A001761 -id:A001761 - cp's OEIS frontend

A349598 E.g.f. satisfies: log(A(x)) = exp(x*A(x)^2) - 1.

1, 1, 6, 71, 1279, 31142, 958127, 35674921, 1560207964, 78410153193, 4453247964775, 282086867840252, 19718661737739301, 1507855981764016549, 125211854842018500134, 11220898483255456505555, 1079389691811367897870339, 110936313685240067472613726

Offset: 0

Seiichi Manyama, Nov 22 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..343

Cf. A001761, A001763, A349599.

Cf. A349524, A349601.

a[n_] := Sum[(2*n + 1)^(k - 1)*StirlingS2[n, k], {k, 0, n}]; Array[a, 18, 0] (* Amiram Eldar, Nov 23 2021 *)

a(n) = sum(k=0, n, (2*n+1)^(k-1)*stirling(n, k, 2));

a(n) = Sum_{k=0..n} (2*n+1)^(k-1) * Stirling2(n,k).
a(n) ~ s * n^(n-1) / (2 * sqrt(1 + r*s^2) * exp(n) * r^n), where r = 0.1513832219344136560178112221696108323993292386502... and s = 1.52429184135463908701026733917578550814344591549... are roots of the system of equations (1 + log(s))*2*r*s^2 = 1, 2*r*s^2*exp(r*s^2) = 1. - Vaclav Kotesovec, Nov 25 2021
Equivalently, a(n) ~ n^(n-1) / (2*sqrt(1 + LambertW(1/2)) * LambertW(1/2)^n * exp(3*n + 1 - (n + 1/2)/LambertW(1/2))). - Vaclav Kotesovec, Nov 26 2021

A365340 a(n) = (4n)!/(3n+1)!.

Original entry on oeis.org

1, 1, 8, 132, 3360, 116280, 5100480, 271252800, 16963914240, 1220096908800, 99225500774400, 9003984596006400, 901928094049382400, 98856066097780992000, 11768525894839633920000, 1512185803617951221760000, 208598907329474462760960000

Offset: 0

Seiichi Manyama, Sep 01 2023

Cf. A001761, A001763, A052795, A365341.
Cf. A004331, A061924.
Cf. A370056, A370057, A370058.
Cf. A000142, A002293.

PARI
```
a(n) = (4*n)!/(3*n+1)!;
```

from sympy import ff
def A365340(n): return ff(n<<2,n-1) # Chai Wah Wu, Sep 01 2023

E.g.f.: exp( 1/4 * Sum_{k>=1} binomial(4*k,k) * x^k/k ). - Seiichi Manyama, Feb 08 2024
a(n) = A000142(n)*A002293(n). - Alois P. Heinz, Feb 08 2024
From Seiichi Manyama, Aug 31 2024: (Start)
E.g.f. satisfies A(x) = 1/(1 - x*A(x)^3).
a(n) = Sum_{k=0..n} (3*n+1)^(k-1) * |Stirling1(n,k)|. (End)

A365341 a(n) = (5n)!/(4n+1)!.

Original entry on oeis.org

1, 1, 10, 210, 6840, 303600, 17100720, 1168675200, 93963542400, 8691104822400, 909171781056000, 106137499051584000, 13679492361575040000, 1929327666754295808000, 295570742023171270656000, 48877281133334949335040000, 8677556868736487617966080000

Offset: 0

Seiichi Manyama, Sep 01 2023

Cf. A001761, A001763, A052795, A365340.
Cf. A004343.
Cf. A000142, A002294.

PARI
```
a(n) = (5*n)!/(4*n+1)!;
```

from sympy import ff
def A365341(n): return ff(5*n,n-1) # Chai Wah Wu, Sep 01 2023

E.g.f.: exp( 1/5 * Sum_{k>=1} binomial(5*k,k) * x^k/k ). - Seiichi Manyama, Feb 08 2024
a(n) = A000142(n)*A002294(n). - Alois P. Heinz, Feb 08 2024
From Seiichi Manyama, Aug 31 2024: (Start)
E.g.f. satisfies A(x) = 1/(1 - x*A(x)^4).
a(n) = Sum_{k=0..n} (4*n+1)^(k-1) * |Stirling1(n,k)|. (End)

