cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A036770 Number of labeled rooted trees with a degree constraint: (2*n)!/(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

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Comments

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

Crossrefs

Programs

  • Magma
    [Factorial(2*n)/(2^n) * Binomial(2*n+1, n): n in [0..15]]; // Vincenzo Librandi, Jan 29 2020
  • Maple
    spec := [S,{S=Union(Z,Prod(Z,Set(S,card=2)))},labeled]: seq(combstruct[count](spec,size=n)
  • Mathematica
    Range[0, 19]! CoefficientList[Series[(1 - (1 - 2 x^2)^(1/2))/x, {x, 0, 20}], x] (* Geoffrey Critzer, Nov 13 2011 *)

Formula

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

A036772 Number of labeled rooted trees with a degree constraint: ((4*n)!/(24^n)) * binomial(4*n+1, n).

Original entry on oeis.org

1, 5, 2520, 9909900, 150089940000, 6217438242015000, 574985352122181000000, 103753754577643425255000000, 33189544956070738228953960000000, 17517292900368819935211385551000000000, 14427024664929016470240101675459976000000000
Offset: 0

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Crossrefs

Programs

  • Mathematica
    Table[(4n)!/24^n Binomial[4n+1,n],{n,0,10}] (* Harvey P. Dale, Aug 10 2011 *)

Formula

From Petros Hadjicostas, Jun 08 2019: (Start)
Recurrence (with no interpolated zeros): -8 * (4*n + 1) * (4*n + 3)^2 * (2*n + 1)^2 * (4*n + 5) * a(n) + (81*n^2 + 162*n + 72) * a(n + 1) = 0 for n >= 0 with a(0) = 1.
E.g.f. (with interpolated zeros): Let G(x) = Sum_{n >= 0} a(n)*x^(4*n + 1)/(4*n + 1)!. Then the e.g.f. satisfies G(x) = x * (1 + G(x)^4/4!).
(End)

A036773 Number of labeled rooted trees with a degree constraint: ((5*n)!/(120^n)) * C(5*n+1, n).

Original entry on oeis.org

1, 6, 13860, 423783360, 70220478968640, 41004669682770393600, 65405789473547026656472320, 240729724316568938710767014707200, 1813083536072533851678174232377806438400, 25541737277107694920826740625991927645705830400
Offset: 0

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Formula

E.g.f. with interpolated zeros: Let G(x) = Sum_{n >= 0} a(n)*x^(5*n + 1)/(5*n + 1)!. Then this e.g.f. satisfies the equation G(x) = x*(1 + G(x)^5/5!). - Petros Hadjicostas, Jun 08 2019

A367625 (A036991(n)-1)/2.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 19, 21, 22, 23, 25, 26, 27, 29, 30, 31, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 51, 53, 54, 55, 57, 58, 59, 61, 62, 63, 71, 75, 77, 78, 79, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95, 99, 101, 102, 103, 105, 106, 107, 109, 110, 111, 115, 117, 118, 119, 121, 122, 123, 125, 126, 127
Offset: 2

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Author

N. J. A. Sloane, Nov 27 2023

Keywords

Crossrefs

Programs

  • Python
    from itertools import count, islice
    def A367625_gen(startvalue=0): # generator of terms >= startvalue
        if startvalue <= 0:
            yield 0
        for n in count(max(startvalue,1)):
            s = bin(n)[2:]
            c, l = 2, len(s)
            for i in range(1,l+1):
                c += int(s[l-i])<<1
                if c <= i:
                    break
            else:
                yield n
    A367625_list = list(islice(A367625_gen(),30)) # Chai Wah Wu, Nov 28 2023
Showing 1-4 of 4 results.