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 1-5 of 5 results.

A007297 Number of connected graphs on n labeled nodes on a circle with straight-line edges that don't cross.

Original entry on oeis.org

1, 1, 4, 23, 156, 1162, 9192, 75819, 644908, 5616182, 49826712, 448771622, 4092553752, 37714212564, 350658882768, 3285490743987, 30989950019532, 294031964658430, 2804331954047160, 26870823304476690, 258548658860327880
Offset: 1

Views

Author

Keywords

Comments

Apart from the initial 1, reversion of g.f. for A162395 (squares with signs): see A263843.

Examples

			G.f. = x*(1 + x + 4*x^2 + 23*x^3 + 156*x^4 + 1162*x^5 + 9192*x^6 + 75819*x^7 + ...).
		

References

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

Crossrefs

Cf. A162395, A000290. 4th row of A107111. Row sums of A089434.
See A263843 for a variant.
Cf. A000108 (non-crossing set partitions), A001006, A001187, A054726 (non-crossing graphs), A054921, A099947, A194560, A293510, A323818, A324167, A324169, A324173.

Programs

  • Maple
    A007297:=proc(n) if n = 1 then 1 else add(binomial(3*n - 3, n + j)*binomial(j - 1, j - n + 1), j = n - 1 .. 2*n - 3)/(n - 1); fi; end;
  • Mathematica
    CoefficientList[ InverseSeries[ Series[(x-x^2)/(1+x)^3, {x, 0, 20}], x], x] // Rest (* From Jean-François Alcover, May 19 2011, after PARI prog. *)
    Table[Binomial[3n, 2n+1] Hypergeometric2F1[1-n, n, 2n+2, -1]/n, {n, 1, 20}] (* Vladimir Reshetnikov, Oct 25 2015 *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse((x-x^2)/(1+x)^3+O(x^(n+2))),n+1)) /* Ralf Stephan */

Formula

Apart from initial term, g.f. is the series reversion of (x-x^2)/(1+x)^3 (A162395). See A263843. - Vladimir Kruchinin, Feb 08 2013
G.f.: (g-z)/z, where g=-1/3+(2/3)*sqrt(1+9z)*sin((1/3)*arcsin((2+27z+54z^2)/2/(1+9*z)^(3/2))). - Emeric Deutsch, Dec 02 2002
a(n) = (1/n)*Sum_{k=0..n} binomial(3n, n-k-1)*binomial(n+k-1, k). - Paul Barry, May 11 2005
a(n) = 4^(n-1)*(Gamma(3*n/2-1)/Gamma(n/2+1)/Gamma(n) -Gamma((3*n-1)/2)/ Gamma( (n+1)/2)/Gamma(n+1)). - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
a(n) = 4^n * binomial(3*n/2, n/2) / (9*n-6) - 4^(n-1) * binomial(3*(n-1)/2, (n-1)/2 ) / n. - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
D-finite with recurrence: n*(n-1)*(3*n-4)*a(n) +36*(n-1)*a(n-1) -12*(3*n-8)*(3*n-1)*(3*n-7)*a(n-2)=0. - Mark van Hoeij, Aug 27 2005, adapted to offset Feb 21 2020 by R. J. Mathar
a(n) = (1/n)*Sum_{k=0..n} C(3n, k)*C(2n-k-2, n-1). - Paul Barry, Sep 27 2005
a(n) ~ (2-sqrt(3)) * 6^n * 3^(n/2) / (sqrt(2*Pi) * n^(3/2)). - Vaclav Kotesovec, Mar 17 2014
a(n) = binomial(3*n,2*n+1)*hypergeom([1-n,n], [2*n+2], -1)/n. - Vladimir Reshetnikov, Oct 25 2015
a(n) = 2*A078531(n) - A085614(n+1). - Vladimir Reshetnikov, Apr 24 2016

Extensions

Better description from Philippe Flajolet, Apr 20 2000
More terms from James Sellers, Aug 21 2000
Definition revised and initial a(1)=1 added by N. J. A. Sloane, Nov 05 2015 at the suggestion of Axel Boldt. Some of the formulas may now need to be adjusted slightly.

A143018 Triangle read by rows: T(n,k) (n >= 2, k >= 1) is the number of non-crossing connected graphs on n nodes on a circle such that the distance from a fixed node (root) to the next node is k. Rows are indexed 2,3,4,...; columns are indexed 1,2,3, ... .

Original entry on oeis.org

1, 3, 1, 16, 6, 1, 105, 41, 9, 1, 768, 306, 75, 12, 1, 6006, 2422, 630, 118, 15, 1, 49152, 19980, 5394, 1104, 170, 18, 1, 415701, 169941, 47061, 10197, 1755, 231, 21, 1, 3604480, 1479786, 417439, 94116, 17425, 2610, 301, 24, 1
Offset: 2

Views

Author

Emeric Deutsch, Jul 30 2008

Keywords

Comments

Row sums yield A007297.
T(n,1) = A085614(n-1).
Sum_{k=1..n-1} k*T(n,k) = A143020(n).

