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 1-6 of 6 results.

A086810 Triangle obtained by adding a leading diagonal 1,0,0,0,... to A033282.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 5, 5, 0, 1, 9, 21, 14, 0, 1, 14, 56, 84, 42, 0, 1, 20, 120, 300, 330, 132, 0, 1, 27, 225, 825, 1485, 1287, 429, 0, 1, 35, 385, 1925, 5005, 7007, 5005, 1430, 0, 1, 44, 616, 4004, 14014, 28028, 32032, 19448, 4862, 0, 1, 54, 936, 7644, 34398, 91728
Offset: 0

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Author

Philippe Deléham, Aug 05 2003

Keywords

Comments

Mirror image of triangle A133336. - Philippe Deléham, Dec 10 2008
From Tom Copeland, Oct 09 2011: (Start)
With polynomials
P(0,t) = 0
P(1,t) = 1
P(2,t) = t
P(3,t) = t + 2 t^2
P(4,t) = t + 5 t^2 + 5 t^3
P(5,t) = t + 9 t^2 + 21 t^3 + 14 t^4
The o.g.f. A(x,t) = {1+x-sqrt[(1-x)^2-4xt]}/[2(1+t)] (see Drake et al.).
B(x,t)= x-t x^2/(1-x)= x-t(x^2+x^3+x^4+...) is the comp. inverse in x.
Let h(x,t) = 1/(dB/dx) = (1-x)^2/(1+(1+t)*x*(x-2)) = 1/(1-t(2x+3x^2+4x^3+...)), an o.g.f. in x for row polynomials in t of A181289. Then P(n,t) is given by (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A = exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t). These results are a special case of A133437 with u(x,t) = B(x,t), i.e., u_1=1 and (u_n)=-t for n > 1. See A001003 for t = 1. (End)
Let U(x,t) = [A(x,t)-x]/t, then U(x,0) = -dB(x,t)/dt and U satisfies dU/dt = UdU/dx, the inviscid Burgers' equation (Wikipedia), also called the Hopf equation (see Buchstaber et al.). Also U(x,t) = U(A(x,t),0) = U(x+tU,0) since U(x,0) = [x-B(x,t)]/t. - Tom Copeland, Mar 12 2012
Diagonals of A132081 are essentially rows of this sequence. - Tom Copeland, May 08 2012
T(r, s) is the number of [0,r]-covering hierarchies with s segments (see Kreweras). - Michel Marcus, Nov 22 2014
From Yu Hin Au, Dec 07 2019: (Start)
T(n,k) is the number of small Schröder n-paths (lattice paths from (0,0) to (2n,0) using steps U=(1,1), F=(2,0), D=(1,-1) with no F step on the x-axis) that has exactly k U steps.
T(n,k) is the number of Schröder trees (plane rooted tree where each internal node has at least two children) with exactly n+1 leaves and k internal nodes. (End)

Examples

			Triangle starts:
  1;
  0,  1;
  0,  1,  2;
  0,  1,  5,  5;
  0,  1,  9, 21, 14;
  ...
		

Crossrefs

The diagonals (except for A000007) are also the diagonals of A033282.
Row sums: A001003 (Schroeder numbers).

Programs

  • Mathematica
    Table[Boole[n == 2] + If[# == -1, 0, Binomial[n - 3, #] Binomial[n + # - 1, #]/(# + 1)] &[k - 1], {n, 2, 12}, {k, 0, n - 2}] // Flatten (* after Jean-François Alcover at A033282, or *)
    Table[If[n == 0, 1, Binomial[n, k] Binomial[n + k, k - 1]/n], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 22 2016 *)
  • PARI
    t(n, k) = if (n==0, 1, binomial(n, k)*binomial(n+k, k-1)/n);
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n,k), ", ");); print(););} \\ Michel Marcus, Nov 22 2014

