A001764 -id:A001764 - cp's OEIS frontend

A188687 Partial binomial sums of binomial(3n,n)/(2n+1) = A001764(n).

1, 2, 6, 25, 126, 704, 4183, 25897, 165166, 1077520, 7156352, 48222354, 328859011, 2265428728, 15740837575, 110187356134, 776336572878, 5501042194580, 39177463572112, 280277949384146, 2013277273220064, 14514764553512488, 104993261648226446

Offset: 0

Emanuele Munarini, Apr 08 2011

Nathaniel Johnston, Table of n, a(n) for n = 0..400
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.

Cf. A005809, A001764, A188675, A188676, A104859, A188678, A188679, A188680, A188681, A188682, A188683, A188684, A188685, A188686.

Table[Sum[Binomial[n,k]Binomial[3k,k]/(2k+1),{k,0,n}],{n,0,22}]

makelist(sum(binomial(n,k)*binomial(3*k,k)/(2*k+1),k,0,n),n,0,20);

a(n) = Sum_{k=0..n} binomial(n,k)*binomial(3k,k)/(2k+1).
G.f.: (2/sqrt(3x*(1-x)))*sin((1/3)*arcsin(3/2*sqrt(3*x/(1-x)))).
Recurrence: 2*n*(2*n+1)*a(n) = (39*n^2-35*n+8)*a(n-1) - 2*(n-1)*(33*n-32)*a(n-2) + 31*(n-2)*(n-1)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 31^(n+3/2)/(3^4*2^(2*n+2)*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x) * A(x)^3. - Ilya Gutkovskiy, Jul 25 2021

A006629 Self-convolution 4th power of A001764, which enumerates ternary trees.

Original entry on oeis.org

1, 4, 18, 88, 455, 2448, 13566, 76912, 444015, 2601300, 15426840, 92431584, 558685348, 3402497504, 20858916870, 128618832864, 797168807855, 4963511449260, 31032552351570, 194743066471800, 1226232861415695

Offset: 0

Simon Plouffe, N. J. A. Sloane

Sum of root degrees of all noncrossing trees on nodes on a circle. - Emeric Deutsch

H. M. Finucan, Some decompositions of generalized Catalan numbers, pp. 275-293 of Combinatorial Mathematics IX. Proc. Ninth Australian Conference (Brisbane, August 1981). Ed. E. J. Billington, S. Oates-Williams and A. P. Street. Lecture Notes Math., 952. Springer-Verlag, 1982.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Andrew Howroyd, Table of n, a(n) for n = 0..200
Emanuele Munarini, Shifting Property for Riordan, Sheffer and Connection Constants Matrices, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.2.
Joris Nieuwveld, Fractions, Functions and Folding. A Novel Link between Continued Fractions, Mahler Functions and Paper Folding, Master's Thesis, arXiv:2108.11382 [math.NT], 2021.
C. H. Pah, Single polygon counting on Cayley Tree of order 3, J. Stat. Phys. 140 (2010) 198-207.
Index entries for sequences related to rooted trees

Column 2 of A092276.

Cf. A000139, A001764, A006013, A006630, A006631, A230547, A233657.

A006629:= func< n | 2*Binomial(3*n+3,n)/(n+2) >;
[A006629(n): n in [0..40]]; // G. C. Greubel, Aug 29 2025

Table[2*Binomial[3*n+3,n]/(n+2), {n,0,40}] (* G. C. Greubel, Aug 29 2025 *)

a(n)=my(m=4);binomial(3*n+m-1,n)*m/(2*n+m) /* 4th power of A001764 with offset n=0 */ \\ Paul D. Hanna, May 10 2008

def A006629(n): return 2*binomial(3*n+3,n)//(n+2)
print([A006629(n) for n in range(41)]) # G. C. Greubel, Aug 29 2025

a(n) = 2*binomial(3*n+3,n)/(n+2). - Emeric Deutsch
a(n) = (n+1) * A000139(n+1). - F. Chapoton, Feb 23 2024
G.f.: hypergeom( [ 2, 5/3, 4/3 ]; [ 3, 5/2 ]; 27*x/4 ).
G.f.: A(x) = G(x)^4 where G(x) = 1 + x*G(x)^3 = g.f. of A001764 giving a(n)=C(3n+m-1,n)*m/(2n+m) at power m=4 with offset n=0. - Paul D. Hanna, May 10 2008
G.f.: (((4*sin(arcsin((3*sqrt(3*x))/2)/3))/(sqrt(3*x))-1)^2-1)/(4*x). - Vladimir Kruchinin, Feb 17 2023
E.g.f.: hypergeom([4/3, 5/3, 2]; [1, 5/2, 3]; 27*x/4). - G. C. Greubel, Aug 29 2025