A256061 Number T(n,k) of 2n-length strings of balanced parentheses of exactly k different types; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 2, 4, 0, 5, 30, 30, 0, 14, 196, 504, 336, 0, 42, 1260, 6300, 10080, 5040, 0, 132, 8184, 71280, 205920, 237600, 95040, 0, 429, 54054, 774774, 3603600, 7207200, 6486480, 2162160, 0, 1430, 363220, 8288280, 58378320, 180180000, 273873600, 201801600, 57657600

Offset: 0

Alois P. Heinz, Mar 13 2015

Also number of binary trees with n inner nodes of exactly k different dimensions.  T(2,2) = 4:
: balanced parentheses :  ([]) :  [()] : ()[]  : []()  :
:----------------------:-------:-------:-------:-------:
:                trees :   (1) :   [2] : (1)   : [2]   :
:                      :   / \ :   / \ : / \   : / \   :
:                      : [2]   : (1)   :   [2] :   (1) :
:                      : / \   : / \   :   / \ :   / \ :

			A(3,2) = 30: (())[], (()[]), (([])), ()()[], ()([]), ()[()], ()[[]], ()[](), ()[][], ([()]), ([[]]), ([]()), ([])(), ([])[], ([][]), [(())], [()()], [()[]], [()](), [()][], [([])], [[()]], [[]()], [[]](), [](()), []()(), []()[], []([]), [][()], [][]().
Triangle T(n,k) begins:
  1;
  0,   1;
  0,   2,     4;
  0,   5,    30,     30;
  0,  14,   196,    504,     336;
  0,  42,  1260,   6300,   10080,    5040;
  0, 132,  8184,  71280,  205920,  237600,   95040;
  0, 429, 54054, 774774, 3603600, 7207200, 6486480, 2162160;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Columns k=0-1 give: A000007, A000108 (for n>0).
Main diagonal gives A001761.
Cf. A253180, A255982, A258427, A290605.

ctln:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
A:= proc(n, k) option remember; k^n*ctln(n) end:
T:= (n, k)-> add(A(n, k-i)*(-1)^i*binomial(k, i), i=0..k):
seq(seq(T(n, k), k=0..n), n=0..10);

A[0, 0] = 1; A[n_, k_] := A[n, k] = k^n*CatalanNumber[n]; T[n_, k_] := Sum[A[n, k-i]*(-1)^i*Binomial[k, i], {i, 0, k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 20 2017, translated from Maple *)

T(n,k) = Sum_{i=0..k} (-1)^i * C(k,i) * (k-i)^n * A000108(n).
T(n,k) = k! * A253180(n,k).
T(n,k) = Sum_{i=0..k} (-1)^i * C(k,i) * A290605(n,k-i). - Alois P. Heinz, Oct 28 2019

A355766 E.g.f. satisfies A(x) = 1/(1 - x*A(x))^(A(x)^2).

Original entry on oeis.org

1, 1, 8, 126, 3028, 98540, 4056948, 202301456, 11855415920, 798682318848, 60823290655680, 5167260183157248, 484519323081722784, 49705696509114472320, 5537956421036240838336, 665926312161296782156800, 85960998514145805006711552

Offset: 0

Seiichi Manyama, Jul 16 2022

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 0..328

Cf. A001761, A349556.

Table[Sum[(n + 2*k + 1)^(k - 1)* Abs[StirlingS1[n, k]], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jul 21 2022 *)

a(n) = sum(k=0, n, (n+2*k+1)^(k-1)*abs(stirling(n, k, 1)));

a(n) = Sum_{k=0..n} (n+2*k+1)^(k-1) * |Stirling1(n,k)|.

a(n) ~ s^(5/2) * n^(n-1) * sqrt((1 - r*s)/(2 - 4*r*s + 2*r^2*s^2 + 3*r*s^3 - 2*r^2*s^4)) / (exp(n) * r^(n - 1/2)), where r = 0.11275303067590951818975824... and s = 1.382171434168172073998532... are real roots of the system of equations (1 - r*s)^(s^2) = 1/s, r*s/(1 - r*s) = 1/s^2 + 2*log(1 - r*s). - Vaclav Kotesovec, Jul 21 2022

A038455 Triangle read by rows: T(n, k) = [x^k] x*Pochhammer(n + x, n)/(n + x).

