A108447 -id:A108447 - cp's OEIS frontend

A367233 G.f. satisfies A(x) = 1 + xA(x)^3 / (1 - xA(x))^3.

1, 1, 6, 39, 284, 2223, 18267, 155445, 1358073, 12111306, 109802183, 1009001571, 9376972698, 87978198364, 832223905371, 7928413841673, 76002832317437, 732578811761670, 7095717550127526, 69029297500888522, 674181392461483212, 6607910786529613248

Offset: 0

Seiichi Manyama, Nov 11 2023

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..983

Cf. A161797, A365113, A365150.

Cf. A108447, A367232, A367235.

a(n, s=3, t=3, u=1) = sum(k=0, n, binomial(t*k+u*(n-k)+1, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+1));

If g.f. satisfies A(x) = 1 + x*A(x)^t / (1 - x*A(x)^u)^s, then a(n) = Sum_{k=0..n} binomial(t*k+u*(n-k)+1,k) * binomial(n+(s-1)*k-1,n-k) / (t*k+u*(n-k)+1).

A367235 G.f. satisfies A(x) = 1 + xA(x)^3 / (1 - xA(x))^4.

Original entry on oeis.org

1, 1, 7, 50, 399, 3422, 30798, 286974, 2744947, 26798010, 265945022, 2674970684, 27209385886, 279412999031, 2892787737002, 30161921520976, 316440334960563, 3338105334701396, 35385133077851602, 376732207920371784, 4026682585718602014

Offset: 0

Seiichi Manyama, Nov 11 2023

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..946

Cf. A321798, A365114, A367234.

Cf. A108447, A367232, A367233.

a(n, s=4, t=3, u=1) = sum(k=0, n, binomial(t*k+u*(n-k)+1, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+1));

If g.f. satisfies A(x) = 1 + x*A(x)^t / (1 - x*A(x)^u)^s, then a(n) = Sum_{k=0..n} binomial(t*k+u*(n-k)+1,k) * binomial(n+(s-1)*k-1,n-k) / (t*k+u*(n-k)+1).

A378685 G.f. A(x) satisfies A(x) = 1 + xA(x)^7/(1 - xA(x)^3).

Original entry on oeis.org

1, 1, 8, 88, 1126, 15716, 232069, 3564835, 56382489, 912031280, 15018257510, 250913307393, 4242722219425, 72470224174650, 1248608968982903, 21673752440979879, 378677335852165297, 6654158090059397480, 117523324766568499072, 2085095374834405245007

Offset: 0

Seiichi Manyama, Dec 04 2024

nonn

Cf. A243659, A300048.
Cf. A108447, A365192.
Cf. A378686.

a(n, r=1, s=1, t=7, u=3) = r*sum(k=0, n, binomial(t*k+u*(n-k)+r, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+r));

G.f. A(x) satisfies A(x) = 1/(1 - x*A(x)^6/(1 - x*A(x)^3)).

If g.f. satisfies A(x) = ( 1 + x*A(x)^(t/r) / (1 - x*A(x)^(u/r))^s )^r, then a(n) = r * Sum_{k=0..n} binomial(t*k+u*(n-k)+r,k) * binomial(n+(s-1)*k-1,n-k)/(t*k+u*(n-k)+r).

A378691 G.f. A(x) satisfies A(x) = 1 + xA(x)^6/(1 - xA(x)).

Original entry on oeis.org

1, 1, 7, 65, 699, 8192, 101538, 1309007, 17373825, 235820907, 3258327727, 45676003435, 648019627185, 9286982935406, 134247731827970, 1955128344950960, 28659409029300490, 422517375650417841, 6260750426764454787, 93191618760715641120, 1392823412892172416996

Offset: 0

Seiichi Manyama, Dec 04 2024

nonn

Cf. A000108, A001003, A108447, A364747, A364748, A378692.

a(n, r=1, s=1, t=6, u=1) = r*sum(k=0, n, binomial(t*k+u*(n-k)+r, k)*binomial(n+(s-1)*k-1, n-k)/(t*k+u*(n-k)+r));

G.f. A(x) satisfies A(x) = 1/(1 - x*A(x)^5/(1 - x*A(x))).

If g.f. satisfies A(x) = ( 1 + x*A(x)^(t/r) / (1 - x*A(x)^(u/r))^s )^r, then a(n) = r * Sum_{k=0..n} binomial(t*k+u*(n-k)+r,k) * binomial(n+(s-1)*k-1,n-k)/(t*k+u*(n-k)+r).