More precise definition from Paul D. Hanna, May 10 2008

A085357 Common residues of binomial(3n,n)/(2n+1) modulo 2: relates ternary trees (A001764) to the infinite Fibonacci word (A003849).

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Paul D. Hanna, Jun 25 2003

nonn

The n-th runs of ones is given by: 3 - A003849(n) (infinite Fibonacci word) = A076662(n+1). Runs of zeros are given by: A085358 and are also directly related to the Fibonacci sequence. Coefficients of A(x)^3 are found in A085359.
a(n) = 0 iff some binary digit of n is 1 while the corresponding binary digit of 3*n is 0. - Robert Israel, Jul 12 2016
The Run Length Transform of [0,1,0,0,0,...], A063524, the characteristic function of 1. (See A227349 for the definition). - Antti Karttunen, Oct 15 2016

Robert Israel, Table of n, a(n) for n = 0..10000
J.-P. Allouche, F. v. Haeseler, H.-O. Peitgen, A. Petersen and G. Skordev, Automaticity of double sequences generated by one-dimensional linear cellular automata, Theoret. Comput. Sci. 186 (1997), 195-209.
J.-P. Allouche, F. v. Haeseler, H.-O. Peitgen and G. Skordev, Linear cellular automata, finite automata and Pascal's triangle, Discrete Appl. Math. 66 (1996), 1-22.
R. Bacher and C. Reutenauer, The number of right ideals of given codimension over a finite field, in Noncommutative Birational Geometry, Representations and Combinatorics, edited by Arkady. Berenstein and Vladimir. Retakha, Contemporary Mathematics, Vol. 592, 2013.
Paul Tarau, Emulating Primality with Multiset Representations of Natural Numbers, in Theoretical Aspects of Computing, ICTAC 2011, Lecture Notes in Computer Science, 2011, Volume 6916/2011, 218-238, DOI: 10.1007/978-3-642-23283-1_15.
Index entries for characteristic functions
Index entries for sequences computed with run length transform

Cf. A001764 (ternary trees), A085358 (runs of zeros), A076662 (runs of ones), A003849 (infinite Fibonacci word), A085359 (A(x)^3).
Cf. also A003714, A005940, A008966, A063524, A227349, A277010, A324964.
Absolute values of A132971.

[Binomial(3*n,n) mod 2: n in [0..100]]; // Vincenzo Librandi, Jul 09 2016

f:= proc(n) local L,Lp;
  L:= convert(n,base,2);
  Lp:= convert(3*n,base,2);
  if has(L-Lp[1..nops(L)],1) then 0 else 1 fi
end proc:
map(f, [$0..100]); # Robert Israel, Jul 12 2016

Table[Mod[Binomial[3 n, n], 2], {n, 0, 120}] (* Michael De Vlieger, Jul 08 2016 *)

A085357(n) = !bitand(n,n<<1); \\ Antti Karttunen, Aug 22 2019

def A085357(n): return int(not n&(n<<1)) # Chai Wah Wu, Jun 25 2025

G.f.: 1 + x*A(x)^3 = A(x) (Mod 2); a(n) = A001764(n) (Mod 2).
a(n) = binomial(3n, n) (mod 2). Characteristic function of Fibbinary numbers (i.e. a(n)=1 iff n is in A003714). - Benoit Cloitre, Nov 15 2003
Recurrence: a(0) = 1, a(2n) = a(4n+1) = a(n), a(4n+3) = 0.
a(n-2) = A000256(n)(mod 2), for n>2. - John M. Campbell, Jul 08 2016
a(n) = A000621(n+1)(mod 2). - John M. Campbell, Jul 15 2016
a(n) = A000625(n)(mod 2). - John M. Campbell, Jul 15 2016
a(n) = A008966(A005940(1+n)). [Follows from the Run Length Transform interpretation, see also A277010.] - Antti Karttunen, Oct 15 2016
a(n) = abs(A132971(n)) = abs(A008683(A005940(1+n))). - Antti Karttunen, May 30 2017

A104859 Partial sums of A001764.

