A005725 -id:A005725 - cp's OEIS frontend

A036765 Number of ordered rooted trees with n non-root nodes and all outdegrees <= three.

1, 1, 2, 5, 13, 36, 104, 309, 939, 2905, 9118, 28964, 92940, 300808, 980864, 3219205, 10626023, 35252867, 117485454, 393133485, 1320357501, 4449298136, 15038769672, 50973266380, 173214422068, 589998043276, 2014026871496, 6889055189032, 23608722350440

Offset: 0

N. J. A. Sloane

Number of Dyck n-paths that avoid UUUU. For example, a(4)=13 counts all 14 Dyck 4-paths except UUUUDDDD. - David Callan, Dec 09 2004
Number of restricted growth strings for Dyck paths with at most 2 consecutive rises (this is equivalent to the comment above, see example). - Joerg Arndt, Oct 31 2012
Let A(x) be the g.f. for the sequence of numbers of Dyck words with at most k consecutive ones (paths with at most k consecutive up-steps 'U', Restricted Growth Strings with at most k-1 consecutive rises), then B(x) := x*A(x) is the series reversion of x/(1+x+x^2+...+x^k). - Joerg Arndt, Oct 31 2012
a(n) is the number of ordered unlabeled rooted trees on n+1 nodes where each node has no more than 3 children. - Geoffrey Critzer, Jan 05 2013
a(n) = number of noncrossing partitions of [n] in which all blocks are of size <= 3. - David Callan, Aug 27 2014