Examples

			T(3,1)=3 and T(3,2)=1 because in the graphs (AB,BC,CA), (AB,AC), (AB,BC) and (AC,BC) the distances from A to B are 1, 1, 1 and 2, respectively.
Triangle starts:
    1;
    3,   1;
   16,   6,  1;
  105,  41,  9,  1;
  768, 306, 75, 12, 1;
  ...
		

Crossrefs

Programs

  • Maple
    L:=proc(p,q,r) options operator, arrow: sum(binomial(q, i)*binomial(r+p-1-i, r-1), i=0..min(p,q)) end proc: T:=proc(n,k) options operator, arrow: k*L(n-k-1, 3*n-k-4, n-1)/(n-1) end proc: for n from 2 to 10 do seq(T(n,k),k=1..n-1) end do; # yields sequence in triangular form
  • Mathematica
    t[n_, k_] := k*L[n - k - 1, 3*n - k - 4, n-1]/(n-1); L[p_, q_, r_] := Sum[ Binomial[q, i]*Binomial[r + p - 1 - i, r-1], {i, 0, Min[p, q]}]; Flatten[ Table[ t[n, k], {n, 2, 10}, {k, 1, n-1}]] (* Jean-François Alcover, Oct 05 2011, Oct 05 2011, after Maple *)
  • PARI
    T(n,k)=k*sum(i=0, min(n-k-1, 3*n-k-4), binomial(3*n-k-4, i)*binomial(2*n-k-i-3, n-2))/(n-1);
    for(n=2, 10, for(k=1, n-1, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Nov 17 2017

Formula

T(n,k) = k*L(n-k-1, 3n-k-4, n-1)/(n-1) (n >= 2, 1 <= k <= n-1), where L(p,q,r) = [u^p](1+u)^q/(1-u)^r = Sum_{i=0..min(p,q)} binomial(q,i)*binomial(r+p-1-i, r-1).
G.f.: G(t,z) = zg/[g - t*(g - z)], where g=g(z), the g.f. for the number of non-crossing connected graphs on n nodes on a circle, satisfies g^3 + g^2 - 3z*g + 2*z^2 = 0 (A007297).
T(n,k) = k*Sum_{i=0..min(n-k-1, 3*n-k-4)} binomial(3*n-k-4, i)*binomial(2*n-k-i-3, n-2)/(n-1). - Andrew Howroyd, Nov 17 2017

A206300 Expand the real root of y^3 - y + x in powers of x, then multiply coefficient of x^n by -4^n to get integers.

Original entry on oeis.org

-1, 2, 6, 32, 210, 1536, 12012, 98304, 831402, 7208960, 63740820, 572522496, 5209363380, 47915728896, 444799488600, 4161823309824, 39209074920090, 371626340253696, 3541117629057540
Offset: 0

Views

Author

Roger L. Bagula, Feb 05 2012

Keywords

Comments

Also coefficients of the series S(u) for which (-sqrt(3u))*S converges to the larger of the two real roots of x^3 - 3ux + 4u for u >= 4. Specifically, S(u)=Sum_{n>=0} a(n)/(27*u)^(n/2). - Dixon J. Jones, Jun 24 2021

References

  • George E. Andrews, Number Theory, 1971, Dover Publications, New York, pp. 41-43.

Crossrefs

Cf. A000108, A048990, A224884 (signed version).
Cf. A085614.

Programs

  • Mathematica
    p[x_] = y /. Solve[y^3 - y + x == 0, y][[1]]
    b = Table[-4^n*FullSimplify[ExpandAll[SeriesCoefficient[ Series[p[x], {x, 0, 30}], n]]], {n, 0, 30}]
    Table[2^(2n - 1) Gamma[(3n - 1)/2]/(Gamma[(n + 1)/2]n!), {n, 0, 20}] (* Dixon J. Jones, Jun 24 2021 *)
    Table[2^(2n - 1) Pochhammer[(n + 1)/2, (n-1)]/n!, {n, 0, 20}] (* Dixon J. Jones, Jun 24 2021 *)
  • PARI
    -x/serreverse((x*sqrt(1-4*x))) \\ Thomas Baruchel, Jul 02 2018

Formula

G.f.: -(12*x)/(2*sin(arcsin(216*x^2-1)/3)+1). - Vladimir Kruchinin, Oct 30 2014
G.f.: -x/Revert((x*sqrt(1-4*x))). - Thomas Baruchel, Jul 02 2018
G.f.: - (1/x) * Revert( x*sqrt(c(4*x)) ), where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108 and sqrt(c(4*x)) is the g.f. of A048990. - Peter Bala, Mar 05 2020
From Dixon J. Jones, Jun 24 2021: (Start)
a(n) = 2*A085614(n) for n>=1.
a(n) = 2^(2*n - 1) Gamma((3*n - 1)/2)/(Gamma((n + 1)/2)*n!).
a(n) = (2^(2*n - 1)*((n + 1)/2)_(n-1))/n!, where (x)_k is the Pochhammer symbol. (End)
a(n) ~ 2^(n-1/2) * 3^(3*n/2-1) / (sqrt(Pi) * n^(3/2)). - Amiram Eldar, Sep 01 2025

Extensions

Edited by N. J. A. Sloane, Feb 09 2012

A371395 Triangle read by rows: T(n, k) = binomial(n + k, k) * binomial(2*n - k, n - k) / (n + 1).