Original entry on oeis.org

1, 2, 5, 17, 72, 345, 1773, 9525, 52788, 299463, 1730178, 10144818, 60211926, 361042498, 2183809018, 13308564682, 81637319641, 503667864976, 3123298907641, 19456221197941, 121696331095636, 764008782313381

Offset: 0

Emeric Deutsch, Apr 24 2005

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

Cf. A001764.

a:=n->add(binomial(3*k,k)/(2*k+1),k=0..n): seq(a(n),n=0..26);

Table[Sum[Binomial[3k,k]/(2k+1),{k,0,n}],{n,0,20}] (* Emanuele Munarini, Apr 08 2011 *)

makelist(sum(binomial(3*k,k)/(2*k+1),k,0,n),n,0,20); /* Emanuele Munarini, Apr 08 2011 */

a(n) = Sum_{k=0..n} binomial(3k, k)/(2k+1).
G.f.: T(z)/(1-z), where T = 1+z*T^3.
G.f.: 2*sin((1/3)*arcsin(sqrt(27*z/4)))/((1-z)*sqrt(3*z)).
Recurrence: 2*(2*n^2 + 9*n + 10)*a(n+2) - (31*n^2 + 99*n + 80)*a(1+n) + 3*(9*n^2 + 27*n + 20)*a(n) = 0. - Emanuele Munarini, Apr 08 2011
a(n) ~ 3^(3*n+7/2)/(23*sqrt(Pi)*2^(2*n+2)*n^(3/2)). - Vaclav Kotesovec, Oct 17 2012
G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * (1 - x)^2 * A(x)^3. - Ilya Gutkovskiy, Jul 25 2021

A213282 G.f. satisfies A(x) = G(x/(1-x)^3) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 6, 36, 236, 1656, 12192, 92960, 727824, 5817696, 47281472, 389533056, 3245867136, 27308274688, 231654031104, 1979205694464, 17016094611712, 147104972637696, 1277988764697600, 11151534242977792, 97692088569096192, 858890594909048832, 7575804347863105536

Offset: 0

Paul D. Hanna, Jun 08 2012

nonn

Compare to the g.f. B(x) of A006319 where B(x) = C(x/(1-x)^2) such that C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

			G.f.: A(x) = 1 + x + 6*x^2 + 36*x^3 + 236*x^4 + 1656*x^5 + 12192*x^6 +...
G.f.: A(x) = G(x/(1-x)^3) where G(x) = 1 + x*G(x)^3 is g.f. of A001764:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...

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

Cf. A213281, A001764; variants: A006319 (royal paths in a lattice), A213336.

series(RootOf(G = 1 + G^3*x/(1-x)^3, G),x=0,30); # Mark van Hoeij, Apr 18 2013

/* G.f. A(x) = G(x/(1-x)^3) where G(x) = 1 + x*G(x)^3: */
{a(n)=local(A,G=1+x);for(i=1,n,G=1+x*G^3+x*O(x^n));A=subst(G,x,x/(1-x+x*O(x^n))^3);polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

/* G.f. A(x) = F(x*A(x)^3) where F(x) = 1 + x/F(-x)^3: */
{a(n)=local(F=1+x+x*O(x^n), A=1); for(i=1, n+1, F=1+x/subst(F^3, x, -x+x*O(x^n))); A=(serreverse(x/F^3)/x)^(1/3); polcoeff(A, n)}
for(n=0, 25, print1(a(n), ", "))

G.f. satisfies: A(x) = F(x*A(x)^3) where F(x) = 1 + x/F(-x)^3 is the g.f. of A213281.
G.f. A(x) satisfies: A(1 - G(-x)) = G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
a(n) = Sum_{k=0..n} binomial(n+2*k-1,n-k) * binomial(3*k,k)/(2*k+1). - Seiichi Manyama, Oct 03 2023

A251573 E.g.f.: exp(3xG(x)^2) / G(x)^2 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 3, 21, 261, 4833, 120303, 3778029, 143531433, 6404711553, 328447585179, 19037277446949, 1230842669484717, 87829738967634849, 6856701559496841159, 581343578623728854397, 53196439113856500195537, 5225543459274294130169601, 548468830470032135590262067, 61258398893626609968686844597