			a(4) = 13 since the top row of M^4 = (13, 8, 4, 1, 1).
From _Joerg Arndt_, Oct 31 2012: (Start)
a(5)=36 because there are 36 Dyck words of length 5 that avoid "1111":
[ #]      RGS                rises         Dyck word
[ 1]    [ . . . . . ]     [ . . . . . ]    1.1.1.1.1.
[ 2]    [ . . . . 1 ]     [ . . . . 1 ]    1.1.1.11..
[ 3]    [ . . . 1 . ]     [ . . . 1 . ]    1.1.11..1.
[ 4]    [ . . . 1 1 ]     [ . . . 1 . ]    1.1.11.1..
[ 5]    [ . . . 1 2 ]     [ . . . 1 2 ]    1.1.111...
[ 6]    [ . . 1 . . ]     [ . . 1 . . ]    1.11..1.1.
[ 7]    [ . . 1 . 1 ]     [ . . 1 . 1 ]    1.11..11..
[ 8]    [ . . 1 1 . ]     [ . . 1 . . ]    1.11.1..1.
[ 9]    [ . . 1 1 1 ]     [ . . 1 . . ]    1.11.1.1..
[10]    [ . . 1 1 2 ]     [ . . 1 . 1 ]    1.11.11...
[11]    [ . . 1 2 . ]     [ . . 1 2 . ]    1.111...1.
[12]    [ . . 1 2 1 ]     [ . . 1 2 . ]    1.111..1..
[13]    [ . . 1 2 2 ]     [ . . 1 2 . ]    1.111.1...
[--]    [ . . 1 2 3 ]     [ . . 1 2 3 ]    1.1111....
[14]    [ . 1 . . . ]     [ . 1 . . . ]    11..1.1.1.
[15]    [ . 1 . . 1 ]     [ . 1 . . 1 ]    11..1.11..
[16]    [ . 1 . 1 . ]     [ . 1 . 1 . ]    11..11..1.
[17]    [ . 1 . 1 1 ]     [ . 1 . 1 . ]    11..11.1..
[18]    [ . 1 . 1 2 ]     [ . 1 . 1 2 ]    11..111...
[19]    [ . 1 1 . . ]     [ . 1 . . . ]    11.1..1.1.
[20]    [ . 1 1 . 1 ]     [ . 1 . . 1 ]    11.1..11..
[21]    [ . 1 1 1 . ]     [ . 1 . . . ]    11.1.1..1.
[22]    [ . 1 1 1 1 ]     [ . 1 . . . ]    11.1.1.1..
[23]    [ . 1 1 1 2 ]     [ . 1 . . 1 ]    11.1.11...
[24]    [ . 1 1 2 . ]     [ . 1 . 1 . ]    11.11...1.
[25]    [ . 1 1 2 1 ]     [ . 1 . 1 . ]    11.11..1..
[26]    [ . 1 1 2 2 ]     [ . 1 . 1 . ]    11.11.1...
[27]    [ . 1 1 2 3 ]     [ . 1 . 1 2 ]    11.111....
[28]    [ . 1 2 . . ]     [ . 1 2 . . ]    111...1.1.
[29]    [ . 1 2 . 1 ]     [ . 1 2 . 1 ]    111...11..
[30]    [ . 1 2 1 . ]     [ . 1 2 . . ]    111..1..1.
[31]    [ . 1 2 1 1 ]     [ . 1 2 . . ]    111..1.1..
[32]    [ . 1 2 1 2 ]     [ . 1 2 . 1 ]    111..11...
[33]    [ . 1 2 2 . ]     [ . 1 2 . . ]    111.1...1.
[34]    [ . 1 2 2 1 ]     [ . 1 2 . . ]    111.1..1..
[35]    [ . 1 2 2 2 ]     [ . 1 2 . . ]    111.1.1...
[36]    [ . 1 2 2 3 ]     [ . 1 2 . 1 ]    111.11....
[--]    [ . 1 2 3 . ]     [ . 1 2 3 . ]    1111....1.
[--]    [ . 1 2 3 1 ]     [ . 1 2 3 . ]    1111...1..
[--]    [ . 1 2 3 2 ]     [ . 1 2 3 . ]    1111..1...
[--]    [ . 1 2 3 3 ]     [ . 1 2 3 . ]    1111.1....
[--]    [ . 1 2 3 4 ]     [ . 1 2 3 4 ]    11111.....
(Dots are used for zeros for readability.)
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..1000 (first 201 terms from T. D. Noe)
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Paul Barry, Riordan arrays, generalized Narayana triangles, and series reversion, Linear Algebra and its Applications, 491 (2016) 343-385.
Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
Paul Barry, d-orthogonal polynomials, Fuss-Catalan matrices and lattice paths, arXiv:2505.16718 [math.CO], 2025. See p. 25.
N. T. Cameron, Random walks, trees and extensions of Riordan group techniques, Dissertation, Howard University, 2002.
Naiomi Cameron and J. E. McLeod, Returns and Hills on Generalized Dyck Paths, Journal of Integer Sequences, Vol. 19, 2016, Article #16.6.1.
Colin Defant and Kai Zheng, Stack-Sorting with Consecutive-Pattern-Avoiding Stacks, arXiv:2008.12297 [math.CO], 2020.
Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
Filippo Disanto and Thomas Wiehe, Some instances of a sub-permutation problem on pattern avoiding permutations, arXiv preprint arXiv:1210.6908 [math.CO], 2012.
M. Dziemianczuk, Enumerations of plane trees with multiple edges and Raney lattice paths, Discrete Mathematics 337 (2014): 9-24.
Petr Gregor, Torsten Mütze, and Namrata, Combinatorial generation via permutation languages. VI. Binary trees, arXiv:2306.08420 [cs.DM], 2023.
Nickolas Hein and Jia Huang, Modular Catalan Numbers, arXiv:1508.01688 [math.CO], 2015.
Nickolas Hein and Jia Huang, Modular Catalan Numbers, European Journal of Combinatorics 61 (2017), 197-218.
JiSun Huh, Sangwook Kim, Seunghyun Seo, and Heesung Shin, Bijections on pattern avoiding inversion sequences and related objects, arXiv:2404.04091 [math.CO], 2024. See p. 22.
L. Pudwell, Pattern avoidance in trees (slides from a talk, mentions many sequences), 2012. - From _N. J. A. Sloane_, Jan 03 2013
A. Sapounakis, I. Tasoulas, and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
L. Takacs, Enumeration of rooted trees and forests, Math. Scientist 18 (1993), 1-10, esp. Eq. (6).
M. Wallner, Lattice Path Combinatorics, Diplomarbeit, Institut für Diskrete Mathematik und Geometrie der Technischen Universität Wien, 2013.
Index entries for sequences related to rooted trees

Right hand column of triangle A064580. The sequence of sequences A000007 (0^n), A000012 (constant 1), A001006 (Motzkin), A036765, A036766, ... tends to A000108 (Catalan).
Cf. A005725, A186241, A198951, A200731.
Column k=3 of A288942.