A108446 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, 4, 5, 1, 20, 32, 13, 1, 113, 223, 135, 26, 1, 688, 1620, 1300, 412, 45, 1, 4404, 12064, 12050, 5350, 1030, 71, 1, 29219, 91335, 109134, 62450, 17575, 2247, 105, 1, 199140, 699689, 973077, 682234, 254625, 49210, 4438, 148, 1, 1385904, 5407744

Offset: 0

Emeric Deutsch, Jun 10 2005

Row sums yield A027307. Column 0 yields A108447. T(n,n-1) = A008778(n-1) = n(n^2+6n-1)/6. Number of ud peaks in all paths from (0,0) to (3n,0) is given by A108448.

			T(2,1) = 5 because we have udUdd, uudd, Uddud, Ududd and Uuddd.
Triangle begins:
1;
1,1;
4,5,1;
20,32,13,1;
113,223,135,26,1;

Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Cf. A027307, A008778, A108447, A108448, A108425, A108426.

T:=proc(n,k) if n=0 and k=0 then 1 elif n=0 then 0 elif k=n then 1 elif k=n then 1 else (1/n)*binomial(n,k)*sum(binomial(n-k,j)*binomial(n+2*j,k+j-1),j=0..n-k) fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[0, 0] = 1; T[n_, k_] := (1/n) Binomial[n, k]*Sum[Binomial[n-k, j]* Binomial[n+2j, k+j-1], {j, 0, n-k}];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 19 2018 *)

T(n,k) = (1/n) binomial(n, k)*sum(binomial(n-k,j)*binomial(n+2j,k+j-1), j=0..n-k).

G.f.: G = G(t,z) satisfies G = 1+z(G-1+t)G+zG^3.

A078623 Number of matched parentheses and brackets of length n, where a closing bracket will close any remaining open parentheses back to the matching open bracket (as in some versions of LISP).

Original entry on oeis.org

1, 0, 2, 1, 9, 11, 56, 106, 421, 1009, 3565, 9736, 32594, 95811, 313535, 961780, 3123577, 9831373, 31915121, 102110314, 332366526, 1075228773, 3513373374, 11456961550, 37590603312, 123327267531, 406246177511, 1339274997451, 4427777075497, 14655559052686

Offset: 0

Brian T. Howard (bhoward(AT)depauw.edu), Dec 11 2002

An unambiguous context-free grammar generating valid strings from S is S -> ( S ) S | [ T ] S | e T -> ( T | ( S ) T | [ T ] T | e

			a(5) = 11 because the valid strings of length 5 are ()[(], [(](), [(][], [][(], ([(]), [(()], [()(], [(((], [([]], [[(]] and [[](].

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for sequences related to parenthesizing

Cf. A000108, A001003, A001006, A108447.

a:= proc(n) option remember; `if`(n<4, [1, 0, 2, 1][n+1],
      (4*(n+1)*(14*n^3-9*n^2-62*n+39) *a(n-1)
      +(140*n^4-160*n^3-401*n^2+469*n-78) *a(n-2)
      -12*(n-2)*(14*n^3-9*n^2-28*n-8) *a(n-3)
      +23*(n-2)*(n-3)*(28*n^2+24*n-43) *a(n-4))/
      ((n+2)*(n+1)*(28*n^2-32*n-39)))
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, May 19 2014

a[n_] := Sum[Binomial[n+1, j]*Sum[(-1)^(i+j)*Binomial[j, i]*Binomial[2*n-2*j-i, n-i-j], {i, 0, n-j}], {j, 0, n+1}]/(n+1); Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Apr 03 2015, after Vladimir Kruchinin *)
nmax = 40; A[] = 1; Do[A[x] = 1 - x*A[x] + x*(1 + 2*x)*A[x]^2 - x^3*A[x]^3 + O[x]^nmax // Normal, {nmax}]; CoefficientList[A[x], x] (* Vaclav Kotesovec, May 09 2019 *)

a(n):=sum(binomial(n+1,j)*sum((-1)^(i+j)*binomial(j,i)*binomial(2*n-2*j-i,n-i-j),i,0,n-j),j,0,n+1)/(n+1); /* Vladimir Kruchinin, May 19 2014 */

