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.

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A000598 Number of rooted ternary trees with n nodes; number of n-carbon alkyl radicals C(n)H(2n+1) ignoring stereoisomers.

Original entry on oeis.org

1, 1, 1, 2, 4, 8, 17, 39, 89, 211, 507, 1238, 3057, 7639, 19241, 48865, 124906, 321198, 830219, 2156010, 5622109, 14715813, 38649152, 101821927, 269010485, 712566567, 1891993344, 5034704828, 13425117806, 35866550869, 95991365288, 257332864506, 690928354105
Offset: 0

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Author

Keywords

Comments

Number of unlabeled rooted trees in which each node has out-degree <= 3.
Ignoring stereoisomers means that the children of a node are unordered. They can be permuted in any way and it is still the same tree. See A000625 for the analogous sequence with stereoisomers counted.
In alkanes every carbon has valence exactly 4 and every hydrogen has valence exactly 1. But the trees considered here are just the carbon "skeletons" (with all the hydrogen atoms stripped off) so now each carbon bonds to 1 to 4 other carbons. The out-degree is then <= 3.
Other descriptions of this sequence: quartic planted trees with n nodes; ternary rooted trees with n nodes and height at most 3.
The number of aliphatic amino acids with n carbon atoms in the side chain, and no rings or double bonds, has the same growth as this sequence. - Konrad Gruetzmann, Aug 13 2012

Examples

			From _Joerg Arndt_, Feb 25 2017: (Start)
The a(5) = 8 rooted trees with 5 nodes and out-degrees <= 3 are:
:         level sequence    out-degrees (dots for zeros)
:     1:  [ 0 1 2 3 4 ]    [ 1 1 1 1 . ]
:  O--o--o--o--o
:
:     2:  [ 0 1 2 3 3 ]    [ 1 1 2 . . ]
:  O--o--o--o
:        .--o
:
:     3:  [ 0 1 2 3 2 ]    [ 1 2 1 . . ]
:  O--o--o--o
:     .--o
:
:     4:  [ 0 1 2 3 1 ]    [ 2 1 1 . . ]
:  O--o--o--o
:  .--o
:
:     5:  [ 0 1 2 2 2 ]    [ 1 3 . . . ]
:  O--o--o
:     .--o
:     .--o
:
:     6:  [ 0 1 2 2 1 ]    [ 2 2 . . . ]
:  O--o--o
:     .--o
:  .--o
:
:     7:  [ 0 1 2 1 2 ]    [ 2 1 . 1 . ]
:  O--o--o
:  .--o--o
:
:     8:  [ 0 1 2 1 1 ]    [ 3 1 . . . ]
:  O--o--o
:  .--o
:  .--o
(End)
		

References

  • N. L. Biggs et al., Graph Theory 1736-1936, Oxford, 1976, p. 62 (quoting Cayley, who is wrong).
  • A. Cayley, On the mathematical theory of isomers, Phil. Mag. vol. 67 (1874), 444-447 (a(6) is wrong).
  • J. L. Faulon, D. Visco and D. Roe, Enumerating Molecules, In: Reviews in Computational Chemistry Vol. 21, Ed. K. Lipkowitz, Wiley-VCH, 2005.
  • R. A. Fisher, Contributions to Mathematical Statistics, Wiley, 1950, 41.397.
  • J. L. Gross and J. Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 529.
  • Handbook of Combinatorics, North-Holland '95, p. 1963.
  • Knop, Mueller, Szymanski and Trinajstich, Computer generation of certain classes of molecules.
  • D. Perry, The number of structural isomers ..., J. Amer. Chem. Soc. 54 (1932), 2918-2920.
  • G. Polya, Mathematical and Plausible Reasoning, Vol. 1 Prob. 4 pp. 85; 233.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Maple
    N := 45; G000598 := 0: i := 0: while i<(N+1) do G000598 := series(1+z*(G000598^3/6+subs(z=z^2,G000598)*G000598/2+subs(z=z^3,G000598)/3)+O(z^(N+1)),z,N+1): t[ i ] := G000598: i := i+1: od: A000598 := n->coeff(G000598,z,n);
    # Another Maple program for g.f. G000598:
    G000598 := 1; f := proc(n) global G000598; coeff(series(1+(1/6)*x*(G000598^3+3*G000598*subs(x=x^2,G000598)+2*subs(x=x^3,G000598)),x, n+1),x,n); end; for n from 1 to 50 do G000598 := series(G000598+f(n)*x^n,x,n+1); od; G000598;
    spec := [S, {Z=Atom, S=Union(Z, Prod(Z, Set(S, card=3)))}, unlabeled]: [seq(combstruct[count](spec, size=n), n=0..20)];
  • Mathematica
    m = 45; Clear[f]; f[1, x_] := 1+x; f[n_, x_] := f[n, x] = Expand[1+x*(f[n-1, x]^3/6 + f[n-1, x^2]*f[n-1, x]/2 + f[n-1, x^3]/3)][[1 ;; n]]; Do[f[n, x], {n, 2, m}]; CoefficientList[f[m, x], x]
    (* second program (after N. J. A. Sloane): *)
    m = 45; gf[] = 0; Do[gf[z] = 1 + z*(gf[z]^3/6 + gf[z^2]*gf[z]/2 + gf[z^3]/3) + O[z]^m // Normal, m]; CoefficientList[gf[z], z]  (* Jean-François Alcover, Sep 23 2014, updated Jan 11 2018 *)
    b[0, i_, t_, k_] = 1; m = 3; (* m = maximum children *)
    b[n_,i_,t_,k_]:= b[n,i,t,k]= If[i<1,0,
      Sum[Binomial[b[i-1, i-1, k, k] + j-1, j]*
      b[n-i*j, i-1, t-j, k], {j, 0, Min[t, n/i]}]];
    Join[{1},Table[b[n-1, n-1, m, m], {n, 1, 35}]] (* Robert A. Russell, Dec 27 2022 *)
  • PARI
    seq(n)={my(g=O(x)); for(n=1, n, g = 1 + x*(g^3/6 + subst(g,x,x^2)*g/2 + subst(g,x,x^3)/3) + O(x^n)); Vec(g)} \\ Andrew Howroyd, May 22 2018
    
  • SageMath
    def seq(n):
        B = PolynomialRing(QQ, 't', n+1);t = B.gens()
        R. = B[[]]
        T = sum([t[i] * z^i for i in range(1,n+1)]) + O(z^(n+1))
        lhs, rhs = T, 1 + z/6 * (T(z)^3 + 3*T(z)*T(z^2) + 2*T(z^3))
        I = B.ideal([lhs.coefficients()[i] - rhs.coefficients()[i] for i in range(n)])
        return [I.reduce(t[i]) for i in range(1,n+1)]
    seq(33) # Chris Grossack, Mar 31 2025

Formula

G.f. A(x) satisfies A(x) = 1 + (1/6)*x*(A(x)^3 + 3*A(x)*A(x^2) + 2*A(x^3)).
a(n) ~ c * d^n / n^(3/2), where d = 1/A261340 = 2.8154600331761507465266167782426995425365065396907..., c = 0.517875906458893536993162356992854345458168348098... . - Vaclav Kotesovec, Aug 15 2015

Extensions

Additional comments from Steve Strand (snstrand(AT)comcast.net), Aug 20 2003

A025192 a(0)=1; a(n) = 2*3^(n-1) for n >= 1.