[&+[Binomial(n+1, n-2*k)*Binomial(n+1, k)/(n+1): k in [0..n]]: n in [0..30]]; // Vincenzo Librandi, Oct 16 2018

r := 3; [ seq((1/n)*add( (-1)^j*binomial(n,j)*binomial(2*n-2-j*(r+1), n-1),j=0..floor((n-1)/(r+1))), n=1..30) ];
# second Maple program:
b:= proc(u, o) option remember; `if`(u+o=0, 1,
      add(b(u-j, o+j-1), j=1..min(1, u))+
      add(b(u+j-1, o-j), j=1..min(3, o)))
    end:
a:= n-> b(0, n):
seq(a(n), n=0..30);  # Alois P. Heinz, Aug 28 2017

InverseSeries[Series[y/(1+y+y^2+y^3), {y, 0, 24}], x] (* then A(x)=y(x)/x *) (* Len Smiley, Apr 11 2000 *)
b[u_, o_, k_] := b[u, o, k] = If[u + o == 0, 1, Sum[b[u - j, o + j - 1, k], {j, 1, Min[1, u]}] + Sum[b[u + j - 1, o - j, k], {j, 1, Min[k, o]}]];
a[n_] := b[0, n, 3];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Nov 05 2017, after Alois P. Heinz *)
Table[HypergeometricPFQ[{-n-1, (1-n)/2, -n/2}, {1, 3/2}, -1], {n, 0, 28}] (* Vladimir Reshetnikov, Oct 15 2018 *)

{a(n)=sum(j=0,n\2,binomial(n+1, n-2*j)*binomial(n+1,j))/(n+1)}

{a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=1+x*A+(x*A)^2+(x*A)^3);polcoeff(A,n)}

{a(n)=local(A=1+x);for(i=1,n,A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*(x*A+x*O(x^n))^j)*x^m/m)));polcoeff(A, n, x)}

{a(n)=local(A=1+x);for(i=1,n,A=exp(sum(m=1, n, sum(j=0, n, binomial(m+j, j)^2*(x*A+x*O(x^n))^j)*(1-x*A)^(2*m+1)*x^m/m)));polcoeff(A, n, x)}

{a(n)=local(A=1+x);for(i=1,n,A=1/(1-x+x*O(x^n))*exp(sum(m=1,n,A^m*sum(k=0,m-1,binomial(m-1,k)*binomial(m,k)*x^k)/(1-x)^(2*m)*x^(2*m)/m) +x*O(x^n)));polcoeff(A,n)} /* Paul D. Hanna */

{a(n)=local(A=1+x);for(i=1,n,A=1/(1-x+x*O(x^n))*exp(sum(m=1,n,A^m*sum(k=0,n,binomial(m+k-1,k)*binomial(m+k,k)*x^k)*x^(2*m)/m) +x*O(x^n)));polcoeff(A,n)} /* Paul D. Hanna */

