A114276 -id:A114276 - cp's OEIS frontend

1, 5, 19, 67, 232, 804, 2806, 9878, 35072, 125512, 452388, 1641028, 5986993, 21954973, 80884423, 299233543, 1111219333, 4140813373, 15478839553, 58028869153, 218123355523, 821908275547, 3104046382351, 11747506651599

Offset: 0

Emeric Deutsch, Nov 20 2005

nonn

Also number of Dyck paths of semilength n+4 having length of second ascent equal to three. Example: a(1)=5 because we have UD(UUU)DUDDD, UD(UUU)DDUDD, UD(UUU)DDDUD, UUD(UUU)DDDD and UUDD(UUU)DDD (second ascents shown between parentheses). Partial sums of A002057. Column 3 of A114276. a(n)=absolute value of A104496(n+3).

Also number of Dyck paths of semilength n+3 that do not start with a pyramid (a pyramid in a Dyck path is a factor of the form U^j D^j (j>0), starting at the x-axis; here U=(1,1) and D=(1,-1); this definition differs from the one in A091866). Equivalently, a(n)=A127156(n+3,0). Example: a(1)=5 because we have UUDUDDUD, UUDUDUDD, UUUDUDDD, UUDUUDDD and UUUDDUDD. - Emeric Deutsch, Feb 27 2007

			a(3)=5 because the total length of the second ascents in UD(U)DUD, UD(UU)DD, UUDD(U)D, UUD(U)DD and UUUDDD (shown between parentheses) is 5.

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Cf. A002057, A114276, A104496, A127156, A279557.

Cf. A014137 (n=1), A014138 (n=2), A001453 (n=3), this sequence (n=4), A143955 (n=5), A323224 (array).

a:=n->4*sum(binomial(2*j+3,j)/(j+4),j=0..n): seq(a(n),n=0..28);

Table[4*Sum[Binomial[2j+3,j]/(j+4),{j,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 19 2012 *)

from functools import cache
@cache
def B(n, k):
    if n <= 0 or k <= 0: return 0
    if n == k: return 1
    return B(n - 1, k) + B(n, k - 1)
def A114277(n): return B(n + 5, n + 1)
print([A114277(n) for n in range(24)]) # Peter Luschny, May 16 2022

a(n) = 4*Sum_{j=0..n} binomial(2*j+3, j)/(j+4).
G.f.: C^4/(1-z), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function.
a(n) = c(n+3) - (c(0) + c(1) + ... + c(n+2)), where c(k)=binomial(2k,k)/(k+1) is a Catalan number (A000108). - Emeric Deutsch, Feb 27 2007
D-finite with recurrence: n*(n+4)*a(n) = (5*n^2 + 14*n + 6)*a(n-1) - 2*(n+1)*(2*n+3)*a(n-2). - Vaclav Kotesovec, Oct 19 2012
a(n) ~ 2^(2*n+7)/(3*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 19 2012
a(n) = exp((2*i*Pi)/3)-4*binomial(2*n+5,n+1)*hypergeom([1,3+n,n+7/2],[n+2,n+6],4)/ (n+5). - Peter Luschny, Feb 26 2017
a(n-1) = Sum_{i+j+k+lA000108 Catalan number. - Yuchun Ji, Jan 10 2019

More terms from Emeric Deutsch, Feb 27 2007

cp's OEIS Frontend

A114277 Sum of the lengths of the second ascents in all Dyck paths of semilength n+2.

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