Original entry on oeis.org

1, 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366, 118098, 354294, 1062882, 3188646, 9565938, 28697814, 86093442, 258280326, 774840978, 2324522934, 6973568802, 20920706406, 62762119218, 188286357654, 564859072962, 1694577218886, 5083731656658, 15251194969974
Offset: 0

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Author

Keywords

Comments

Warning: there is considerable overlap between this entry and the essentially identical A008776.
Shifts one place left when plus-convolved (PLUSCONV) with itself. a(n) = 2*Sum_{i=0..n-1} a(i). - Antti Karttunen, May 15 2001
Let M = { 0, 1, ..., 2^n-1 } be the set of all n-bit numbers. Consider two operations on this set: "sum modulo 2^n" (+) and "bitwise exclusive or" (XOR). The results of these operations are correlated.
To give a numerical measure, consider the equations over M: u = x + y, v = x XOR y and ask for how many pairs (u,v) is there a solution? The answer is exactly a(n) = 2*3^(n-1) for n >= 1. The fraction a(n)/4^n of such pairs vanishes as n goes to infinity. - Max Alekseyev, Feb 26 2003
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+2, s(0) = 3, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004
Number of compositions of n into parts of two kinds. For a string of n objects, before the first, choose first kind or second kind; before each subsequent object, choose continue, first kind, or second kind. For example, compositions of 3 are 3; 2,1; 1,2; and 1,1,1. Using parts of two kinds, these produce respectively 2, 4, 4 and 8 compositions, 2+4+4+8 = 18. - Franklin T. Adams-Watters, Aug 18 2006
In the compositions the kinds of parts are ordered inside a run of identical parts, see example. Replacing "ordered" by "unordered" gives A052945. - Joerg Arndt, Apr 28 2013
Number of permutations of {1, 2, ..., n+1} such that no term is more than 2 larger than its predecessor. For example, a(3) = 18 because all permutations of {1, 2, 3, 4} are valid except 1423, 1432, 2143, 3142, 2314, 3214, in which 1 is followed by 4. Proof: removing (n + 1) gives a still-valid sequence. For n >= 2, can insert (n + 1) either at the beginning or immediately following n or immediately following (n - 1), but nowhere else. Thus the number of such permutations triples when we increase the sequence length by 1. - Joel B. Lewis, Nov 14 2006
Antidiagonal sums of square array A081277. - Philippe Deléham, Dec 04 2006
Equals row sums of triangle A160760. - Gary W. Adamson, May 25 2009
Let M = a triangle with (1, 2, 4, 8, ...) as the left border and all other columns = (0, 1, 2, 4, 8, ...). A025192 = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n with no 3-element antichain. ("Uniform" used in the sense of Retakh, Serconek and Wilson. By "graded" we mean that all maximal chains have the same length n.) - David Nacin, Feb 13 2012
Equals partial sums of A003946 prefaced with a 1: (1, 1, 4, 12, 36, 108, ...). - Gary W. Adamson, Feb 15 2012
Number of vertices (or sides) of the (n-1)-th iteration of a Gosper island. - Arkadiusz Wesolowski, Feb 07 2013
Row sums of triangle in A035002. - Jon Perry, May 30 2013
a(n) counts walks (closed) on the graph G(1-vertex; 1-loop, 1-loop, 2-loop, 2-loop, 3-loop, 3-loop, ...). - David Neil McGrath, Jan 01 2015
From Tom Copeland, Dec 03 2015: (Start)
For n > 0, a(n) are the traces of the even powers of the adjacency matrix M of the simple Lie algebra B_3, tr(M^(2n)) where M = Matrix(row 1; row 2; row 3) = Matrix[0,1,0; 1,0,2; 0,1,0], same as the traces of Matrix[0,2,0; 1,0,1; 0,1,0] (cf. Damianou). The traces of the odd powers vanish.
The characteristic polynomial of M equals determinant(x*I - M) = x^3 - 3x = A127672(3,x), so 1 - 3*x^2 = det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n), implying Sum_{n>=1} a(n+1) x^(2n) / (2n) = -log(1 - 3*x^2), giving a logarithmic generating function for the aerated sequence, excluding a(0) and a(1).
a(n+1) = tr(M^(2n)), where tr(M^n) = 3^(n/2) + (-1)^n * 3^(n/2) = 2^n*(cos(Pi/6)^n + cos(5*Pi/6)^n) = n-th power sum of the eigenvalues of M = n-th power sum of the zeros of the characteristic polynomial.
The relation det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n) = Sum_{n>=0} P_n(-tr(M), -tr(M^2), ..., -tr(M^n)) x^n/n! = exp(P.(-tr(M), -tr(M^2), ...)x), where P_n(x(1), ..., x(n)) are the partition polynomials of A036039 implies that with x(2n) = -tr(M^(2n)) = -a(n+1) for n > 0 and x(n) = 0 otherwise, the partition polynomials evaluate to zero except for P_2(x(1), x(2)) = P_2(0,-6) = -6.
Because of the inverse relation between the partition polynomials of A036039 and the Faber polynomials F_k(b1,b2,...,bk) of A263916, F_k(0,-3,0,0,...) = tr(M^k) gives aerated a(n), excluding n=0,1. E.g., F_2(0,-3) = -2(-3) = 6, F_4(0,-3,0,0) = 2 (-3)^2 = 18, and F_6(0,-3,0,0,0,0) = -2(-3)^3 = 54. (Cf. A265185.)
(End)
Number of permutations of length n > 0 avoiding the partially ordered pattern (POP) {1>2, 1>3, 1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is the largest. - Sergey Kitaev, Dec 08 2020
For n > 0, a(n) is the number of 3-colorings of the grid graph P_2 X P_(n-1). More generally, for q > 1, the number of q-colorings of the grid graph P_2 X P_n is given by q*(q - 1)*((q - 1)*(q - 2) + 1)^(n - 1). - Sela Fried, Sep 25 2023
For n > 1, a(n) is the largest solution to the equation phi(x) = a(n-1). - M. Farrokhi D. G., Oct 25 2023
Number of dotted compositions of degree n. - Diego Arcis, Feb 01 2024

Examples

			There are a(3)=18 compositions of 3 into 2 kinds of parts. Here p:s stands for "part p of sort s":
01:  [ 1:0  1:0  1:0  ]
02:  [ 1:0  1:0  1:1  ]
03:  [ 1:0  1:1  1:0  ]
04:  [ 1:0  1:1  1:1  ]
05:  [ 1:0  2:0  ]
06:  [ 1:0  2:1  ]
07:  [ 1:1  1:0  1:0  ]
08:  [ 1:1  1:0  1:1  ]
09:  [ 1:1  1:1  1:0  ]
10:  [ 1:1  1:1  1:1  ]
11:  [ 1:1  2:0  ]
12:  [ 1:1  2:1  ]
13:  [ 2:0  1:0  ]
14:  [ 2:0  1:1  ]
15:  [ 2:1  1:0  ]
16:  [ 2:1  1:1  ]
17:  [ 3:0  ]
18:  [ 3:1  ]
- _Joerg Arndt_, Apr 28 2013
G.f. = 1 + 2*x + 6*x^2 + 18*x^3 + 54*x^4 + 162*x^5 + 486*x^6 + 1458*x^7 + ...
		

References

  • Richard P. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Crossrefs

First differences of 3^n (A000244). Other self-convolved sequences: A000108, A007460, A007461, A007462, A007463, A007464, A061922.
Apart from initial term, same as A008776.