Vec(serreverse(x/(1+x+x^2+x^3)+O(x^66))/x) /* Joerg Arndt, Jun 10 2011 */

a(n) = (1/(n+1))*Sum_{j=0..floor(n/2)} binomial(n+1, n-2*j)*binomial(n+1, j). G.f. A(z) satisfies A=1+z*A+(z*A)^2+(z*A)^3. - Emeric Deutsch, Nov 29 2003
G.f.: F(x)/x where F(x) is the reversion of x/(1+x+x^2+x^3). - Joerg Arndt, Jun 10 2011
From Paul D. Hanna, Feb 13 2011: (Start)
O.g.f.: A(x) = exp( Sum_{n>=1} (Sum_{k=0..n} C(n,k)^2*x^k*A(x)^k) * x^n/n ).
O.g.f.: A(x) = exp( Sum_{n>=1} (Sum_{k>=0} C(n+k,k)^2*x^k*A(x)^k)*(1-x*A(x))^(2*n+1)* x^n/n ). (End)
From Paul D. Hanna, Feb 24 2011: (Start)
O.g.f.: A(x) = (1/(1-x))*exp( Sum_{n>=1} A(x)^n*(Sum_{k=0..n-1} C(n-1,k)*C(n,k)*x^k)/(1-x)^(2*n) * x^(2*n)/n ).
O.g.f.: A(x) = (1/(1-x))*exp( Sum_{n>=1} A(x)^n*(Sum_{k>=0} C(n+k-1,k)*C(n+k,k)*x^k) * x^(2*n)/n ). (End)
Let M = an infinite quadradiagonal matrix with all 1's in every diagonal: (sub, main, super, and super-super), and the rest zeros. V = vector [1,0,0,0,...]. The sequence = left column terms of M*V iterates. - Gary W. Adamson, Jun 06 2011
An infinite square production matrix M for the sequence is:
  1, 1, 0, 0, 0, 0, ...
  1, 0, 1, 0, 0, 0, ...
  2, 1, 0, 1, 0, 0, ...
  3, 2, 1, 0, 1, 0, ...
  4, 3, 2, 1, 0, 1, ...
  5, 4, 3, 2, 1, 0, ...
  ..., such that a(n) is the top left term of M^n. - Gary W. Adamson, Feb 21 2012
D-finite with recurrence: 2*(n+1)*(2*n+3)*(13*n-1)*a(n) = (143*n^3 + 132*n^2 - 17*n - 18)*a(n-1) + 4*(n-1)*(26*n^2 + 11*n - 6)*a(n-2) + 16*(n-2)*(n-1)*(13*n + 12)*a(n-3). - Vaclav Kotesovec, Sep 09 2013
a(n) ~ c*d^n/n^(3/2), where d = 1/12*((6371+624*sqrt(78))^(2/3)+11*(6371+624*sqrt(78))^(1/3)+217)/(6371+624*sqrt(78))^(1/3) = 3.610718613276... is the root of the equation -16-8*d-11*d^2+4*d^3=0 and c = sqrt(f/Pi) = 0.9102276936417..., where f = 1/9984*(9295 + (13*(45085576939 - 795629568*sqrt(78)))^(1/3) + (13*(45085576939 + 795629568*sqrt(78)))^(1/3)) is the root of the equation -128+1696*f-9295*f^2+3328*f^3=0. - Vaclav Kotesovec, Sep 10 2013
From Peter Bala, Jun 21 2015: (Start)
The coefficient of x^n in A(x)^r equals r/(n + r)*Sum_{k = 0..floor(n/2)} binomial(n + r,k)*binomial(n + r,n - 2*k) by the Lagrange-Bürmann formula.
O.g.f. A(x) = exp(Sum_{n >= 1} A005725(n)*x^n/n), where A005725(n) = Sum_{k = 0..floor(n/2)} binomial(n,k)*binomial(n,n - 2*k). Cf. A186241, A198951, A200731. (End)
a(n) = hypergeom([-n-1, (1-n)/2, -n/2], [1, 3/2], -1). - Vladimir Reshetnikov, Oct 15 2018

Name clarified by Andrew Howroyd, Dec 04 2017

A305161 Number A(n,k) of compositions of n into exactly n nonnegative parts <= k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 3, 1, 0, 1, 1, 3, 7, 1, 0, 1, 1, 3, 10, 19, 1, 0, 1, 1, 3, 10, 31, 51, 1, 0, 1, 1, 3, 10, 35, 101, 141, 1, 0, 1, 1, 3, 10, 35, 121, 336, 393, 1, 0, 1, 1, 3, 10, 35, 126, 426, 1128, 1107, 1, 0, 1, 1, 3, 10, 35, 126, 456, 1520, 3823, 3139, 1, 0

Offset: 0

Alois P. Heinz, Aug 17 2018

			A(3,1) = 1: 111.
A(3,2) = 7: 012, 021, 102, 111, 120, 201, 210.
A(3,3) = 10: 003, 012, 021, 030, 102, 111, 120, 201, 210, 300.
A(4,2) = 19: 0022, 0112, 0121, 0202, 0211, 0220, 1012, 1021, 1102, 1111, 1120, 1201, 1210, 2002, 2011, 2020, 2101, 2110, 2200.
A(4,3) = 31: 0013, 0022, 0031, 0103, 0112, 0121, 0130, 0202, 0211, 0220, 0301, 0310, 1003, 1012, 1021, 1030, 1102, 1111, 1120, 1201, 1210, 1300, 2002, 2011, 2020, 2101, 2110, 2200, 3001, 3010, 3100.
Square array A(n,k) begins:
  1, 1,    1,    1,    1,    1,    1,    1,    1, ...
  0, 1,    1,    1,    1,    1,    1,    1,    1, ...
  0, 1,    3,    3,    3,    3,    3,    3,    3, ...
  0, 1,    7,   10,   10,   10,   10,   10,   10, ...
  0, 1,   19,   31,   35,   35,   35,   35,   35, ...
  0, 1,   51,  101,  121,  126,  126,  126,  126, ...
  0, 1,  141,  336,  426,  456,  462,  462,  462, ...
  0, 1,  393, 1128, 1520, 1667, 1709, 1716, 1716, ...
  0, 1, 1107, 3823, 5475, 6147, 6371, 6427, 6435, ...