Offset: 0

Paul D. Hanna, Dec 06 2014

nonn

			E.g.f.: A(x) = 1 + x + 3*x^2/2! + 21*x^3/3! + 261*x^4/4! + 4833*x^5/5! +...
such that A(x) = exp(3*x*G(x)^2) / G(x)^2
where G(x) = 1 + x*G(x)^3 is the g.f. of A001764:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...
The e.g.f. satisfies:
A(x) = 1 + x/A(x) + 5*x^2/(2!*A(x)^2) + 54*x^3/(3!*A(x)^3) + 945*x^4/(4!*A(x)^4) + 23328*x^5/(5!*A(x)^5) + 750141*x^6/(6!*A(x)^6) + 29859840*x^7/(7!*A(x)^7) +...+ 3^(n-1)*(n+1)^(n-3)*(n+3) * x^n/(n!*A(x)^n) +...
Note that
A'(x) = exp(3*x*G(x)^2) = 1 + 3*x + 21*x^2/2! + 261*x^3/3! + 4833*x^4/4! +...
LOGARITHMIC DERIVATIVE.
The logarithm of the e.g.f. begins:
log(A(x)) = x + 2*x^2/2 + 7*x^3/3 + 30*x^4/4 + 143*x^5/5 +...
and so A'(x)/A(x) = G(x)^2.
TABLE OF POWERS OF E.G.F.
Form a table of coefficients of x^k/k! in A(x)^n as follows.
n=1: [1, 1,  3,   21,   261,   4833,  120303,   3778029, ...];
n=2: [1, 2,  8,   60,   744,  13536,  330912,  10232928, ...];
n=3: [1, 3, 15,  123,  1557,  28179,  680427,  20771235, ...];
n=4: [1, 4, 24,  216,  2832,  51552, 1237248,  37404288, ...];
n=5: [1, 5, 35,  345,  4725,  87285, 2094975,  62949825, ...];
n=6: [1, 6, 48,  516,  7416, 139968, 3378528, 101278944, ...];
n=7: [1, 7, 63,  735, 11109, 215271, 5250987, 157613463, ...];
n=8: [1, 8, 80, 1008, 16032, 320064, 7921152, 238878720, ...]; ...
in which the main diagonal begins (see A251583):
[1, 2, 15, 216, 4725, 139968, 5250987, 238878720, ...]
and is given by the formula:
[x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3) for n>=0.

Paul D. Hanna, Table of n, a(n) for n = 0..200

Cf. A251583, A251663, A001764, A006013.

Cf. Variants: A243953, A251574, A251575, A251576, A251577, A251578, A251579, A251580.

Flatten[{1,1,Table[Sum[3^k * n!/k! * Binomial[3*n-k-3, n-k] * (k-1)/(n-1),{k,0,n}],{n,2,20}]}] (* Vaclav Kotesovec, Dec 07 2014 *)
Flatten[{1,1,RecurrenceTable[{27*(n-2)*a[n-2]-3*(3*n-8)*(15-13*n+3*n^2)*a[n-1]+2*(n-3)*(2*n-3)*a[n]==0,a[2]==3,a[3]==21},a,{n,20}]}] (* Vaclav Kotesovec, Dec 07 2014 *)

{a(n) = local(G=1);for(i=1,n,G=1+x*G^3 +x*O(x^n)); n!*polcoeff(exp(3*x*G^2)/G^2, n)}
for(n=0, 20, print1(a(n), ", "))

{a(n) = if(n==0||n==1, 1, sum(k=0, n, 3^k * n!/k! * binomial(3*n-k-3,n-k) * (k-1)/(n-1) ))}
for(n=0, 20, print1(a(n), ", "))