Programs

  • Haskell
    a025192 0 = 1
    a025192 n = 2 * 3 ^ (n -1)
    a025192_list = 1 : iterate (* 3) 2  -- Reinhard Zumkeller, Nov 27 2012
  • Maple
    A025192 := proc(n): if n=0 then 1 else 2*3^(n-1) fi: end: seq(A025192(n),n=0..26);
  • Mathematica
    Join[{1},2*3^(Range[30]-1)]  (* Harvey P. Dale, Mar 22 2011 *)
  • PARI
    a(n)=max(1,2*3^(n-1)) \\ Charles R Greathouse IV, Jul 25 2011
    
  • PARI
    Vec((1-x)/(1-3*x) + O(x^100)) \\ Altug Alkan, Dec 05 2015
    
  • Python
    [1]+[2*3**(n-1) for n in range(1,30)] # David Nacin, Mar 04 2012
    

Formula

G.f.: (1-x)/(1-3*x).
E.g.f.: (2*exp(3*x) + exp(0))/3. - Paul Barry, Apr 20 2003
a(n) = phi(3^n) = A000010(A000244(n)). - Labos Elemer, Apr 14 2003
a(0) = 1, a(n) = Sum_{k=0..n-1} (a(k) + a(n-k-1)). - Benoit Cloitre, Jun 24 2003
a(n) = A002326((3^n-1)/2). - Vladimir Shevelev, May 26 2008
a(1) = 2, a(n) = 3*a(n-1). - Vincenzo Librandi, Jan 01 2011
a(n) = lcm(a(n-1), Sum_{k=1..n-1} a(k)) for n >= 3. - David W. Wilson, Sep 27 2011
a(n) = ((2*n-1)*a(n-1) + (3*n-6)*a(n-2))/(n-1); a(0)=1, a(1)=2. - Sergei N. Gladkovskii, Jul 16 2012
From Sergei N. Gladkovskii, Jul 17 2012: (Start)
For the e.g.f. E(x) = (2/3)*exp(3*x) + exp(0)/3 we have
E(x) = 2*G(0)/3 where G(k) = 1 + k!/(3*(9*x)^k - 3*(9*x)^(2*k+1)/((9*x)^(k+1) + (k+1)!/G(k+1))); (continued fraction, 3rd kind, 3-step).
E(x) = 1+2*x/(G(0)-3*x) where G(k) = 3*x + 1 + k - 3*x*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). (End)
a(n) = A114283(0,0). - Reinhard Zumkeller, Nov 27 2012
G.f.: 1 + ((1/2)/G(0) - 1)/x where G(k) = 1 - 2^k/(2 - 4*x/(2*x - 2^k/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 22 2012
G.f.: 1 + x*W(0), where W(k) = 1 + 1/(1 - x*(2*k+3)/(x*(2*k+4) + 1/W(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
G.f.: 1 / (1 - 2*x / (1 - x)). - Michael Somos, Apr 03 2014
Construct the power matrix T(n,j) = [A(n)^*j]*[S(n)^*(j-1)] where A(n)=(2,2,2,...) and S(n)=(0,1,0,0,...). (* is convolution operation.) Then a(n) = Sum_{j=1..n} T(n,j). - David Neil McGrath, Jan 01 2015
G.f.: 1 + 2*x/(1 + 2*x)*( 1 + 5*x/(1 + 5*x)*( 1 + 8*x/(1 + 8*x)*( 1 + 11*x/(1 + 11*x)*( 1 + .... - Peter Bala, May 27 2017
Sum_{n>=0} 1/a(n) = 7/4. - Bernard Schott, Oct 02 2021
From Amiram Eldar, May 08 2023: (Start)
Sum_{n>=0} (-1)^n/a(n) = 5/8.
Product_{n>=1} (1 - 1/a(n)) = A132019. (End)

Extensions

Additional comments from Barry E. Williams, May 27 2000
a(22) corrected by T. D. Noe, Feb 08 2008
Maple programs simplified by Johannes W. Meijer, Jun 02 2011

A055277 Triangle T(n,k) of number of rooted trees with n nodes and k leaves, 1 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 4, 3, 1, 0, 1, 6, 8, 4, 1, 0, 1, 9, 18, 14, 5, 1, 0, 1, 12, 35, 39, 21, 6, 1, 0, 1, 16, 62, 97, 72, 30, 7, 1, 0, 1, 20, 103, 212, 214, 120, 40, 8, 1, 0, 1, 25, 161, 429, 563, 416, 185, 52, 9, 1, 0, 1, 30, 241, 804, 1344, 1268, 732, 270, 65, 10, 1, 0
Offset: 1

Views

Author

Christian G. Bower, May 09 2000

Keywords

Comments

Harary denotes the g.f. as P(x, y) on page 33 "... , and let P(x,y) = Sum Sum P_{nm} x^ny^m where P_{nm} is the number of planted trees with n points and m endpoints, in which again the plant has not been counted either as a point or as an endpoint." - Michael Somos, Nov 02 2014

Examples

			From _Joerg Arndt_, Aug 18 2014: (Start)
Triangle starts:
01: 1
02: 1    0
03: 1    1    0
04: 1    2    1    0
05: 1    4    3    1    0
06: 1    6    8    4    1    0
07: 1    9   18   14    5    1    0
08: 1   12   35   39   21    6    1    0
09: 1   16   62   97   72   30    7    1    0
10: 1   20  103  212  214  120   40    8    1    0
11: 1   25  161  429  563  416  185   52    9    1    0
12: 1   30  241  804 1344 1268  732  270   65   10    1    0
13: 1   36  348 1427 2958 3499 2544 1203  378   80   11    1    0
...
The trees with n=5 nodes, as (preorder-) level sequences, together with their number of leaves, and an ASCII rendering, are:
:
:     1:  [ 0 1 2 3 4 ]   1
:  O--o--o--o--o
:
:     2:  [ 0 1 2 3 3 ]   2
:  O--o--o--o
:        .--o
:
:     3:  [ 0 1 2 3 2 ]   2
:  O--o--o--o
:     .--o
:
:     4:  [ 0 1 2 3 1 ]   2
:  O--o--o--o
:  .--o
:
:     5:  [ 0 1 2 2 2 ]   3
:  O--o--o
:     .--o
:     .--o
:
:     6:  [ 0 1 2 2 1 ]   3
:  O--o--o
:     .--o
:  .--o
:
:     7:  [ 0 1 2 1 2 ]   2
:  O--o--o
:  .--o--o
:
:     8:  [ 0 1 2 1 1 ]   3
:  O--o--o
:  .--o
:  .--o
:
:     9:  [ 0 1 1 1 1 ]   4
:  O--o
:  .--o
:  .--o
:  .--o
:
This gives [1, 4, 3, 1, 0], row n=5 of the triangle.
(End)
G.f. = x*(y + x*y + x^2*(y + y^2) + x^3*(y + 2*y^2 + y^3) + x^4*(y + 4*y^2 + 3*x^3 + y^4) + ...).
		

References

  • F. Harary, Recent results on graphical enumeration, pp. 29-36 of Graphs and Combinatorics (Washington, Jun 1973), Ed. by R. A. Bari and F. Harary. Lect. Notes Math., Vol. 406. Springer-Verlag, 1974.