Alois P. Heinz, Antidiagonals n = 0..200

Columns k=0-10 give: A000007, A000012, A002426, A005725, A187925, A318113, A318114, A318115, A318116, A167403, A318117.
Rows n=0-1 give: A000012, A057427.
Main diagonal gives A088218 or A001700(n-1) for n>0.
A(n+1,n) gives A048775.
Cf. A180281.

A:= (n, k)-> coeff(series(((x^(k+1)-1)/(x-1))^n, x, n+1), x, n):
seq(seq(A(n, d-n), n=0..d), d=0..12);
# second Maple program:
b:= proc(n, i, k) option remember; `if`(n=0, 1,
      `if`(i=0, 0, add(b(n-j, i-1, k), j=0..min(n, k))))
    end:
A:= (n, k)-> b(n$2, k):
seq(seq(A(n, d-n), n=0..d), d=0..12);

b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i == 0, 0, Sum[b[n - j, i - 1, k], {j, 0, Min[n, k]}]]];
A[n_, k_] := b[n, n, k];
Table[A[n, d - n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 05 2019, after Alois P. Heinz *)

A(n,k) = [x^n] ((x^(k+1)-1)/(x-1))^n.
A(n,k) - A(n,k-1) = A180281(n,k) for n,k > 0.
A(n,k) = A(n,n) for all k >= n.

A213742 Triangle of numbers C^(3)(n,k) of combinations with repetitions from n different elements over k for each of them not more than three appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 31, 1, 5, 15, 35, 65, 101, 1, 6, 21, 56, 120, 216, 336, 1, 7, 28, 84, 203, 413, 728, 1128, 1, 8, 36, 120, 322, 728, 1428, 2472, 3823, 1, 9, 45, 165, 486, 1206, 2598, 4950, 8451, 13051, 1, 10

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

The left side of triangle consists of 1's, while the right side is formed by A005725. Further, T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A005718(n) for n>=2, T(n,5)=A005719(n) for n>=5, T(n,6)=A005720(n) for n>=6, T(n,7)=A001919(n) for n>=7, T(n,8)=A064055(n) for n>=5.

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....3
.3..|..1.....3.....6....10
.4..|..1.....4....10....20....31
.5..|..1.....5....15....35....65....101
.6..|..1.....6....21....56...120....216...336
.7..|..1.....7....28....84...203....413...728....1128

Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Cf. A007318, A005725, A111808, A000217, A000292, A005718, A005719, A005720, A001919, A064055.

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[4] (* Peter J. C. Moses, Apr 16 2013 *)

C^(3)(n,k)=sum{r=0,...,floor(k/4)}(-1)^r*C(n,r)*C(n-4*r+k-1, n-1)

A213743 Triangle T(n,k), read by rows, of numbers T(n,k)=C^(4)(n,k) of combinations with repetitions from n different elements over k for each of them not more than four appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 121, 1, 6, 21, 56, 126, 246, 426, 1, 7, 28, 84, 210, 455, 875, 1520, 1, 8, 36, 120, 330, 784, 1652, 3144, 5475, 1, 9, 45, 165, 495, 1278, 2922, 6030, 11385, 19855, 1, 10, 55, 220, 715, 1992, 4905, 10890, 22110, 41470, 72403

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

The left side of triangle consists of 1's, while the right side is formed by A187925. Further, T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A027659(n) for n>=3, T(n,6)=A064056(n) for n>=4, T(n,7)=A064057(n) for n>=5, T(n,8)=A064058(n) for n>=6, T(n,9)=A000575(n) for n>=6.