Let G(x) = 1 + x*G(x)^3 be the g.f. of A001764, then the e.g.f. A(x) of this sequence satisfies:
(1) A'(x)/A(x) = G(x)^2.
(2) A'(x) = exp(3*x*G(x)^2).
(3) A(x) = exp( Integral G(x)^2 dx ).
(4) A(x) = exp( Sum_{n>=1} A006013(n-1)*x^n/n ), where A006013(n-1) = binomial(3*n-2,n)/(2*n-1).
(5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251583.
(6) A(x) = Sum_{n>=0} A251583(n)*(x/A(x))^n/n! where A251583(n) = 3^(n-1) * (n+1)^(n-3) * (n+3).
(7) [x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3).
a(n) = Sum_{k=0..n} 3^k * n!/k! * binomial(3*n-k-3, n-k) * (k-1)/(n-1) for n>1.
Recurrence (for n>3): 2*(n-3)*(2*n-3)*a(n) = 3*(3*n-8)*(3*n^2 - 13*n + 15)*a(n-1) - 27*(n-2)*a(n-2). - Vaclav Kotesovec, Dec 07 2014
a(n) ~ 3^(3*n-7/2) * n^(n-2) / (2^(2*n-5/2) * exp(n-1)). - Vaclav Kotesovec, Dec 07 2014

A251663 E.g.f.: exp( 3xG(x)^2 ) / G(x), where G(x) = 1 + x*G(x)^3 is the g.f. A001764.

Original entry on oeis.org

1, 2, 11, 120, 2061, 48918, 1487151, 55188108, 2419385625, 122367255498, 7014349322739, 449405251066368, 31826192109186789, 2468711973793223070, 208159999898813165079, 18957203713618483723092, 1854424578467714146269489, 193922780991931737971748882, 21588348501840566333913576795

Offset: 0

Paul D. Hanna, Dec 07 2014

nonn

			E.g.f.: A(x) = 1 + 2*x + 11*x^2/2! + 120*x^3/3! + 2061*x^4/4! + 48918*x^5/5! +...
such that A(x) = exp(3*x*G(x)^2) / G(x)
where G(x) = 1 + x*G(x)^3 is the g.f. of A001764:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...
The e.g.f. satisfies:
A(x) = 1 + 2*x/A(x)^2 + 27*x^2/(2!*A(x)^4) + 756*x^3/(3!*A(x)^6) + 32805*x^4/(4!*A(x)^8) + 1940598*x^5/(5!*A(x)^10) + 145746783*x^6/(6!*A(x)^12) + 13286025000*x^7/(7!*A(x)^14) +...+ (n+1)*(2*n+1)^(n-2)*3^n * x^n/(n!*A(x)^(2*n)) +...

Cf. A251573, A251693, A001764.

Cf. Variants: A243953, A251664, A251665, A251666, A251667, A251668, A251669, A251670.

Table[Sum[3^k * n!/k! * Binomial[3*n-k-2, n-k] * (2*k-1)/(2*n-1),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 07 2014 *)

{a(n)=local(G=1); for(i=0, n, G=1+x*G^3 +x*O(x^n)); n!*polcoeff(exp(3*x*G^2)/G, n)}
for(n=0, 20, print1(a(n), ", "))

{a(n) = sum(k=0, n, 3^k * n!/k! * binomial(3*n-k-2,n-k) * (2*k-1)/(2*n-1) )}
for(n=0, 20, print1(a(n), ", "))

Let G(x) = 1 + x*G(x)^3 be the g.f. of A001764, then the e.g.f. A(x) of this sequence satisfies:
(1) A'(x)/A(x) = G(x)^2 + G'(x)/G(x).
(2) A(x) = F(x/A(x)^2) where F(x) is the e.g.f. of A251693.
(3) A(x) = Sum_{n>=0} A251693(n)*(x/A(x)^2)^n/n! where A251693(n) = (n+1) * (2*n+1)^(n-2) * 3^n.
(4) [x^n/n!] A(x)^(2*n+1) = (n+1) * (2*n+1)^(n-1) * 3^n.
a(n) = Sum_{k=0..n} 3^k * n!/k! * binomial(3*n-k-2, n-k) * (2*k-1)/(2*n-1) for n>=0.
Recurrence: 2*(2*n-1)*(9*n^2 - 30*n + 19)*a(n) = 3*(81*n^4 - 432*n^3 + 756*n^2 - 393*n - 88)*a(n-1) - 27*(9*n^2 - 12*n - 2)*a(n-2). - Vaclav Kotesovec, Dec 07 2014
a(n) ~ 3^(3*n-3/2) * n^(n-1) / (2^(2*n-1/2) * exp(n-1)). - Vaclav Kotesovec, Dec 07 2014

A064017 Number of ternary trees (A001764) with n nodes and maximal diameter.