Original entry on oeis.org

1, 3, 1, 20, 9, 1, 210, 107, 18, 1, 3024, 1650, 335, 30, 1, 55440, 31594, 7155, 805, 45, 1, 1235520, 725592, 176554, 22785, 1645, 63, 1, 32432400, 19471500, 4985316, 705649, 59640, 3010, 84, 1, 980179200, 598482000, 159168428, 24083892, 2267769, 136080, 5082, 108, 1

Offset: 1

Wolfdieter Lang

Original name: A Jabotinsky-triangle related to A006963.
i) This triangle gives the nonvanishing entries of the Jabotinsky matrix for F(z) = c(z) with c(z) the g.f. for the Catalan numbers A000108. (Notation of F(z) as in Knuth's paper).
ii) E(n, x) = Sum_{m=1..n} a(n, m)*x^m, E(0, x) = 1, are exponential convolution polynomials: E(n, x + y) = Sum_{k=0..n} binomial(n, k)*E(k, x)*E(n-k, y), (cf. Knuth's paper with E(n, x)= n!*F(n, x).)
iii) Explicit formula: see Knuth's paper for f(n, m) formula with f(k) = A006963(k + 1).
Bell polynomial of second kind for log(A000108(x)). - Vladimir Kruchinin, Mar 26 2013
Also the Bell transform of A006963(n+2). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 28 2016

			Triangle starts:
  [1]       1;
  [2]       3,      1;
  [3]      20,      9,      1;
  [4]     210,    107,     18,     1;
  [5]    3024,   1650,    335,    30,    1;
  [6]   55440,  31594,   7155,   805,   45,  1;
  [7] 1235520, 725592, 176554, 22785, 1645, 63, 1;

Priyavrat Deshpande and Krishna Menon, A statistic for regions of braid deformations, Séminaire Lotharingien de Combinatoire (2022) Vol. 86, Issue B, Art. No. 23.
D. E. Knuth, Convolution polynomials, arXiv:math/9207221 [math.CA], 1992; Mathematica J. 2.1 (1992), no. 4, 67-78.
J.-C. Novelli and J.-Y. Thibon, Noncommutative Symmetric Functions and Lagrange Inversion, arXiv:math/0512570 [math.CO], 2005-2006.

Cf. A006963, A000108, A001761, A039619, A039646.

# The function BellMatrix is defined in A264428.
# Adds (1,0,0,0, ..) as column 0.
BellMatrix(n -> (2*n+1)!/(n+1)!, 9); # Peter Luschny, Jan 28 2016
gf := n -> x*pochhammer(n + x, n)/(n + x):
ser := n -> series(gf(n), x, n + 2):
seq(seq(coeff(ser(n), x, k), k = 1..n), n = 1..9);  # Peter Luschny, Jun 27 2024

BellMatrix[f_Function, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 11; M = BellMatrix[(2#+1)!/(#+1)!&, rows];
Table[M[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten
(* Jean-François Alcover, Jun 24 2018, after Peter Luschny *)

a(n,m):=(n-1)!*(sum((stirling1(k,m)*binomial(2*n,n-k))/(k-1)!,k,m,n)); /* Vladimir Kruchinin, Mar 26 2013 */

a(n, 1) = A006963(n + 1) = (2*n - 1)!/n!, n >= 1;
a(n, m) = Sum_{j=1..n-m+1} binomial(n - 1, j - 1)*A006963(j + 1)*a(n - j, m - 1), n >= m >= 2.
E.g.f.: ((1 - sqrt(1 - 4*x))/(2*x))^y. - Vladeta Jovovic, May 02 2003
a(n, m) = (n - 1)!*(Sum_{k=m..n} Stirling1(k, m)*binomial(2*n, n-k)/(k-1)!). - Vladimir Kruchinin, Mar 26 2013

New name by Peter Luschny, Jun 27 2024

A262033 Number of permutations of [n] beginning with at least floor(n/2) ascents.