			Triangle begins
  n/k.|..0.....1.....2.....3.....4.....5.....6.....7
  ==================================================
  .0..|..1
  .1..|..1.....1
  .2..|..1.....2.....3
  .3..|..1.....3.....6....10
  .4..|..1.....4....10....20....35
  .5..|..1.....5....15....35....70....121
  .6..|..1.....6....21....56...126....246...426
  .7..|..1.....7....28....84...210....455...875....1520
T(4,2)=C^(4)(4,2): From 4 elements {1,2,3,4}, we have the following 10 allowed combinations of 2 elements: {1,1}, {1,2}, {1,3}, {1,4}, {2,2}, {2,3}, {2,4}, {3,3}, {3,4}, {4,4}.

Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Cf. A007318, A005725, A111808, A187925, A213742, A000217, A000292, A000332, A027659, A064056, A064057, A064058, A000575.

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[5] (* Peter J. C. Moses, Apr 16 2013 *)

C^(4)(n,k) = Sum_{r=0...floor(k/5)} (-1)^r*C(n,r)*C(n-5*r+k-1, n-1).

A213744 Triangle of numbers C^(5)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 5 appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 456, 1, 7, 28, 84, 210, 462, 917, 1667, 1, 8, 36, 120, 330, 792, 1708, 3368, 6147, 1, 9, 45, 165, 495, 1287, 2994, 6354, 12465, 22825, 1, 10

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

For k<=4, the triangle coincides with triangle A213743.

We have over columns of the triangle: T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A000389(n) for n>=9, T(n,6)=A062989(n) for n>=5, T(n,7)=A063262 for n>=5, T(n,8)=A063263 for n>=6, T(n,9)=A063264 for n>=7.

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....3
.3..|..1.....3.....6....10
.4..|..1.....4....10....20....35
.5..|..1.....5....15....35....70....126
.6..|..1.....6....21....56...126....252...456
.7..|..1.....7....28....84...210....462...917....1667

Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Cf. A007318, A005725, A111808, A187925, A213742, A213743, A000217, A000292, A000332, A000389, A062989, A063262, A063263, A063264.

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[6] (* Peter J. C. Moses, Apr 16 2013 *)

C^(5)(n,k)=sum{r=0,...,floor(k/6)}(-1)^r*C(n,r)*C(n-6*r+k-1, n-1)

A213745 Triangle of numbers C^(6)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 6 appearances allowed.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 10, 20, 35, 1, 5, 15, 35, 70, 126, 1, 6, 21, 56, 126, 252, 462, 1, 7, 28, 84, 210, 462, 924, 1709, 1, 8, 36, 120, 330, 792, 1716, 3424, 6371, 1, 9, 45, 165, 495, 1287, 3003, 6426, 12789, 23905, 1, 10

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Jun 19 2012

For k<=5, the triangle coincides with triangle A213744.

We have over columns of the triangle: T(n,0)=1, T(n,1)=n, T(n,2)=A000217(n) for n>1, T(n,3)=A000292(n) for n>=3, T(n,4)=A000332(n) for n>=7, T(n,5)=A000389(n) for n>=9, T(n,6)=A000579(n) for n>=11, T(n,7)=A063267 for n>=5, T(n,8)=A063417 for n>=6, T(n,9)=A063418 for n>=7.

			Triangle begins
n/k.|..0.....1.....2.....3.....4.....5.....6.....7
==================================================
.0..|..1
.1..|..1.....1
.2..|..1.....2.....3
.3..|..1.....3.....6....10
.4..|..1.....4....10....20....35
.5..|..1.....5....15....35....70....126
.6..|..1.....6....21....56...126....252...462
.7..|..1.....7....28....84...210....462...924....1709

Peter J. C. Moses, Rows n = 0..50 of triangle, flattened

Cf. A007318, A005725, A059481, A111808, A187925, A213742, A213743, A213744, A000217, A000292, A000332, A000389, A000579, A063267, A063417, A063418.