Crossrefs

Programs

  • Mathematica
    rut[n_]:=rut[n]=If[n===1,{{}},Join@@Function[c,Union[Sort/@Tuples[rut/@c]]]/@IntegerPartitions[n-1]];
    Table[Length[Select[rut[n],Count[#,{},{-2}]===k&]],{n,13},{k,n}] (* Gus Wiseman, Mar 19 2018 *)
  • PARI
    {T(n, k) = my(A = O(x)); if(k<1 || k>n, 0, for(j=1, n, A = x*(y - 1  + exp( sum(i=1, j, 1/i * subst( subst( A + x * O(x^(j\i)), x, x^i), y, y^i) ) ))); polcoeff( polcoeff(A, n), k))}; /* Michael Somos, Aug 24 2015 */

Formula

G.f. satisfies A(x, y) = x*y + x*EULER(A(x, y)) - x. Shifts up under EULER transform.
G.f. satisfies A(x, y) = x*y - x + x * exp(Sum_{i>0} A(x^i, y^i) / i). [Harary, p. 34, equation (10)]. - Michael Somos, Nov 02 2014
Sum_k T(n, k) = A000081(n). - Michael Somos, Aug 24 2015

A003603 Fractal sequence obtained from Fibonacci numbers (or Wythoff array).

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 1, 4, 3, 2, 5, 1, 6, 4, 3, 7, 2, 8, 5, 1, 9, 6, 4, 10, 3, 11, 7, 2, 12, 8, 5, 13, 1, 14, 9, 6, 15, 4, 16, 10, 3, 17, 11, 7, 18, 2, 19, 12, 8, 20, 5, 21, 13, 1, 22, 14, 9, 23, 6, 24, 15, 4, 25, 16, 10, 26, 3, 27, 17, 11, 28, 7, 29, 18, 2, 30, 19, 12, 31, 8, 32, 20, 5, 33
Offset: 1

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Author

Keywords

Comments

Length of n-th row = A000045(n); last term of n-th row = A094967(n-1); sum of n-th row = A033192(n-1). - Reinhard Zumkeller, Jan 26 2012

Examples

			In the recurrence for making new rows, we get row 5 from row 4 thus: write row 4: 1,3,2, and then place 4 right after 1, and place 5 right after 2, getting 1,4,3,2,5. - _Clark Kimberling_, Oct 29 2009
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    -- according to Kimberling, see formula section.
    a003603 n k = a003603_row n !! (k-1)
    a003603_row n = a003603_tabl !! (n-1)
    a003603_tabl = [1] : [1] : wythoff [2..] [1] [1] where
       wythoff is xs ys = f is xs ys [] where
          f js     []     []     ws = ws : wythoff js ys ws
          f js     []     [v]    ws = f js [] [] (ws ++ [v])
          f (j:js) (u:us) (v:vs) ws
            | u == v = f js us vs (ws ++ [v,j])
            | u /= v = f (j:js) (u:us) vs (ws ++ [v])
    -- Reinhard Zumkeller, Jan 26 2012
  • Maple
    A003603 := proc(n::posint)
        local r,c,W ;
        for r from 1 do
            for c from 1 do
                W := A035513(r,c) ;
                if W = n then
                    return r ;
                elif W > n then
                    break ;
                end if;
            end do:
        end do:
    end proc:
    seq(A003603(n),n=1..100) ; # R. J. Mathar, Aug 13 2021
  • Mathematica
    num[n_, b_] := Last[NestWhile[{Mod[#[[1]], Last[#[[2]]]], Drop[#[[2]], -1], Append[#[[3]], Quotient[#[[1]], Last[#[[2]]]]]} &, {n, b, {}}, #[[2]] =!= {} &]];
    left[n_, b_] := If[Last[num[n, b]] == 0, Dot[num[n, b], Rest[Append[Reverse[b], 0]]], n];
    fractal[n_, b_] := # - Count[Last[num[Range[#], b]], 0] &@
       FixedPoint[left[#, b] &, n];
    Table[fractal[n, Table[Fibonacci[i], {i, 2, 12}]], {n, 30}] (* Birkas Gyorgy, Apr 13 2011 *)
    row[1] = row[2] = {1};
    row[n_] := row[n] = Module[{ro, pos, lp, ins}, ro = row[n-1]; pos = Position[ro, Alternatives @@ Intersection[ro, row[n-2]]] // Flatten; lp = Length[pos]; ins = Range[lp] + Max[ro]; Do[ro = Insert[ro, ins[[i]], pos[[i]] + i], {i, 1, lp}]; ro];
    Array[row, 9] // Flatten (* Jean-François Alcover, Jul 12 2016 *)

Formula

Vertical para-budding sequence: says which row of Wythoff array (starting row count at 1) contains n.
If one deletes the first occurrence of 1, 2, 3, ... the sequence is unchanged.
From Clark Kimberling, Oct 29 2009: (Start)
The fractal sequence of the Wythoff array can be constructed without reference to the Wythoff array or Fibonacci numbers. Write initial rows:
Row 1: .... 1
Row 2: .... 1
Row 3: .... 1..2
Row 4: .... 1..3..2
For n>4, to form row n+1, let k be the least positive integer not yet used; write row n, and right after the first number that is also in row n-1, place k; right after the next number that is also in row n-1, place k+1, and continue. A003603 is the concatenation of the rows. (End)
Conjecture: a(n) = abs(floor(n/phi) - floor(n*(1/phi + 1/(-phi)^(A035612(n) + 1)))) where phi = (1+sqrt(5))/2. - Alan Michael Gómez Calderón, Oct 27 2023

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003
Keyword tabf added by Reinhard Zumkeller, Jan 26 2012

A004211 Shifts one place left under 2nd-order binomial transform.

Original entry on oeis.org

1, 1, 3, 11, 49, 257, 1539, 10299, 75905, 609441, 5284451, 49134923, 487026929, 5120905441, 56878092067, 664920021819, 8155340557697, 104652541401025, 1401572711758403, 19546873773314571, 283314887789276721, 4259997696504874817, 66341623494636864963
Offset: 0

Views

Author

Keywords

Comments

Equals the eigensequence of A038207, the square of Pascal's triangle. - Gary W. Adamson, Apr 10 2009
The g.f. of the second binomial transform is 1/(1-2x-x/(1-2x/(1-2x-x/(1-4x/(1-2x-x/(1-6x/(1-2x-x/(1-8x/(1-... (continued fraction). - Paul Barry, Dec 04 2009
Length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(k)<=F(k)+2 where F(0)=0 and F(k+1)=s(k+1) if s(k+1)-s(k)=2, otherwise F(k+1)=F(k); see example and Fxtbook link. - Joerg Arndt, Apr 30 2011
It appears that the infinite set of "Shifts 1 place left under N-th order binomial transform" sequences has a production matrix of:
1, N, 0, 0, 0, ...
1, 1, N, 0, 0, ...
1, 2, 1, N, 0, ...
1, 3, 3, 1, N, ...
... (where a diagonal of (N,N,N,...) is appended to the right of Pascal's triangle). a(n) in each sequence is the upper left term of M^n such that N=1 generates A000110, then (N=2 - A004211), (N=3 - A004212), (N=4 - A004213), (N=5 - A005011), ... - Gary W. Adamson, Jul 29 2011
Number of "unlabeled" hierarchical orderings on set partitions of {1..n}, see comments on A034691. - Gus Wiseman, Mar 03 2016
From Lorenzo Sauras Altuzarra, Jun 17 2022: (Start)
Number of n-variate noncontradictory conjunctions of logical equalities of literals (modulo logical equivalence).
Equivalently, number of n-variate noncontradictory Krom formulas with palindromic truth-vector (modulo logical equivalence).
a(n) <= A109457(n). (End)