Original entry on oeis.org

1, 1, 1, 3, 4, 20, 30, 210, 336, 3024, 5040, 55440, 95040, 1235520, 2162160, 32432400, 57657600, 980179200, 1764322560, 33522128640, 60949324800, 1279935820800, 2346549004800, 53970627110400, 99638080819200, 2490952020480000, 4626053752320000

Offset: 0

Alois P. Heinz, Sep 08 2015

nonn

			a(4) = 4: 1234, 1243, 1342, 2341.
a(5) = 20: 12345, 12354, 12435, 12453, 12534, 12543, 13425, 13452, 13524, 13542, 14523, 14532, 23415, 23451, 23514, 23541, 24513, 24531, 34512, 34521.

Alois P. Heinz, Table of n, a(n) for n = 0..732

Cf. A001761, A006963, A262034, A262035.

a:= proc(n) option remember; `if`(n<2, 1,
      2*n*(n*(n-1)*a(n-2)-a(n-1))/((n+2)*(n-1)))
    end:
seq(a(n), n=0..30);

a[n_] := n!/Ceiling[(n + 1)/2]!; Array[a, 30, 0] (* Amiram Eldar, Dec 04 2022 *)

E.g.f.: (x+1)*(exp(x^2)-1)/x^2.
a(n) = 2*n*(n*(n-1)*a(n-2)-a(n-1))/((n+2)*(n-1)) for n>1, a(0)=a(1)=1.
a(n) = n!/ceiling((n+1)/2)!.
a(2n) = A262034(2n) = A001761(n).
a(2n+1) = A006963(n+2).
Sum_{n>=0} 1/a(n) = 7/4 + 13*exp(1/4)*sqrt(Pi)*erf(1/2)/8, where erf is the error function. - Amiram Eldar, Dec 04 2022

A291699 a(n) = n^n(2n)!/(n!*(n + 1)!).

Original entry on oeis.org

1, 1, 8, 135, 3584, 131250, 6158592, 353299947, 23991418880, 1883638417518, 167960000000000, 16772331868538246, 1854655886442627072, 225005916687384753700, 29718395534545380311040, 4245313393689422607421875, 652233889532678001886494720, 107247390031799133661006687830

Offset: 0

Ilya Gutkovskiy, Aug 30 2017

nonn

Robert Israel, Table of n, a(n) for n = 0..322

Main diagonal of A290605.

Cf. A000108, A000312, A001761, A061711.

seq(n^n*(2*n)!/n!/(n+1)!, n=0..50); # Robert Israel, Aug 30 2017

Join[{1}, Table[n^n (2 n)!/(n! (n + 1)!), {n, 1, 17}]]
Table[SeriesCoefficient[2/(1 + Sqrt[1 - 4 n x]), {x, 0, n}], {n, 0, 17}]
Table[SeriesCoefficient[1/(1 + ContinuedFractionK[-n x, 1, {i, 1, n}]), {x, 0, n}], {n, 0, 17}]

a(n)=binomial(2*n,n)/(n+1)*n^n \\ Charles R Greathouse IV, Oct 23 2023

a(n) = [x^n] 2/(1 + sqrt(1 - 4*n*x)).
a(n) = [x^n] 1/(1 - n*x/(1 - n*x/(1 - n*x/(1 - n*x/(1 - n*x/(1 - ...)))))), a continued fraction.
a(n) = n! * [x^n] (BesselI(0,2*n*x) - BesselI(1,2*n*x))*exp(2*n*x).
a(n) = n^n*binomial(2*n,n)/(n + 1).
a(n) = A000312(n)*A000108(n).
a(n) = A290605(n,n).
a(n) ~ 4^n*n^(n-3/2)/sqrt(Pi).

A093049 n-1 minus exponent of 2 in n, a(0) = 0.