Flatten[Table[Sum[(-1)^r Binomial[n,r] Binomial[n-# r+k-1,n-1],{r,0,Floor[k/#]}],{n,0,15},{k,0,n}]/.{0}->{1}]&[7] (* Peter J. C. Moses, Apr 16 2013 *)

C^(6)(n,k)=sum{r=0,...,floor(k/7)}(-1)^r*C(n,r)*C(n-7*r+k-1, n-1).
A generalization. The numbers C^(t)(n,k) of combinations with repetitions from n different elements over k, for each of them not more than t>=1 appearances allowed, are enumerated by the formula:
C^(t)(n,k)=sum{r=0,...,floor(k/(t+1))}(-1)^r*C(n,r)*C(n-(t+1)*r+k-1, n-1).
In case t=1, it is binomial coefficient C^(t)(n,k)=C(n,k), and we have the combinatorial identity: sum{r=0,...,floor(k/2)}(-1)^r*C(n,r)*C(n-2*r+k-1, n-1)=C(n,k). On the other hand, if t=n, then r=0, and for the corresponding numbers of combinations with repetitions without a restriction on appearances of elements we obtain a well known formula C(n+k-1, n-1) (cf. triangle A059481).
In addition, note that, if k<=t, then C^(t)(n,k)=C(n+k-1, n-1). Therefore, triangle {C^(t+1)(n,k)} coincides with the previous triangle {C^(t)(n,k)} for k<=t.

A245195 a(n) = 2^A014081(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 2, 2, 2, 4, 2

Offset: 0

N. J. A. Sloane, Jul 24 2014

This sequence provides a bridge between A245180 (and, presumably, A160239) and A014081.
See A245196 for more about this class of sequences.
Run length transform of A011782: 1,1,2,4,8,16,32,64,... - Chai Wah Wu, Oct 19 2016

Chai Wah Wu and Robert Israel, Table of n, a(n) for n = 0..10000
Robert Israel, Proof that A277560 is the same as A245195 [This will be modified to reflect the fact that the two sequences have now been merged]
Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166 [math.CO], 2016.

Cf. A014081, A245180, A160239, A048896, A245196, A005725, A005940, A011782, A106737, A162510.

# This Maple program applies more generally to a sequence where the recurrence across a block is as follows. The parameters to be set are the sequence G(0), G(1), G(2), ... (the final terms in the blocks), and the multiplier m.
# For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
# (if j=0) a(2^k-2^r) = G(k-r-1),
# (if j>0) a(2^k-2^r+j) = m*G(k-r-1)*a(j).
# Since Maple gives its lists an offset of 1, it is necessary to add 1 to the arguments of G.
# For the present sequence, G(n)=2^n and m=1.
G:=[seq(2^n,n=0..30)];
m:=1;
f:=proc(n) option remember; global m,G; local k,r,j,np;
if n <= 2 then G[0+1] elif n=3 then G[1+1]
elif n=4 then G[0+1] elif n=5 then m*G[0+1] elif n=6 then G[1+1] elif n=7 then G[2+1]
else
   k:=1+floor(log[2](n)); np:=2^k-n;
   if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
   if j=0 then G[k-r-1+1]; else m*G[k-r-1+1]*f(j); fi;
fi;
end;
[seq(f(n),n=1..520)]:
# Setting G(n) = A083424(n) and m = 8 gives A245180. Setting G(n) = 2^n and m = 2 gives A048896.
A245195:=n->add(binomial(n,2*k)*binomial(n,k) mod 2, k=0..floor(n/2)): seq(A245195(n), n=0..200); # Wesley Ivan Hurt, Nov 01 2016

Table[Sum[Mod[Binomial[n, 2 k] Binomial[n, k], 2], {k, 0, n}], {n, 0, 85}] (* Michael De Vlieger, Oct 21 2016 *)

a(n) = 2^hammingweight(bitand(n, n>>1)) \\ Charles R Greathouse IV, Jul 16 2016

a(n) = sum(k=0, n, binomial(n, 2*k)*binomial(n,k) % 2); \\ Michel Marcus, Oct 21 2016

from _future_ import division
def A277560(n):
    return sum(int(not (~n & 2*k) | (~n & k)) for k in range(n//2+1))