Original entry on oeis.org

1, 3, 12, 45, 162, 567, 1944, 6561, 21870, 72171, 236196, 767637, 2480058, 7971615, 25509168, 81310473, 258280326, 817887699, 2582803260, 8135830269, 25569752274, 80196041223, 251048476872, 784526490225, 2447722649502

Offset: 1

Danail Bonchev (bonchevd(AT)aol.com), Sep 07 2001

A problem important for polymer science because it counts the trees having unbranched branches; they are called "combs".
Equals (1, 3, 9, 27, 81, ...) convolved with (1, 0, 3, 9, 27, 81, ...). Example: a(5) = 162 = (81, 27, 9, 3, 1) dot (1, 0, 3, 9, 27) = 81 + 3*27. - Gary W. Adamson, Jul 31 2010
Floretion Algebra Multiplication Program, FAMP Code: lesforseq[ - 'i + 'j - 'kk' - 'ki' - 'kj' ], vesforseq(n) = 3^n, tesforseq = A006234

			a(5) = 162 because we can write (5+1)*3^(5-2) = 6*3^3 = 6*27.

Harry J. Smith, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (6,-9).

Cf. A001764, A006234, A014915, A025192, A027261, A079272, A196389.

a:=n->ceil(sum(3^(n-2),j=0..n)): seq(a(n), n=1..26); # Zerinvary Lajos, Jun 05 2008

Join[{1},Table[(n+1)3^(n-2),{n,2,30}]] (* or *) Join[{1}, LinearRecurrence[ {6,-9},{3,12},30]] (* Harvey P. Dale, Feb 07 2012 *)

{ for (n=1, 200, if (n>1, a=(n + 1)*p; p*=3, a=p=1); write("b064017.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 06 2009

a(n)=if(n==1, 1, (n+1)*3^(n-2)); \\ Joerg Arndt, May 06 2013

@CachedFunction
def BB(n, k, x):  # modified cardinal B-splines
    if n == 1: return 0 if (x < 0) or (x >= k) else 1
    return x*BB(n-1, k, x) + (n*k-x)*BB(n-1, k, x-k)
def EulerianPolynomial(n, k, x):
    if n == 0: return 1
    return add(BB(n+1, k, k*m+1)*x^m for m in (0..n))
def A064017(n) : return 3^(n-1)*EulerianPolynomial(1,n-1,1/3) if n != 1 else 1
[A064017(n) for n in (1..25)]  # Peter Luschny, May 04 2013

a(n) = 3*a(n-1) + 3^(n-2).
a(n) = (n+1)*3^(n-2), for n > 1.
From Paul Barry, Sep 05 2003: (Start)
a(n) = (n+2)3^(n-1) + 0^n/3 (offset 0).
a(n) = A025192(n) + A027471(n). (End)
A006234(n+4) - a(n+2) = 3^n. - Creighton Dement, Mar 01 2005
a(n+1) = Sum_{k=0..n} A196389(n,k)*3^k. - Philippe Deléham, Oct 31 2011
G.f.: (1 - 3*x + 3*x^2)*x/(1 - 3*x)^2. - Philippe Deléham, Oct 31 2011
a(n) = 6*a(n-1) - 9*a(n-2), with a(1)=1, a(2)=3, a(3)=12. - Harvey P. Dale, Feb 07 2012
E.g.f.: (exp(3*x)*(1 + 3*x) - 1)/9. - Stefano Spezia, Mar 05 2020
From Amiram Eldar, Jan 18 2021: (Start)
Sum_{n>=1} 1/a(n) = 27*log(3/2) - 19/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 17/2 - 27*log(4/3). (End)

A226751 G.f.: 1 / (1 + 6xG(x) - 7xG(x)^2), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

Original entry on oeis.org

1, 1, 9, 48, 289, 1761, 10932, 68664, 435201, 2777763, 17829489, 114968052, 744178716, 4832624044, 31469746632, 205422018288, 1343734578561, 8806130111847, 57805893969531, 380013533789928, 2501507255441049, 16486378106441697, 108773240389894056

Offset: 0

Paul D. Hanna, Jun 16 2013

nonn

			G.f.: A(x) = 1 + x + 9*x^2 + 48*x^3 + 289*x^4 + 1761*x^5 + 10932*x^6 +...
A related series is G(x) = 1 + x*G(x)^3, where
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...
G(x)^2 = 1 + 2*x + 7*x^2 + 30*x^3 + 143*x^4 + 728*x^5 + 3876*x^6 +...
such that A(x) = 1/(1 + 6*x*G(x) - 7*x*G(x)^2).