Examples

			From _Joerg Arndt_, Apr 30 2011: (Start)
Restricted growth strings: a(0)=1 corresponds to the empty string;
a(1)=1 to [0]; a(2)=3 to [00], [01], and [02]; a(3) = 11 to
        RGS           F
[ 1]  [ 0 0 0 ]    [ 0 0 0 ]
[ 2]  [ 0 0 1 ]    [ 0 0 0 ]
[ 3]  [ 0 0 2 ]    [ 0 0 2 ]
[ 4]  [ 0 1 0 ]    [ 0 0 0 ]
[ 5]  [ 0 1 1 ]    [ 0 0 0 ]
[ 6]  [ 0 1 2 ]    [ 0 0 2 ]
[ 7]  [ 0 2 0 ]    [ 0 2 2 ]
[ 8]  [ 0 2 1 ]    [ 0 2 2 ]
[ 9]  [ 0 2 2 ]    [ 0 2 2 ]
[10]  [ 0 2 3 ]    [ 0 2 2 ]
[11]  [ 0 2 4 ]    [ 0 2 4 ]. (End)
From _Lorenzo Sauras Altuzarra_, Jun 17 2022: (Start)
The 11 trivariate noncontradictory conjunctions of logical equalities of literals are (x <-> y) /\ (y <-> z), (~ x <-> y) /\ (y <-> z), (x <-> ~ y) /\ (~ y <-> z), (x <-> y) /\ (y <-> ~ z), (x <-> y) /\ (z <-> z), (~ x <-> y) /\ (z <-> z), (x <-> z) /\ (y <-> y), (~ x <-> z) /\ (y <-> y), (y <-> z) /\ (x <-> x), (~ y <-> z) /\ (x <-> x), and (x <-> x) /\ (y <-> y) /\ (z <-> z) (modulo logical equivalence).
The third complete Bell polynomial is x^3 + 3 x y + z; and note that (2^0)^3 + 3*2^0*2^1 + 2^2 = 11.
The truth-vector of (x <-> y) /\ (y <-> z), 10000001, is palindromic. (End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A075497 (row sums).
Cf. A038207.
Cf. A000110 (RGS where s(k) <= F(k) + 1), A004212 (RGS where s(k) <= F(k) + 3), A004213 (s(k) <= F(k) + 4), A005011 (s(k) <= F(k) + 5), A005012 (s(k) <= F(k) + 6), A075506 (s(k) <= F(k) + 7), A075507 (s(k) <= F(k) + 8), A075508 (s(k) <= F(k) + 9), A075509 (s(k) <= F(k) + 10).
Main diagonal of A261275.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, add(
          a(n-j)*binomial(n-1, j-1)*2^(j-1), j=1..n))
        end:
    seq(a(n), n=0..23);  # Alois P. Heinz, May 30 2021
    # second Maple program:
    a:= n -> CompleteBellB(n, seq(2^k, k=0..n)):
    seq(a(n), n=0..23);  # Lorenzo Sauras Altuzarra, Jun 17 2022
  • Mathematica
    Table[ Sum[ StirlingS2[ n, k ] 2^(-k+n), {k, n} ], {n, 16} ] (* Wouter Meeussen *)
    Table[2^n BellB[n, 1/2], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(2^(n-k)*binomial(n-1,k-1)*a(k-1),k,1,n); /* Vladimir Kruchinin, Nov 28 2011 */
    
  • PARI
    x='x+O('x^66);
    egf=exp(intformal(exp(2*x))); /* = 1 + x + 3/2*x^2 + 11/6*x^3 + ... */
    /* egf=exp(1/2*(exp(2*x)-1)) */ /* alternative e.g.f. */
    Vec(serlaplace(egf))  /* Joerg Arndt, Apr 30 2011 */
    
  • SageMath
    def A004211(n): return sum(2^(n-k)*stirling_number2(n, k) for k in (0..n))
    print([A004211(n) for n in range(21)]) # Peter Luschny, Apr 15 2020

Formula

E.g.f. A(x) satisfies A'(x)/A(x) = e^(2x).
E.g.f.: exp(sinh(x)*exp(x)) = exp(Integral_{t = 0..x} exp(2*t)) = exp((exp(2*x)-1)/2). - Joerg Arndt, Apr 30 2011 and May 13 2011
a(n) = Sum_{k=0..n} 2^(n-k)*Stirling2(n, k). - Emeric Deutsch, Feb 11 2002
G.f.: Sum_{k >= 0} x^k/Product_{j = 1..k} (1-2*j*x). - Ralf Stephan, Apr 18 2004
Stirling transform of A000085. - Vladeta Jovovic May 14 2004
O.g.f.: A(x) = 1/(1-x-2*x^2/(1-3*x-4*x^2/(1-5*x-6*x^2/(1-... -(2*n-1)*x-2*n*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
Define f_1(x), f_2(x), ... such that f_1(x)=e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n) = e^(-1/2)*2^{n-1}*f_n(1/2). - Milan Janjic, May 30 2008
G.f.: 1/(1-x/(1-2x/(1-x/(1-4x/(1-x/(1-6x/(1-x/(1-8x/(1-... (continued fraction). - Paul Barry, Dec 04 2009
a(n) = upper left term in M^n, M = an infinite square production matrix with an appended diagonal of (2,2,2,...) to the right of Pascal's triangle:
1, 2, 0, 0, 0, ...
1, 1, 2, 0, 0, ...
1, 2, 1, 2, 0, ...
1, 3, 3, 1, 2, ...
... - Gary W. Adamson, Jul 29 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+2*x)*d/dx. Cf. A000110. - Peter Bala, Nov 25 2011
G.f. A(x) satisfies A(x)=1+x/(1-2*x)*A(x/(1-2*x)), a(n) = Sum_{k=1..n} binomial(n-1,k-1)*2^(n-k)*a(k-1), a(0)=1. - Vladimir Kruchinin, Nov 28 2011 [corrected by Ilya Gutkovskiy, May 02 2019]
From Peter Bala, May 16 2012: (Start)
Recurrence equation: a(n+1) = Sum_{k = 0..n} 2^(n-k)*C(n,k)*a(k).
Written umbrally this is a(n+1) = (a + 2)^n (expand the binomial and replace a^k with a(k)). More generally, a*f(a) = f(a+2) holds umbrally for any polynomial f(x). An inductive argument then establishes the umbral recurrence a*(a-2)*(a-4)*...*(a-2*(n-1)) = 1 with a(0) = 1. Compare with the Bell numbers B(n) = A000110(n), which satisfy the umbral recurrence B*(B-1)*...*(B-(n-1)) = 1 with B(0) = 1. Cf. A009235.
Touchard's congruence holds: for odd prime p, a(p+k) == (a(k) + a(k+1)) (mod p) for k = 0,1,2,... (adapt the proof of Theorem 10.1 in Gessel). In particular, a(p) == 2 (mod p) for odd prime p. (End)
G.f.: (2/E(0) - 1)/x where E(k) = 1 + 1/(1 + 2*x/(1 - 4*(k+1)*x/E(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 20 2012
G.f.: (1/E(0)-1)/x where E(k) = 1 - x/(1 + 2*x - 2*x*(k+1)/E(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Sep 21 2012
a(n) = -1 + 2*Sum_{k=0..n} C(n,k)*A166922(k). - Peter Luschny, Nov 01 2012
G.f.: G(0)- 1/x where G(k) = 1 - (4*x*k-1)/(x - x^4/(x^3 - (4*x*k-1)*(4*x*k+2*x-1)*(4*x*k+4*x-1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 08 2013.
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-2*k*x)/(1-x/(x-1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 15 2013
G.f.: -G(0) where G(k) = 1 + 2*(1-k*x)/(2*k*x - 1 - x*(2*k*x - 1)/(x - 2*(1-k*x)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jan 29 2013
G.f.: 1/Q(0), where Q(k) = 1 - x/(1 - 2*x*(2*k+1)/( 1 - x/(1 - 4*x*(k+1)/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Apr 15 2013
G.f.: 1 + x/Q(0), where Q(k)= 1 - x*(2*k+3) - x^2*(2*k+2)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 05 2013
For n > 0, a(n) = exp(-1/2)*Sum_{k > 0} (2*k)^n/(k!*2^k). - Vladimir Reshetnikov, May 09 2013
G.f.: -(1+(2*x+1)/G(0))/x, where G(k)= 2*x*k - x - 1 - 2*(k+1)*x^2/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jul 20 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)/( 2*x^2*(k+1) - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
Sum_{k=0..n} C(n,k)*a(k)*a(n-k) = 2^n*Bell(n) = A055882(n). - Vaclav Kotesovec, Apr 03 2016
a(n) ~ 2^n * n^n * exp(n/LambertW(2*n) - n - 1/2) / (sqrt(1 + LambertW(2*n)) * LambertW(2*n)^n). - Vaclav Kotesovec, Jan 07 2019, simplified Oct 01 2022
a(n) = B_n(2^0, ..., 2^(n - 1)), where B_n(x_1, ..., x_n) is the n-th complete Bell polynomial. - Lorenzo Sauras Altuzarra, Jun 17 2022