Original entry on oeis.org

0, 0, 0, 2, 1, 4, 4, 6, 4, 8, 8, 10, 9, 12, 12, 14, 11, 16, 16, 18, 17, 20, 20, 22, 20, 24, 24, 26, 25, 28, 28, 30, 26, 32, 32, 34, 33, 36, 36, 38, 36, 40, 40, 42, 41, 44, 44, 46, 43, 48, 48, 50, 49, 52, 52, 54, 52, 56, 56, 58, 57, 60, 60, 62, 57, 64, 64, 66, 65, 68

Offset: 0

Ralf Stephan, Mar 16 2004

nonn

			G.f. = 2*x^3 + x^4 + 4*x^5 + 4*x^6 + 6*x^7 + 4*x^8 + 8*x^9 + 8*x^10 + ... - _Michael Somos_, Jan 25 2020

a(n) = n - A007814(n) - 1 = A093048(n) - 1, n>0.

a(n) is the exponent of 2 in A001761(n+1), A002105(n), A002682(n-1), A006963(n), A036770(n-1), A059837(n), A084623(n), |A003707(n)|, |A011859(n)|.

a[ n_] := If[ n == 0, 0, n - 1 - IntegerExponent[n, 2]]; (* Michael Somos, Jan 25 2020 *)

a(n)=if(n<1,0,if(n%2==0,a(n/2)+n/2-1,n-1))

{a(n) = if( n, n - 1 - valuation(n, 2))}; /* Michael Somos, Jan 25 2020 */

def A093049(n): return n-1-(~n& n-1).bit_length() if n else 0 # Chai Wah Wu, Jul 07 2022

Recurrence: a(2n) = a(n) + n - 1, a(2n+1) = 2n.

G.f.: sum(k>=0, t^3(t+2)/(1-t^2)^2, t=x^2^k).

A262034 Number of permutations of [n] beginning with at least ceiling(n/2) ascents.

Original entry on oeis.org

1, 0, 1, 1, 4, 5, 30, 42, 336, 504, 5040, 7920, 95040, 154440, 2162160, 3603600, 57657600, 98017920, 1764322560, 3047466240, 60949324800, 106661318400, 2346549004800, 4151586700800, 99638080819200, 177925144320000, 4626053752320000, 8326896754176000

Offset: 0

Alois P. Heinz, Sep 08 2015

nonn

			a(4) = 4: 1234, 1243, 1342, 2341.
a(5) = 5: 12345, 12354, 12453, 13452, 23451.

Alois P. Heinz, Table of n, a(n) for n = 0..733

Cf. A001761, A102693, A262033, A262035.

a:= proc(n) option remember; `if`(n<4, [1, 0, 1$2][n+1],
      2*((n^2-1)*a(n-2)-a(n-1))/(n+3))
    end:
seq(a(n), n=0..30);

np=Rest[With[{nn=30},CoefficientList[Series[(Exp[x^2](x+1)-x^4/2+x^2+x+1)/ x^3,{x,0,nn}],x] Range[0,nn]!]//Quiet];Join[{1},np] (* Harvey P. Dale, May 18 2019 *)

E.g.f.: (exp(x^2)*(x+1)-(x^4/2+x^2+x+1))/x^3.
a(n) = 2*((n^2-1)*a(n-2)-a(n-1))/(n+3) for n>3, a(0)=a(2)=a(3)=1, a(1)=0.
a(n) = n!/(n/2+1)! if n even, a(n) = floor(C(n+1,(n+1)/2)/(n+3)*((n-1)/2)!) if n odd.
a(2n) = A262033(2n) = A001761(n).
a(2n+1) = A102693(n+1).
Sum_{n>=2} 1/a(n) = (39*exp(1/4)*sqrt(Pi)*erf(1/2) - 6)/16, where erf is the error function. - Amiram Eldar, Dec 04 2022

cp's OEIS Frontend

A349598 E.g.f. satisfies: log(A(x)) = exp(x*A(x)^2) - 1.

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A365340 a(n) = (4*n)!/(3*n+1)!.

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A365341 a(n) = (5*n)!/(4*n+1)!.

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A256061 Number T(n,k) of 2n-length strings of balanced parentheses of exactly k different types; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

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A355766 E.g.f. satisfies A(x) = 1/(1 - x*A(x))^(A(x)^2).

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A038455 Triangle read by rows: T(n, k) = [x^k] x*Pochhammer(n + x, n)/(n + x).

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A262033 Number of permutations of [n] beginning with at least floor(n/2) ascents.

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A291699 a(n) = n^n*(2*n)!/(n!*(n + 1)!).

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A365340 a(n) = (4n)!/(3n+1)!.

A365341 a(n) = (5n)!/(4n+1)!.

A291699 a(n) = n^n(2n)!/(n!*(n + 1)!).