a(0) = 1, a(n) = Sum_{i=0..n-2} a(n-2-i)*(a(i) + b(i)), where b(0) = 1, b(n) = b(n-1) + Sum_{i=0..n-2} b(n-2-i)*(a(i) + b(i)).
a(n) = (Sum_{j=0..n+1} C(n+1,j)*Sum_{i=0..n-j} (-1)^(i+j)*C(j,i)*C(2*n-2*j-i,n-i-j)) / (n+1). - Vladimir Kruchinin, May 19 2014
a(n) ~ c * ((1+sqrt(13+16*sqrt(2)))/2)^n / n^(3/2), where c = sqrt(1 + 9/(8*sqrt(2)) - sqrt(211/224 + 43/(7*sqrt(2)))/2) / sqrt(Pi) = 0.453452365404498112381472576661214848447318569684502125279149391488... . - Vaclav Kotesovec, Aug 25 2014, updated May 09 2019
From Peter Bala, Oct 23 2015: (Start)
Conjecturally, a(n-1) = (-1)^(n-1)*(1/n)*Sum_{k=1..n} binomial(n,k)*binomial(n - 2*k,k - 1).
The formula (1/n)*Sum_{k=1..n} binomial(n,k)*binomial(n + m*k,k - 1) gives A001006 (m = -1), A000108 (m = 0), A001003 (m = 1) and A108447 (m = 2).
(End)
G.f. A(x) satisfies -1 + (1+x)*A(x) - x*(1+2*x)*A(x)^2 + x^3*A(x)^3 = 0. - Vaclav Kotesovec, May 09 2019

A262410 a(n) = (n+1)Sum_{k=1..n} binomial(n-1,k-1)binomial(n+2*k+2,k+1)/(n+k+2).

Original entry on oeis.org

5, 37, 275, 2071, 15781, 121395, 940915, 7337560, 57507892, 452598884, 3574599205, 28316957579, 224901946395, 1790287826789, 14279629073403, 114097695427295, 913103420246956, 7317725618907700, 58719917176448820, 471733089071984376

Offset: 1

Vladimir Kruchinin, Sep 22 2015

nonn

Cf. A108447.

Table[(n + 1) Sum[Binomial[n - 1, k - 1] Binomial[n + 2 k + 2, k + 1]/(n + k + 2), {k, n}], {n, 20}] (* Michael De Vlieger, Sep 22 2015 *)

b(n):=sum((binomial(n-1,n-k)*binomial(2*k+n,k))/(n+k+1),k,0,n);
B(x):=sum(b(n)*x^n,n,0,30);
taylor(diff(B(x),x,1)/B(x)-x*(diff(B(x),x,1))-B(x),x,0,10);

a(n) = (n+1)*sum(k=1,n,((binomial(n-1,k-1) *binomial(n+2*k+2,k+1))/(n+k+2))) \\ Anders Hellström, Sep 22 2015

G.f.: B'(x)/B(x)-x*B'(x)-B(x), where B(x) is g.f. of A108447.

Recurrence: 2*n*(n+1)*(2*n + 1)*(74*n^3 - 183*n^2 + 129*n - 18)*a(n) = 2*n*(1184*n^5 - 1744*n^4 - 242*n^3 + 857*n^2 - 201*n + 26)*a(n-1) + 2*(n-2)*(296*n^5 + 8*n^4 - 410*n^3 + 35*n^2 + 29*n - 18)*a(n-2) - 5*(n-3)*(n-2)*(n+2)*(74*n^3 + 39*n^2 - 15*n + 2)*a(n-3). - Vaclav Kotesovec, Sep 22 2015

A263917 Riordan array (f(x)^3, f(x)), where 1 + xf^3(x)/(1 - xf(x)) = f(x).

Original entry on oeis.org

1, 3, 1, 15, 4, 1, 85, 22, 5, 1, 519, 132, 30, 6, 1, 3330, 837, 190, 39, 7, 1, 22135, 5516, 1250, 260, 49, 8, 1, 151089, 37404, 8461, 1773, 343, 60, 9, 1, 1052805, 259280, 58550, 12324, 2422, 440, 72, 10, 1, 7458236, 1829018, 412375, 87045, 17283, 3214, 552, 85, 11, 1

Offset: 0

Peter Bala, Oct 29 2015

Riordan arrays of the form (f(x)^(m+1), f(x)), where f(x) satisfies 1 + x*f^(m+1)(x)/(1 - x*f(x)) = f(x) include (modulo differences of offset) the Motzkin triangle A091836 (m = -1), the Catalan triangle A033184 (m = 0) and the Schroder triangle A091370 (m = 1). This is the case m = 2. See A263918 for the case m = 3.
The coefficients of the power series solution of the equation 1 + x*f^(m+1)(x)/(1 - x*f(x)) = f(x) appear to be given by [x^0] f(x) = 1 and [x^n] f(x) = 1/n * Sum_{k = 1..n} binomial(n,k)*binomial(n + m*k, k - 1) for n >= 1.
This triangle appears in Novelli et al., Figure 8, p. 24, where a combinatorial interpretation is given in terms of trees.