A048768 Numbers n such that A048767(n) = n.

Original entry on oeis.org

1, 2, 9, 12, 18, 40, 112, 125, 250, 352, 360, 675, 832, 1008, 1125, 1350, 1500, 2176, 2250, 2401, 3168, 3969, 4802, 4864, 7488, 7938, 11776, 14000, 19584, 21609, 28812, 29403, 29696, 43218, 43776, 44000, 58806, 63488, 75600, 96040, 104000, 105984, 123201, 126000
Offset: 1

Views

Author

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so these are Heinz numbers of integer partitions that are fixed points under the map described in A217605 (which interchanges the parts with their multiplicities). The enumeration of these partitions by sum is given by A217605. - Gus Wiseman, May 04 2019

Examples

			12 = (2^2)*(3^1) = (2nd prime)^pi(2) * (first prime)^pi(3).
From _Gus Wiseman_, May 04 2019: (Start)
The sequence of terms together with their prime indices begins:
     1: {}
     2: {1}
     9: {2,2}
    12: {1,1,2}
    18: {1,2,2}
    40: {1,1,1,3}
   112: {1,1,1,1,4}
   125: {3,3,3}
   250: {1,3,3,3}
   352: {1,1,1,1,1,5}
   360: {1,1,1,2,2,3}
   675: {2,2,2,3,3}
   832: {1,1,1,1,1,1,6}
  1008: {1,1,1,1,2,2,4}
  1125: {2,2,3,3,3}
  1350: {1,2,2,2,3,3}
  1500: {1,1,2,3,3,3}
  2176: {1,1,1,1,1,1,1,7}
  2250: {1,2,2,3,3,3}
  2401: {4,4,4,4}
(End)
		

Crossrefs

Programs

  • Mathematica
    wt[n_]:=Times@@Cases[FactorInteger[n],{p_,k_}:>Prime[k]^PrimePi[p]];
    Select[Range[1000],wt[#]==#&] (* Gus Wiseman, May 04 2019 *)
  • PARI
    is(n) = {my(f = factor(n), p = f[, 1], e = f[, 2]); #Set(e) == #e && prod(i = 1, #e, prime(e[i])^primepi(p[i])) == n;} \\ Amiram Eldar, Oct 20 2023

Extensions

a(1) inserted and more terms added by Amiram Eldar, Oct 20 2023

A030266 Shifts left under COMPOSE transform with itself.

Original entry on oeis.org

0, 1, 1, 2, 6, 23, 104, 531, 2982, 18109, 117545, 808764, 5862253, 44553224, 353713232, 2924697019, 25124481690, 223768976093, 2062614190733, 19646231085928, 193102738376890, 1956191484175505, 20401540100814142, 218825717967033373, 2411606083999341827
Offset: 0

Views

Author

Keywords

Examples

			G.f.: A(x) = x + x^2 + 2*x^3 + 6*x^4 + 23*x^5 + 104*x^6 + ...
A(A(x)) = x + 2*x^2 + 6*x^3 + 23*x^4 + 104*x^5 + 531*x^6 + ...
		

Crossrefs

Programs

  • Maple
    A:= proc(n) option remember;
          unapply(`if`(n=0, x,
          A(n-1)(x)+coeff(A(n-1)(A(n-1)(x)), x, n) *x^(n+1)), x)
        end:
    a:= n-> coeff(A(n)(x),x,n):
    seq(a(n), n=0..20);  # Alois P. Heinz, Feb 24 2012
  • Mathematica
    A[0] = Identity; A[n_] := A[n] = Function[x, Evaluate[A[n-1][x]+Coefficient[A[n-1][A[n-1][x]], x, n]*x^(n+1)]]; a[n_] := Coefficient[A[n][x], x, n]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Feb 17 2014, after Alois P. Heinz *)
    CoefficientList[Nest[x + x (# /. x -> #) &, O[x], 30], x] (* Vladimir Reshetnikov, Aug 08 2019 *)
  • PARI
    {a(n)=local(A=1+x);for(i=0,n,A=1+x*A*subst(A,x,x*A+x*O(x^n)));polcoeff(A,n)} \\ Paul D. Hanna, Mar 10 2007
    
  • PARI
    {a(n)=local(A=sum(i=1,n-1,a(i)*x^i)+x*O(x^n));if(n==0,0,polcoeff((1+A)^n/n,n-1))} \\ Paul D. Hanna, Nov 18 2008
    
  • PARI
    {a(n,m=1)=if(n==0,1,if(m==0,0^n,sum(k=0,n,m*binomial(n+m,k)/(n+m)*a(n-k,k))))} \\ Paul D. Hanna, Jul 09 2009
    
  • PARI
    {a(n)=local(A=1+sum(i=1,n-1,a(i)*x^i+x*O(x^n)));
    for(i=1,n,A=exp(sum(m=1,n,sum(k=0,n-m,binomial(m+k-1,k)*polcoeff(A^(2*m),k)*x^k)*x^m/m)+x*O(x^n)));polcoeff(A,n)} \\ Paul D. Hanna, Dec 15 2010

