A052510 -id:A052510 - cp's OEIS frontend

A101921 a(2n) = a(n) + 2n - 1, a(2n+1) = 4n.

0, 1, 4, 4, 8, 9, 12, 11, 16, 17, 20, 20, 24, 25, 28, 26, 32, 33, 36, 36, 40, 41, 44, 43, 48, 49, 52, 52, 56, 57, 60, 57, 64, 65, 68, 68, 72, 73, 76, 75, 80, 81, 84, 84, 88, 89, 92, 90, 96, 97, 100, 100, 104, 105, 108, 107, 112, 113, 116, 116, 120, 121, 124, 120, 128

Offset: 1

Ralf Stephan, Dec 21 2004

nonn

Exponent of 2 in tangent numbers A000182.
Also, exponent of 2 in the sequences A008775, A009670, A009764, A009798, A012227, A024236, A024277, A024299, A052510.
Also, exponent of 2 in 4^(n-1)/n. [David Brink, Aug 08 2013]

			G.f. = x^2 + 4*x^3 + 4*x^4 + 8*x^5 + 9*x^6 + 12*x^7 + 11*x^8 + 16*x^9 + 17*x^10 + ...

Iain Fox, Table of n, a(n) for n = 1..10000

Cf. A000182, A007814.

Cf. A008775, A009670, A009764, A009798, A012227, A024236, A024277, A024299, A052510.

a[n_]:= If[n<1, 0, 2n -2 - IntegerExponent[n, 2]]; (* Michael Somos, Mar 02 2014 *)

a(n)=valuation(4^(n-1)/n,2); \\ Joerg Arndt, Aug 13 2013

def A101921(n): return (n-1<<1)-(~n & n-1).bit_length() # Chai Wah Wu, Apr 14 2023

[2*n-2 -valuation(n,2) for n in (1..100)] # G. C. Greubel, Nov 29 2021

a(n) = 2n - 2 - A007814(n).
a(n) = A007814(A000182(n)).
G.f.: Sum_{k>=0} t^2*(1+4*t+t^2)/(1-t^2)^2 where t=x^2^k.

A036770 Number of labeled rooted trees with a degree constraint: (2n)!/(2^n) C(2*n+1, n).

Original entry on oeis.org

1, 3, 60, 3150, 317520, 52390800, 12843230400, 4382752374000, 1986847742880000, 1155153277710432000, 838011196011749760000, 742058914068404412480000, 787724078011075453248000000, 987468397792455300321600000000, 1443283810213452666950050560000000

Offset: 0

N. J. A. Sloane

nonn

a(n) is the number of rooted labeled strictly binary trees (each vertex has exactly two children or none) on 2*n + 1 vertices. - Geoffrey Critzer, Nov 13 2011

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 46.
Lajos Takacs, Enumeration of rooted trees and forests, Math. Scientist 18 (1993), 1-10; see Eqs. (12) and (13) on p. 4.
Eric Weisstein's World of Mathematics, Strongly Binary Tree.
Index entries for sequences related to rooted trees

Cf. A001190, A036771, A036772, A036773, A052510.

[Factorial(2*n)/(2^n) * Binomial(2*n+1, n): n in [0..15]]; // Vincenzo Librandi, Jan 29 2020

spec := [S,{S=Union(Z,Prod(Z,Set(S,card=2)))},labeled]: seq(combstruct[count](spec,size=n)

Range[0, 19]! CoefficientList[Series[(1 - (1 - 2 x^2)^(1/2))/x, {x, 0, 20}], x] (* Geoffrey Critzer, Nov 13 2011 *)

Recurrence (with interpolated zeros): Define (b(n): n >= 0) by b(2*m+1) = a(m) and b(2*m) = 0 for m >= 0. Then the sequence (b(n): n >= 0) satisfies the recurrence (-2*n^3 - 6*n^2 - 4*n)*b(n) + (n + 3)*b(n+2) = 0 for n >= 0 with b(0) = 0 and b(1) = 1. [Corrected by Petros Hadjicostas, Jun 07 2019]
E.g.f. with interpolated zeros: G(x) = Sum_{n >= 0} b(n)*x^n/n! = Sum_{m >= 0} a(m)*x^(2*m + 1)/(2*m + 1)! = 1/x * (1 - (1 - 2*x^2)^(1/2)) for 0 < |x| < 1/sqrt(2). [Edited by Petros Hadjicostas, Jun 07 2019]
E.g.f. with interpolated zeros satisfies G(x)= x*(1 + G(x)^2/2). - Geoffrey Critzer, Nov 13 2011
D-finite with recurrence (with no interpolated zeros): -2*a(n)*(n + 1)*(2*n + 1)*(2*n + 3) + (n + 2)*a(n+1) = 0 with a(0) = 1. - Petros Hadjicostas, Jun 08 2019
G.f.: 4F1(1,1,1/2,3/2;2;8*x). - R. J. Mathar, Jan 28 2020

A174449 Triangle T(n, k, q) = n!(n+1)!q^k/((n-k)!(n-k+1)!) if floor(n/2) > k-1, otherwise n!(n+1)!q^(n-k)/(k!*(k+1)!) for q = 1, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 12, 12, 1, 1, 20, 240, 20, 1, 1, 30, 600, 600, 30, 1, 1, 42, 1260, 25200, 1260, 42, 1, 1, 56, 2352, 70560, 70560, 2352, 56, 1, 1, 72, 4032, 169344, 5080320, 169344, 4032, 72, 1, 1, 90, 6480, 362880, 15240960, 15240960, 362880, 6480, 90, 1

Offset: 0

Roger L. Bagula, Mar 20 2010

			Triangle begins as:
  1;
  1,   1;
  1,   6,    1;
  1,  12,   12,      1;
  1,  20,  240,     20,        1;
  1,  30,  600,    600,       30,          1;
  1,  42, 1260,  25200,     1260,         42,        1;
  1,  56, 2352,  70560,    70560,       2352,       56,      1;
  1,  72, 4032, 169344,  5080320,     169344,     4032,     72,    1;
  1,  90, 6480, 362880, 15240960,   15240960,   362880,   6480,   90,   1;
  1, 110, 9900, 712800, 39916800, 1676505600, 39916800, 712800, 9900, 110,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. this sequence (q=1), A174450 (q=2), A174451 (q=3).