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 5, 10, 10, 5, 14, 35, 45, 35, 14, 42, 126, 196, 196, 126, 42, 132, 462, 840, 1008, 840, 462, 132, 429, 1716, 3564, 4950, 4950, 3564, 1716, 429, 1430, 6435, 15015, 23595, 27225, 23595, 15015, 6435, 1430
Offset: 0

Views

Author

F. Chapoton, Mar 21 2024

Keywords

Comments

The terms can be seen as graded dimensions of a non-symmetric operad. The Koszul dual operad has Hilbert series x*(1 + x)*(1 + tx). So the current table has as Hilbert series the reverse of x*(1-x)*(1-t*x) w.r.t to x (see Sage below).
The triangle is symmetric under the exchange of k with n - k.

Examples

			Triangle begins:
  [0] [ 1],
  [1] [ 1,   1],
  [2] [ 2,   3,   2],
  [3] [ 5,  10,  10,   5],
  [4] [14,  35,  45,  35,  14],
  [5] [42, 126, 196, 196, 126, 42].
		

Crossrefs

Column 0 and main diagonal are A000108.
Column 1 and subdiagonal are A001700.
Row sums are A006013.
The even bisection of the alternating row sums is A001764.
The central terms are A188681.

Programs

  • Maple
    T := (n, k) -> binomial(n + k, k)*binomial(2*n - k, n)/(n + 1):
    seq(print(seq(T(n, k), k = 0..n)), n = 0..7);  # Peter Luschny, Mar 21 2024
  • Mathematica
    T[n_, k_] := (Hypergeometric2F1[-n, -k, 1, 1] Hypergeometric2F1[-n, k - n, 1, 1]) /(n + 1); Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten
    (* Peter Luschny, Mar 21 2024 *)
  • SageMath
    def Trow(n):
        return [binomial(n+k, k) * binomial(2*n-k, n-k) / (n+1) for k in range(n+1)]
    
  • SageMath
    # As the reverse of x*(1-x)*(1-t*x) w.r.t variable x.
    t = polygen(QQ, 't')
    x = LazyPowerSeriesRing(t.parent(), 'x').0
    gf = x*(1-x)*(1-t*x)
    coeffs = gf.revert() / x
    for n in range(6):
        print(coeffs[n].list())

Formula

From Peter Luschny, Mar 21 2024: (Start)
T(n, k) = hypergeom([-n, -k], [1], 1)*hypergeom([-n, k - n], [1], 1)/(n + 1).
2^n*Sum_{k=0..n} T(n, k)*(1/2)^k = A085614(n + 1).
2^n*Sum_{k=0..n} T(n, k)*(-1/2)^k = A250886(n + 1). (End)

A372506 Coefficient of x^n in the expansion of 1 / ( (1-x) * (1-2*x) )^n.

Original entry on oeis.org

1, 3, 23, 198, 1795, 16758, 159446, 1537308, 14967843, 146833830, 1449054178, 14369723316, 143072565454, 1429331585724, 14320668653580, 143838879376248, 1447883909314851, 14602334949928710, 147518977428892010, 1492559101878005700, 15121898521185194970
Offset: 0

Views

Author

Seiichi Manyama, May 04 2024

Keywords

Crossrefs

Programs

  • Maple
    A372506 := proc(n)
        add(binomial(n+k-1,k)*binomial(3*n-1,n-k),k=0..n) ;
    end proc:
    seq(A372506(n),n=0..80) ; # R. J. Mathar, Oct 24 2024
  • Mathematica
    Table[SeriesCoefficient[1/((1 - x)*(1 - 2*x))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, May 04 2024 *)
    Table[Binomial[3*n - 1, n] * Hypergeometric2F1[-n, n, 2*n, -1], {n, 0, 20}] (* Vaclav Kotesovec, May 04 2024 *)
  • PARI
    a(n, s=1, t=1, u=1) = sum(k=0, n\s, binomial(t*n+k-1, k)*binomial((t+u+1)*n-(s-1)*k-1, n-s*k));

Formula

a(n) = Sum_{k=0..n} binomial(n+k-1,k) * binomial(3*n-1,n-k).
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x * (1-x) * (1-2*x) ).
a(n) ~ (1 + sqrt(3)) * 2^(n - 3/2) * 3^((3*n-1)/2) / sqrt(Pi*n). - Vaclav Kotesovec, May 04 2024
D-finite with recurrence 5*n*(n-1)*a(n) +18*(n-1)*(n-3)*a(n-1) +12*(-45*n^2+90*n-22)*a(n-2) -216*(3*n-7)*(3*n-8)*a(n-3)=0. - R. J. Mathar, Oct 24 2024
Showing 1-5 of 5 results.