Formula

G.f. A(x) satisfies the functional equation: A(x)-x = x*A(A(x)). - Paul D. Hanna, Aug 04 2002
G.f.: A(x/(1+A(x))) = x. - Paul D. Hanna, Dec 04 2003
Suppose the functions A=A(x), B=B(x), C=C(x), etc., satisfy: A = 1 + xAB, B = 1 + xABC, C = 1 + xABCD, D = 1 + xABCDE, etc., then B(x)=A(x*A(x)), C(x)=B(x*A(x)), D(x)=C(x*A(x)), etc., where A(x) = 1 + x*A(x)*A(x*A(x)) and x*A(x) is the g.f. of this sequence (see table A128325). - Paul D. Hanna, Mar 10 2007
G.f. A(x) = x*F(x,1) where F(x,n) satisfies: F(x,n) = F(x,n-1)*(1 + x*F(x,n+1)) for n>0 with F(x,0)=1. - Paul D. Hanna, Apr 16 2007
a(n) = [x^(n-1)] [1 + A(x)]^n/n for n>=1 with a(0)=0; i.e., a(n) equals the coefficient of x^(n-1) in [1+A(x)]^n/n for n >= 1. - Paul D. Hanna, Nov 18 2008
From Paul D. Hanna, Jul 09 2009: (Start)
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n with a(0,m)=1, then
a(n,m) = Sum_{k=0..n} m*C(n+m,k)/(n+m) * a(n-k,k).
(End)
G.f. satisfies:
* A(x) = x*exp( Sum_{m>=0} {d^m/dx^m A(x)^(m+1)/x} * x^(m+1)/(m+1)! );
* A(x) = x*exp( Sum_{m>=1} [Sum_{k>=0} C(m+k-1,k)*{[y^k] A(y)^m/y^m}*x^k]*x^m/m );
which are equivalent. - Paul D. Hanna, Dec 15 2010
The g.f. satisfies:
log(A(x)/x) = A(x) + {d/dx A(x)^2/x}*x^2/2! + {d^2/dx^2 A(x)^3/x}*x^3/3! + {d^3/dx^3 A(x)^4/x}*x^4/4! + ... - Paul D. Hanna, Dec 15 2010

A001858 Number of forests of trees on n labeled nodes.

Original entry on oeis.org

1, 1, 2, 7, 38, 291, 2932, 36961, 561948, 10026505, 205608536, 4767440679, 123373203208, 3525630110107, 110284283006640, 3748357699560961, 137557910094840848, 5421179050350334929, 228359487335194570528, 10239206473040881277575, 486909744862576654283616
Offset: 0

Views

Author

Keywords

Comments

The number of integer lattice points in the permutation polytope of {1,2,...,n}. - Max Alekseyev, Jan 26 2010
Equals the number of score sequences for a tournament on n vertices. See Prop. 7 of the article by Bartels et al., or Example 3.1 in the article by Stanley. - David Radcliffe, Aug 02 2022
Number of labeled acyclic graphs on n vertices. The unlabeled version is A005195. The covering case is A105784, connected A000272. - Gus Wiseman, Apr 29 2024

Examples

			From _Gus Wiseman_, Apr 29 2024: (Start)
Edge-sets of the a(4) = 38 forests:
  {}  {12}  {12,13}  {12,13,14}
      {13}  {12,14}  {12,13,24}
      {14}  {12,23}  {12,13,34}
      {23}  {12,24}  {12,14,23}
      {24}  {12,34}  {12,14,34}
      {34}  {13,14}  {12,23,24}
            {13,23}  {12,23,34}
            {13,24}  {12,24,34}
            {13,34}  {13,14,23}
            {14,23}  {13,14,24}
            {14,24}  {13,23,24}
            {14,34}  {13,23,34}
            {23,24}  {13,24,34}
            {23,34}  {14,23,24}
            {24,34}  {14,23,34}
                     {14,24,34}
(End)
		

References

  • B. Bollobas, Modern Graph Theory, Springer, 1998, p. 290.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

The connected case is A000272, rooted A000169.
The unlabeled version is A005195, connected A000055.
The covering case is A105784, unlabeled A144958.
Row sums of A138464.
For triangles instead of cycles we have A213434, covering A372168.
For a unique cycle we have A372193, covering A372195.
A002807 counts cycles in a complete graph.
A006125 counts simple graphs, unlabeled A000088.
A006129 counts covering graphs, unlabeled A002494.

Programs

  • Maple
    exp(x+x^2+add(n^(n-2)*x^n/n!, n=3..50));
    # second Maple program:
    a:= proc(n) option remember; `if`(n=0, 1, add(
          binomial(n-1, j-1)*j^(j-2)*a(n-j), j=1..n))
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Sep 15 2008
    # third Maple program:
    F:= exp(-LambertW(-x)*(1+LambertW(-x)/2)):
    S:= series(F,x,51):
    seq(coeff(S,x,j)*j!, j=0..50); # Robert Israel, May 21 2015
  • Mathematica
    nn=20;t=Sum[n^(n-1)x^n/n!,{n,1,nn}];Range[0,nn]!CoefficientList[ Series[Exp[t-t^2/2],{x,0,nn}],x] (* Geoffrey Critzer, Sep 05 2012 *)
    nmax = 20; CoefficientList[Series[-LambertW[-x]/(x*E^(LambertW[-x]^2/2)), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Jul 19 2019 *)
  • PARI
    a(n)=if(n<0,0,sum(m=0,n,sum(j=0,m,binomial(m,j)*binomial(n-1,n-m-j)*n^(n-m-j)*(m+j)!/(-2)^j)/m!)) /* Michael Somos, Aug 22 2002 */

Formula

E.g.f.: exp( Sum_{n>=1} n^(n-2)*x^n/n! ). This implies (by a theorem of Wright) that a(n) ~ exp(1/2)*n^(n-2). - N. J. A. Sloane, May 12 2008 [Corrected by Philippe Flajolet, Aug 17 2008]
E.g.f.: exp(T - T^2/2), where T = T(x) = Sum_{n>=1} n^(n-1)*x^n/n! is Euler's tree function (see A000169). - Len Smiley, Dec 12 2001
Shifts 1 place left under the hyperbinomial transform (cf. A088956). - Paul D. Hanna, Nov 03 2003
a(0) = 1, a(n) = Sum_{j=0..n-1} C(n-1,j) (j+1)^(j-1) a(n-1-j) if n>0. - Alois P. Heinz, Sep 15 2008

Extensions

More terms from Michael Somos, Aug 22 2002

A007556 Number of 8-ary trees with n vertices.

Original entry on oeis.org

1, 1, 8, 92, 1240, 18278, 285384, 4638348, 77652024, 1329890705, 23190029720, 410333440536, 7349042994488, 132969010888280, 2426870706415800, 44627576949364104, 826044435409399800, 15378186970730687400, 287756293703544823872, 5409093674555090316300
Offset: 0

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Author

Keywords

Comments

Shifts left when convolved three times.
From Wolfdieter Lang, Sep 14 2007: (Start)
a(n), n >= 1, enumerates octic (8-ary) trees (rooted, ordered, incomplete) with n vertices (including the root).
Pfaff-Fuss-Catalan sequence C^{m}_n for m = 8. See the Graham et al. reference, p. 347. eq. 7.66. See also the Pólya-Szegő reference.
Also 8-Raney sequence. See the Graham et al. reference, p. 346-7.
(End)
This is instance k = 8 of the generalized Catalan family {C(k, n)}A130564.%20-%20_Wolfdieter%20Lang">{n>=0} given in a comment of A130564. - _Wolfdieter Lang, Feb 05 2024

Examples

			There are a(2) = 8 octic trees (vertex degree less than or equal to 8 and 8 possible branchings) with 2 vertices (one of them the root). Adding one more branch (one more vertex) to these 8 trees yields 8*8 + binomial(8, 2) = 92 = a(3) such trees.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, pp. 200, 347.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, Heidelberg, New York, 2 vols., 1972, Vol. 1, problem 211, p. 146 with solution on p. 348.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Seventh column of triangle A062993.
Cf. A130564.

