else binomial(n, k)^2*hypergeom([1, -k, -n + k], [-n, -n], 1) fi: for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
# Peter Luschny, May 13 2024

Flatten[Table[Sum[Binomial[j,k] Binomial[j,n-k],{j,0,n}],{n,0,10},{k,0,n}]] (* Indranil Ghosh, Mar 03 2017 *)

tabl(nn)={for (n=0, nn, for(k=0, n, print1(sum(j=0, n, binomial(j,k)*binomial(j,n-k)),", ");); print(););};
tabl(10); \\ Indranil Ghosh, Mar 03 2017

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

T(n, k) = binomial(n, k)^2*hypergeom([1, -k, -n + k], [-n, -n], 1) for k=0..n-1. - Peter Luschny, May 13 2024

cp's OEIS Frontend

A172064 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=7.

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A172065 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=8.

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A172066 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=9.

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A172067 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=10.

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A333564 a(n) = [x^n] ( c(x)/c(-x) )^n, where c(x) = (1 - sqrt( 1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

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A167772 Riordan array (c(x)/(1+x*c(x)), x*c(x)), c(x) the g.f. of A000108.

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A172064 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=7.

A172065 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=8.

A172066 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=9.

A172067 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=10.

A333564 a(n) = [x^n] ( c(x)/c(-x) )^n, where c(x) = (1 - sqrt( 1 - 4x))/(2x) is the o.g.f. of the Catalan numbers A000108.

A167772 Riordan array (c(x)/(1+xc(x)), xc(x)), c(x) the g.f. of A000108.