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

Cf. A147855, A001764.
Cf. A005809, A006256, A160906, A183160.
Cf. A226705, A226733.

Table[Sum[Binomial[n+2*k,n-k]*Binomial[2*n-2*k,k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 17 2013 *)

{a(n)=sum(k=0, n, binomial(n+2*k, n-k)*binomial(2*n-2*k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(2*k, n-k)*binomial(3*n-2*k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=local(G=1+x); for(i=0, n, G=1+x*G^3+x*O(x^n)); polcoeff(1/(1+6*x*G-7*x*G^2), n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=local(G=1+x); for(i=0, n,G=1+x*G^3+x*O(x^n)); polcoeff(1/(1-x*G-7*x^2*G^4), n)}
for(n=0, 30, print1(a(n), ", "))

a(n) = Sum_{k=0..n} C(2*k, n-k) * C(3*n-2*k, k).
a(n) = Sum_{k=0..n} C(n+2*k, n-k) * C(2*n-2*k, k).
a(n) = Sum_{k=0..n} C(2*n+2*k, n-k) * C(n-2*k, k).
a(n) = Sum_{k=0..n} C(3*n+2*k, n-k) * C(-2*k, k).
G.f.: 1/(1 - x*G(x) - 7*x^2*G(x)^4), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.
a(n) ~ 3^(3*n+3/2)/(5*sqrt(Pi*n)*2^(2*n+1)). - Vaclav Kotesovec, Jun 17 2013
Conjecture: 18*n*(2*n-1)*(55*n-76)*a(n) +(-11605*n^3+28521*n^2-20870*n+4536)*a(n-1) -24*(55*n-21)*(3*n-4)*(3*n-2)*a(n-2)=0. - R. J. Mathar, Jun 14 2016
From Seiichi Manyama, Aug 05 2025: (Start)
a(n) = [x^n] 1/((1+2*x) * (1-x)^(2*n+1)).
a(n) = Sum_{k=0..n} (-3)^(n-k) * binomial(3*n+1,k).
a(n) = Sum_{k=0..n} (-2)^(n-k) * binomial(2*n+k,k). (End)
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} (-2)^k * 3^(n-k) * binomial(3*n+1,k) * binomial(3*n-k,n-k).
G.f.: G(x)^2/((-2+3*G(x)) * (3-2*G(x))) where G(x) = 1+x*G(x)^3 is the g.f. of A001764. (End)
G.f.: B(x)^2/(1 + 5*(B(x)-1)/3), where B(x) is the g.f. of A005809. - Seiichi Manyama, Aug 15 2025

A110616 A convolution triangle of numbers based on A001764.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 12, 7, 3, 1, 55, 30, 12, 4, 1, 273, 143, 55, 18, 5, 1, 1428, 728, 273, 88, 25, 6, 1, 7752, 3876, 1428, 455, 130, 33, 7, 1, 43263, 21318, 7752, 2448, 700, 182, 42, 8, 1, 246675, 120175, 43263, 13566, 3876, 1020, 245, 52, 9, 1

Offset: 0

Philippe Deléham, Sep 14 2005, Jun 15 2007

Reflected version of A069269. - Vladeta Jovovic, Sep 27 2006
With offset 1 for n and k, T(n,k) = number of Dyck paths of semilength n for which all descents are of even length (counted by A001764) with no valley vertices at height 1 and with k returns to ground level. For example, T(3,2)=2 counts U^4 D^4 U^2 D^2, U^2 D^2 U^4 D^4 where U=upstep, D=downstep and exponents denote repetition. - David Callan, Aug 27 2009
Riordan array (f(x), x*f(x)) with f(x) = (2/sqrt(3*x))*sin((1/3)*arcsin(sqrt(27*x/4))). - Philippe Deléham, Jan 27 2014
Antidiagonals of convolution matrix of Table 1.4, p. 397, of Hoggatt and Bicknell. - Tom Copeland, Dec 25 2019