Programs

  • Haskell
    a007556 0 = 1
    a007556 n = a007318' (8 * n) (n - 1) `div` n
    -- Reinhard Zumkeller, Jul 30 2013
    
  • Magma
    [Binomial(8*n, n)/(7*n+1): n in [0..20]]; // Vincenzo Librandi, Apr 02 2015
    
  • Maple
    seq(binomial(8*n+1,n)/(8*n+1),n=0..30); # Robert FERREOL, Apr 01 2015
    n:=30: G:=series(RootOf(g = 1+x*g^8, g),x=0,n+1): seq(coeff(G,x,k),k=0..n); # Robert FERREOL, Apr 01 2015
  • Mathematica
    Table[Binomial[8n, n]/(7n + 1), {n, 0, 20}] (* Harvey P. Dale, Dec 24 2012 *)
  • PARI
    vector(100, n, n--; binomial(8*n, n)/(7*n+1)) \\ Altug Alkan, Oct 14 2015

Formula

a(n) = binomial(8*n, n)/(7*n+1) = binomial(8*n+1, n)/(8*n+1) = A062993(n+6,6).
O.g.f.: A(x) = 1 + x*A(x)^8 = 1/(1-x*A(x)^7).
a(0) = 1; a(n) = Sum_{i1 + i2 + .. i8 = n - 1} a(i1)*a(i2)*...*a(i8) for n >= 1. - Robert FERREOL, Apr 01 2015
a(n) = binomial(8*n, n - 1)/n for n >= 1, a(0) = 1 (from the Lagrange series of the o.g.f. A(x) with its above given implicit equation).
From Karol A. Penson, Mar 26 2015: (Start)
In Maple notation,
e.g.f.: hypergeom([1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8], [2/7, 3/7, 4/7, 5/7, 6/7, 1, 8/7],(2^24/7^7)*z);
o.g.f.: hypergeom([1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8], [2/7, 3/7, 4/7, 5/7, 6/7, 8/7],(2^24/7^7)*z);
a(n) are special values of Jacobi polynomials, in Maple notation:
a(n) = JacobiP(n - 1, 7*n + 1, -n, 1)/n, n = 1, 2, ...
(End)
From Peter Bala, Oct 14 2015: (Start)
A(x)^2 is o.g.f. for A234461; A(x)^3 is o.g.f. for A234462;
A(x)^4 is o.g.f. for A234463; A(x)^5 is o.g.f. for A234464;
A(x)^6 is o.g.f. for A234465; A(x)^7 is o.g.f. for A234466;
A(x)^9 is o.g.f. for A234467. (End)
a(n) ~ 2^(24*n + 1)/(sqrt(Pi)*7^(7*n + 3/2)*n^(3/2)). - Ilya Gutkovskiy, Feb 07 2017
D-finite with recurrence: 7*n*(7*n-3)*(7*n+1)*(7*n-2)*(7*n-5)*(7*n-1)*(7*n-4)*a(n) -128*(8*n-5)*(4*n-1)*(8*n-7)*(2*n-1)*(8*n-1)*(4*n-3)*(8*n-3)*a(n-1)=0. - R. J. Mathar, Feb 20 2020
G.f. A(x) satisfies A(x) = 1/A(-x*A(x)^15). - Seiichi Manyama, Jun 16 2025

A300660 Number of unlabeled rooted phylogenetic trees with n (leaf-) nodes such that for each inner node all children are either leaves or roots of distinct subtrees.

Original entry on oeis.org

0, 1, 1, 2, 3, 6, 13, 30, 72, 182, 467, 1222, 3245, 8722, 23663, 64758, 178459, 494922, 1380105, 3867414, 10884821, 30756410, 87215419, 248117618, 707952902, 2025479210, 5809424605, 16700811214, 48113496645, 138884979562, 401645917999, 1163530868090
Offset: 0

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Author

Alois P. Heinz, Jun 18 2018

Keywords

Comments

From Gus Wiseman, Jul 31 2018 and Feb 06 2020: (Start)
a(n) is the number of lone-child-avoiding rooted identity trees whose leaves form an integer partition of n. For example, the following are the a(6) = 13 lone-child-avoiding rooted identity trees whose leaves form an integer partition of 6.
6,
(15),
(24),
(123), (1(23)), (2(13)), (3(12)),
(1(14)),
(1(1(13))),
(12(12)), (1(2(12))), (2(1(12))),
(1(1(1(12)))).
(End)

Examples

			:   a(3) = 2:        :   a(4) = 3:                      :
:      o       o     :        o         o        o      :
:     / \     /|\    :       / \       / \     /( )\    :
:    o   N   N N N   :      o   N     o   N   N N N N   :
:   ( )              :     / \       /|\                :
:   N N              :    o   N     N N N               :
:                    :   ( )                            :
:                    :   N N                            :
From _Gus Wiseman_, Feb 06 2020: (Start)
The a(2) = 1 through a(6) = 13 unlabeled rooted phylogenetic semi-identity trees:
  (oo) (ooo)     (oooo)         (ooooo)             (oooooo)
       ((o)(oo)) ((o)(ooo))     ((o)(oooo))         ((o)(ooooo))
                 ((o)((o)(oo))) ((oo)(ooo))         ((oo)(oooo))
                                ((o)((o)(ooo)))     ((o)(oo)(ooo))
                                ((oo)((o)(oo)))     (((o)(oo))(ooo))
                                ((o)((o)((o)(oo)))) ((o)((o)(oooo)))
                                                    ((o)((oo)(ooo)))
                                                    ((oo)((o)(ooo)))
                                                    ((o)(oo)((o)(oo)))
                                                    ((o)((o)((o)(ooo))))
                                                    ((o)((oo)((o)(oo))))
                                                    ((oo)((o)((o)(oo))))
                                                    ((o)((o)((o)((o)(oo)))))
(End)
		

Crossrefs

Programs

  • Maple
    b:= proc(n,i) option remember; `if`(n=0, 1, `if`(i<1, 0,
          add(b(n-i*j, i-1)*binomial(a(i), j), j=0..n/i)))
        end:
    a:= n-> `if`(n=0, 0, 1+b(n, n-1)):
    seq(a(n), n=0..30);
  • Mathematica
    b[0, ] = 1; b[, _?NonPositive] = 0;
    b[n_, i_] := b[n, i] = Sum[b[n-i*j, i-1]*Binomial[a[i], j], {j, 0, n/i}];
    a[0] = 0; a[n_] := a[n] = 1 + b[n, n-1];
    Table[a[n], {n, 0, 31}] (* Jean-François Alcover, May 03 2019, from Maple *)
    ursit[n_]:=Prepend[Join@@Table[Select[Union[Sort/@Tuples[ursit/@ptn]],UnsameQ@@#&],{ptn,Select[IntegerPartitions[n],Length[#]>1&]}],n];
    Table[Length[ursit[n]],{n,10}] (* Gus Wiseman, Feb 06 2020 *)

Formula

a(n) ~ c * d^n / n^(3/2), where d = 3.045141208159736483720243229947630323380565686... and c = 0.2004129296838557718008171812000512670126... - Vaclav Kotesovec, Aug 27 2018
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