Cf. A052510.

F:= Factorial; // T = A174449
T:= func< n,k,q | Floor(n/2) gt k-1 select F(n)*F(n+1)*q^k/(F(n-k)*F(n-k+1)) else F(n)*F(n+1)*q^(n-k)/(F(k)*F(k+1)) >;
[T(n,k,1): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 29 2021

T[n_, k_, q_]:= If[Floor[n/2]>k-1, n!*(n+1)!*q^k/((n-k)!*(n-k+1)!), n!*(n+1)!*q^(n-k)/(k!*(k+1)!)];
Table[T[n, k, 1], {n,0,12}, {k,0,n}]//Flatten

f=factorial
def A174449(n,k,q):
    if ((n//2)>k-1): return f(n)*f(n+1)*q^k/(f(n-k)*f(n-k+1))
    else: return f(n)*f(n+1)*q^(n-k)/(f(k)*f(k+1))
flatten([[A174449(n,k,1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Nov 29 2021

T(n, k, q) = n!*(n+1)!*q^k/((n-k)!(n-k+1)!) if floor(n/2) > k-1, otherwise n!*(n+1)!*q^(n-k)/(k!*(k+1)!) for q = 1.
T(n, n-k, q) = T(n, k, q).
From G. C. Greubel, Nov 29 2021: (Start)
T(2*n, n, 1) = A052510(n+1).
T(2*n, n, q) = q^n*(2*n+1)!*Catalan(n) for q = 1.
T(n, k, q) = binomial(n, k)*binomial(n+1, k+1) * ( k!*(k+1)!*q^k/(n-k+1) if (floor(n/2) >= k), otherwise ((n-k)!)^2*q^(n-k) ), for q = 1. (End)

Edited by G. C. Greubel, Nov 29 2021

A329965 a(n) = ((1+n)floor(1+n/2))(n!/floor(1+n/2)!)^2.

Original entry on oeis.org

1, 2, 6, 72, 240, 7200, 25200, 1411200, 5080320, 457228800, 1676505600, 221298739200, 821966745600, 149597947699200, 560992303872000, 134638152929280000, 508633022177280000, 155641704786247680000, 591438478187741184000, 224746621711341649920000

Offset: 0

Peter Luschny, Dec 04 2019

nonn

Cf. A212303, A057977, A052510, A123072, A093005, A262033, A329964.

A329965 := n -> ((1+n)*floor(1+n/2))*(n!/floor(1+n/2)!)^2:
seq(A329965(n), n=0..19);

ser := Series[(1 - Sqrt[1 - 4 x^2] - 4 x^2 (1 - x - Sqrt[1 - 4 x^2]))/(2 x^2 (1 - 4 x^2)^(3/2)), {x, 0, 22}]; Table[n! Coefficient[ser, x, n], {n, 0, 20}]
Table[(1+n)Floor[1+n/2](n!/Floor[1+n/2]!)^2,{n,0,30}] (* Harvey P. Dale, Oct 01 2023 *)

from fractions import Fraction
def A329965():
    x, n = 1, Fraction(1)
    while True:
        yield int(x)
        m = n if n % 2 else 4/(n+2)
        n += 1
        x *= m * n
a = A329965(); [next(a) for i in range(36)]

a(n) = n!*A212303(n+1).
a(n) = (n+1)!*A057977(n).
a(n) = A093005(n+1)*A262033(n)^2.
a(n) = A093005(n+1)*A329964(n).
a(2*n) = A052510(n) (n >= 0).
a(2*n+1) = A123072(n+1) (n >= 0).
a(n) = n! [x^n] (1 - sqrt(1 - 4*x^2) - 4*x^2*(1 - x - sqrt(1 - 4*x^2)))/(2*x^2*(1 - 4*x^2)^(3/2)).

cp's OEIS Frontend

A101921 a(2n) = a(n) + 2n - 1, a(2n+1) = 4n.

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A036770 Number of labeled rooted trees with a degree constraint: (2*n)!/(2^n) * C(2*n+1, n).

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Magma

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A174449 Triangle T(n, k, q) = n!*(n+1)!*q^k/((n-k)!(n-k+1)!) if floor(n/2) > k-1, otherwise n!*(n+1)!*q^(n-k)/(k!*(k+1)!) for q = 1, read by rows.

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Programs

Magma

Mathematica

Sage

Formula

Extensions

A329965 a(n) = ((1+n)*floor(1+n/2))*(n!/floor(1+n/2)!)^2.

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Author

Keywords

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A036770 Number of labeled rooted trees with a degree constraint: (2n)!/(2^n) C(2*n+1, n).

A174449 Triangle T(n, k, q) = n!(n+1)!q^k/((n-k)!(n-k+1)!) if floor(n/2) > k-1, otherwise n!(n+1)!q^(n-k)/(k!*(k+1)!) for q = 1, read by rows.

A329965 a(n) = ((1+n)floor(1+n/2))(n!/floor(1+n/2)!)^2.