def A245195(n): return 1<<(n&(n>>1)).bit_count() # Chai Wah Wu, Feb 11 2023

The entries may be arranged into blocks of sizes 1,2,4,8,...:
B_0: 1,
B_1: 1, 2,
B_2: 1, 1, 2, 4,
B_3: 1, 1, 1, 2, 2, 2, 4, 8,
B_4: 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16,
B_5: 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32,
...
Consider the block B_{k-1} containing terms a(2^(k-1)), a(2^(k-1)+1), ..., a(2^k-1). It is convenient to index the terms working backwards from the next, 2^k-th, term. For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
(if j=0) a(2^k-2^r) = 2^(k-r-1),
(if j>0) a(2^k-2^r+j) = 2^(k-r-1)*a(j).
a(n) = A162510(A005940(1+n)). - Antti Karttunen, Oct 29 2016
From Robert Israel, Nov 02 2016: (Start)
a(2*k)   = a(k).
a(4*k+1) = a(k).
a(4*k+3) = 2*a(2*k+1).
G.f. g(x) satisfies g(x) = x + (2*x+1)*g(x^2) - x*g(x^4). (End)
Also, a(n) = Sum_{k=0..floor(n/2)} ((binomial(n,2k)*binomial(n,k)) mod 2). - Chai Wah Wu, Oct 19 2016 and Robert Israel, Nov 04 2016. For proof, see the article by Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166, or the Robert Israel link.

Changed offset to 0, merged former entry A277560 from Chai Wah Wu (Oct 19 2016) with this sequence. - N. J. A. Sloane, Nov 05 2016

A350406 a(n) = [x^n] 1/(1 + x + x^2 + x^3)^n.

Original entry on oeis.org

1, -1, 1, -1, 5, -26, 91, -246, 597, -1540, 4576, -14521, 44915, -132328, 380290, -1102076, 3268437, -9838428, 29616364, -88538500, 263489380, -785026110, 2348923875, -7053379710, 21204016275, -63716916276, 191394838116, -575200476046, 1730575897202

Offset: 0

Seiichi Manyama, Dec 29 2021

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A005725, A348410, A350383, A350407.

a := proc (n) option remember; if n = 0 then 1 elif n = 1 then -1 elif n = 2 then 1 else -(4*n*(n-1)*(5*n-9)*(550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3+33771*n^2-20232*n+4032)*a(n-2) + (16*n-24)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3))/(16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)) end if; end:
seq(a(n), n = 0..30); # Peter Bala, Mar 18 2023
a := n -> (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1): seq(simplify(a(n)), n = 0..28); # Peter Luschny, Mar 19 2023

a[n_] := Coefficient[Series[1/(1 + x + x^2 + x^3)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)

a(n) = (-1)^n*sum(k=0, n, binomial(n-1+k, k)*binomial(n, 4*k));

a(n) = (-1)^n * Sum_{k=0..n} binomial(n-1+k,k) * binomial(n,4*k).
From Peter Bala, Mar 17 2023: (Start)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^(n-k)*binomial(n+k-1,k)*binomial(2*n-2*k-1,n-2*k).
16*n*(n+1)*(n-1)*(5*n-4)*(5*n-8)*(5*n-9)*a(n) = - ( 4*n*(n-1)*(5*n-9)* (550*n^3-1045*n^2+164*n+223)*a(n-1) + (n-1)*(5*n+1)*(5075*n^4-22330*n^3 +33771*n^2-20232*n+4032)*a(n-2) + 8*(2*n-3)*(4*n-5)*(4*n-7)*(5*n+1)*(5*n-3)*(5*n-4)*a(n-3) ).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. (End)
a(n) = (-1)^n*hypergeom([1/4 - n/4, 1/2 - n/4, 3/4 - n/4, -n/4, n], [1/4, 1/2, 3/4, 1], 1). - Peter Luschny, Mar 19 2023

cp's OEIS Frontend

A386650 Product of first n quadrinomial coefficients (A005725) for n > 0 with a(0) = 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A036765 Number of ordered rooted trees with n non-root nodes and all outdegrees <= three.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

PARI

PARI

PARI

Formula

Extensions

A305161 Number A(n,k) of compositions of n into exactly n nonnegative parts <= k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A213742 Triangle of numbers C^(3)(n,k) of combinations with repetitions from n different elements over k for each of them not more than three appearances allowed.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A213743 Triangle T(n,k), read by rows, of numbers T(n,k)=C^(4)(n,k) of combinations with repetitions from n different elements over k for each of them not more than four appearances allowed.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A213744 Triangle of numbers C^(5)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 5 appearances allowed.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A213745 Triangle of numbers C^(6)(n,k) of combinations with repetitions from n different elements over k for each of them not more than 6 appearances allowed.

Views