			Triangle begins:
       1
       3     1
      15     4     1
      85    22     5    1
     519   132    30    6   1
    3330   837   190   39   7  1
   22135  5516  1250  260  49  8 1
  151089 37404  8461 1773 343 60 9 1

J.-C. Novelli and J.-Y. Thibon, Noncommutative Symmetric Functions and Lagrange Inversion, arXiv:math/0512570 [math.CO], 2005-2006.

Cf. A108447 (row sums), A118342 (column 0).

Cf. A033184, A091370, A091836, A263918.

# For the function TreesByArityOfTheRoot_Row(m, n) see A263918.
A263917_row := n -> TreesByArityOfTheRoot_Row(2,n):
seq(A263917_row(n), n=0..9); # Peter Luschny, Oct 31 2015

rows = 9;
f[] = 1; Do[f[x] = 1 + x*f[x]*(f[x]^2 + f[x] - 1) + O[x]^(rows+1) // Normal, {rows+1}];
coes = CoefficientList[f[x]^3/(1 - x*t*f[x]) + O[x]^(rows+1), x];
row[n_] := CoefficientList[coes[[n+1]], t];
Table[row[n], {n, 0, rows}] // Flatten (* Jean-François Alcover, Jul 19 2018 *)

O.g.f. f^3(x)/(1 - x*t*f(x)), where f(x)  = 1 + x + 4*x^2 + 20*x^3 + 113*x^4 + ... satisfies 1 + x*f^3(x)/(1 - x*f(x)) = f(x);
f(x) is the o.g.f. for A108447.
First column o.g.f f(x)^3 is the o.g.f. for A118342.
f(x) - 1 is the g.f. for the row sums of the array.

A371888 G.f. A(x) satisfies A(x) = 1 - x/A(x) * (1 - A(x) - A(x)^2).

Original entry on oeis.org

1, 1, 2, 3, 3, 1, -2, -1, 10, 25, 12, -65, -151, -7, 588, 1083, -437, -5247, -7732, 7943, 47503, 53793, -105312, -430117, -343042, 1249801, 3866558, 1730019, -13996095, -34243895, -1947202, 150962375, 296101866, -121857183, -1582561868, -2468098041, 2529520767

Offset: 0

Seiichi Manyama, Apr 11 2024

sign

Cf. A002212, A108447, A364792, A365193.

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

a(n) = (1/n) * Sum_{k=0..n} binomial(n,k) * binomial(n-2*k,n-k-1) for n > 0.
a(n) = (1/2) * Sum_{k=0..n} 4^k * binomial(k/2+1/2,k) * binomial(n-1,n-k)/(k+1) for n > 0.
G.f.: A(x) = 2*x/(1+x - sqrt(1-2*x+5*x^2)).
D-finite with recurrence n*a(n) +3*(-n+1)*a(n-1) +(7*n-18)*a(n-2) +5*(-n+3)*a(n-3)=0. - R. J. Mathar, Apr 22 2024

cp's OEIS Frontend

A367233 G.f. satisfies A(x) = 1 + x*A(x)^3 / (1 - x*A(x))^3.

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A367235 G.f. satisfies A(x) = 1 + x*A(x)^3 / (1 - x*A(x))^4.

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A378685 G.f. A(x) satisfies A(x) = 1 + x*A(x)^7/(1 - x*A(x)^3).

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A378691 G.f. A(x) satisfies A(x) = 1 + x*A(x)^6/(1 - x*A(x)).

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A108446 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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A078623 Number of matched parentheses and brackets of length n, where a closing bracket will close any remaining open parentheses back to the matching open bracket (as in some versions of LISP).

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A262410 a(n) = (n+1)*Sum_{k=1..n} binomial(n-1,k-1)*binomial(n+2*k+2,k+1)/(n+k+2).

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A263917 Riordan array (f(x)^3, f(x)), where 1 + x*f^3(x)/(1 - x*f(x)) = f(x).

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A367233 G.f. satisfies A(x) = 1 + xA(x)^3 / (1 - xA(x))^3.

A367235 G.f. satisfies A(x) = 1 + xA(x)^3 / (1 - xA(x))^4.

A378685 G.f. A(x) satisfies A(x) = 1 + xA(x)^7/(1 - xA(x)^3).

A378691 G.f. A(x) satisfies A(x) = 1 + xA(x)^6/(1 - xA(x)).

A262410 a(n) = (n+1)Sum_{k=1..n} binomial(n-1,k-1)binomial(n+2*k+2,k+1)/(n+k+2).

A263917 Riordan array (f(x)^3, f(x)), where 1 + xf^3(x)/(1 - xf(x)) = f(x).