			Triangle begins:
       1;
       1,      1;
       3,      2,     1;
      12,      7,     3,     1;
      55,     30,    12,     4,    1;
     273,    143,    55,    18,    5,    1;
    1428,    728,   273,    88,   25,    6,   1;
    7752,   3876,  1428,   455,  130,   33,   7,  1;
   43263,  21318,  7752,  2448,  700,  182,  42,  8, 1;
  246675, 120175, 43263, 13566, 3876, 1020, 245, 52, 9, 1;
  ...
From _Peter Bala_, Feb 04 2025: (Start)
The transposed array factorizes as an infinite product of upper triangular arrays:
  / 1               \^T   /1             \^T /1             \^T / 1            \^T
  | 1    1           |   | 1   1          | | 0  1           |  | 0  1          |
  | 3    2   1       | = | 2   1   1      | | 0  1   1       |  | 0  0  1       | ...
  |12    7   3   1   |   | 5   2   1  1   | | 0  2   1  1    |  | 0  0  1  1    |
  |55   30  12   4  1|   |14   5   2  1  1| | 0  5   2  1  1 |  | 0  0  2  1  1 |
  |...               |   |...             | |...             |  |...            |
where T denotes transposition and [1, 1, 2, 5, 14,...] is the sequence of Catalan numbers A000108. (End)

Peter Bala, Factorisations of some Riordan arrays as infinite products
Paul Barry, Notes on Riordan arrays and lattice paths, arXiv:2504.09719 [math.CO], 2025. See pp. 27, 29.
Paul Barry, d-orthogonal polynomials, Fuss-Catalan matrices and lattice paths, arXiv:2505.16718 [math.CO], 2025. See p. 21.
Naiomi Cameron and J. E. McLeod, Returns and Hills on Generalized Dyck Paths, Journal of Integer Sequences, Vol. 19, 2016, #16.6.1.
V. E. Hoggatt, Jr. and M. Bicknell, Catalan and related sequences arising from inverses of Pascal's triangle matrices, Fib. Quart., 14 (1976), 395-405.
Sheng-Liang Yang and L. J. Wang, Taylor expansions for the m-Catalan numbers, Australasian Journal of Combinatorics, Volume 64(3) (2016), Pages 420-431.

Successive columns: A001764, A006013, A001764, A006629, A102893, A006630, A102594, A006631; row sums: A098746; see also A092276.

Table[(k + 1) Binomial[3 n - 2 k, 2 n - k]/(2 n - k + 1), {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jun 28 2017 *)

T(n,k):=((k+1)*binomial(3*n-2*k,2*n-k))/(2*n-k+1); /* Vladimir Kruchinin, Nov 01 2011 */

T(n, k) = Sum_{j>=0} T(n-1, k-1+j)*A000108(j); T(0, 0) = 1; T(n, k) = 0 if k < 0 or if k > n.
G.f.: 1/(1 - x*y*TernaryGF) = 1 + (y)x + (y+y^2)x^2 + (3y+2y^2+y^3)x^3 +... where TernaryGF = 1 + x + 3x^2 + 12x^3 + ... is the GF for A001764. - David Callan, Aug 27 2009
T(n, k) = ((k+1)*binomial(3*n-2*k,2*n-k))/(2*n-k+1). - Vladimir Kruchinin, Nov 01 2011

cp's OEIS Frontend

A188687 Partial binomial sums of binomial(3n,n)/(2n+1) = A001764(n).

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A006629 Self-convolution 4th power of A001764, which enumerates ternary trees.

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A085357 Common residues of binomial(3n,n)/(2n+1) modulo 2: relates ternary trees (A001764) to the infinite Fibonacci word (A003849).

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A104859 Partial sums of A001764.

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A213282 G.f. satisfies A(x) = G(x/(1-x)^3) where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

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A251573 E.g.f.: exp(3*x*G(x)^2) / G(x)^2 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

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A251663 E.g.f.: exp( 3*x*G(x)^2 ) / G(x), where G(x) = 1 + x*G(x)^3 is the g.f. A001764.

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A251573 E.g.f.: exp(3xG(x)^2) / G(x)^2 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

A251663 E.g.f.: exp( 3xG(x)^2 ) / G(x), where G(x) = 1 + x*G(x)^3 is the g.f. A001764.

A226751 G.f.: 1 / (1 + 6xG(x) - 7xG(x)^2), where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.