Formula

Triangle T(n, k) read by rows; given by [0, 1, 0, 1, 0, 1, ...] DELTA [1, 1, 1, 1, 1, 1, 1, 1, 1, ...] where DELTA is Deléham's operator defined in A084938.
For k>0, T(n, k) = binomial(n-1, k-1)*binomial(n+k, k)/(n+1); T(0, 0) = 1 and T(n, 0) = 0 if n > 0. [corrected by Marko Riedel, May 04 2023]
Sum_{k>=0} T(n, k)*2^k = A107841(n). - Philippe Deléham, May 26 2005
Sum_{k>=0} T(n-k, k) = A005043(n). - Philippe Deléham, May 30 2005
T(n, k) = A108263(n+k, k). - Philippe Deléham, May 30 2005
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A001003(n), A107841(n), A131763(n), A131765(n), A131846(n), A131926(n), A131869(n), A131927(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, respectively. - Philippe Deléham, Nov 05 2007
Sum_{k=0..n} T(n,k)*5^k*(-2)^(n-k) = A152601(n). - Philippe Deléham, Dec 10 2008
Sum_{k=0..n} T(n,k)*(-1)^k*3^(n-k) = A154825(n). - Philippe Deléham, Jan 17 2009
Umbrally, P(n,t) = Lah[n-1,-t*a.]/n! = (1/n)*Sum_{k=1..n-1} binomial(n-2,k-1)a_k t^k/k!, where (a.)^k = a_k = (n-1+k)!/(n-1)!, the rising factorial, and Lah(n,t) = n!*Laguerre(n,-1,t) are the Lah polynomials A008297 related to the Laguerre polynomials of order -1. - Tom Copeland, Oct 04 2014
T(n, k) = binomial(n, k)*binomial(n+k, k-1)/n, for k >= 0; T(0, 0) = 1 (see Kreweras, p. 21). - Michel Marcus, Nov 22 2014
P(n,t) = Lah[n-1,-:Dt:]/n! t^(n-1) with (:Dt:)^k = (d/dt)^k t^k = k! Laguerre(k,0,-:tD:) with (:tD:)^j = t^j D^j. The normalized Laguerre polynomials of 0 order are given in A021009. - Tom Copeland, Aug 22 2016

Extensions

Typo in a(60) corrected by Michael De Vlieger, Nov 21 2019

A119370 G.f. satisfies A(x) = 1 + x*A(x)^2 + x^2*(A(x)^2 - A(x)).

Original entry on oeis.org

1, 1, 2, 6, 19, 64, 225, 816, 3031, 11473, 44096, 171631, 675130, 2679728, 10719237, 43168826, 174885089, 712222799, 2914150406, 11973792218, 49385167369, 204386777160, 848530495383, 3532844222611, 14747626307436, 61712139464939
Offset: 0

Views

Author

Paul D. Hanna, May 16 2006

Keywords

Comments

Equals base sequence of pendular trinomial triangle A119369; iterated convolutions of this sequence with the central terms (A119371) generates all diagonals of A119369. For example: A119372 = A119370 * A119371; A119373 = A119370^2 * A119371.
Diagonal sums of number array A133336. - Philippe Deléham, Nov 09 2009

Examples

			A(x) = 1 + x + 2*x^2 + 6*x^3 + 19*x^4 + 64*x^5 + 225*x^6 + 816*x^7 +...
x*A(x)^2 = x + 2*x^2 + 5*x^3 + 16*x^4 + 54*x^5 + 190*x^6 + 690*x^7 +...
x^2*( A(x)^2 - A(x) ) = 1*x^3 + 3*x^4 + 10*x^5 + 35*x^6 + 126*x^7 +...
		

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30);
    Coefficients(R!( (1+x^2 -Sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)) )); // G. C. Greubel, Mar 17 2021
  • Maple
    m:= 30;
    S:= series( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)), x, m+1);
    seq(coeff(S, x, j), j = 0..m); # G. C. Greubel, Mar 17 2021
  • Mathematica
    CoefficientList[Series[((1+x^2)-Sqrt[(1+x^2)^2-4*x*(1+x)])/(2*x*(1+x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 11 2013 *)
  • PARI
    {a(n)=polcoeff(2/((1+x^2)+sqrt((1+x^2)^2-4*x*(1+x)+x*O(x^n))),n)}
    
  • Sage
    def A119370_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( (1+x^2 -sqrt(1-4*x-2*x^2+x^4))/(2*x*(1+x)) ).list()
    A119370_list(30) # G. C. Greubel, Mar 17 2021
    

Formula

G.f.: A(x) = ((1+x^2) - sqrt( (1+x^2)^2 - 4*x*(1+x) ))/(2*x*(1+x)). Equals the inverse binomial transform of A104547.
Recurrence: (n+1)*a(n) = 3*(n-1)*a(n-1) + 6*(n-1)*a(n-2) + 2*(n-2)*a(n-3) - (n-5)*a(n-4) - (n-5)*a(n-5). - Vaclav Kotesovec, Sep 11 2013
a(n) ~ sqrt(-z^2-3*z+1)*(4+2*z-z^3)^(n+1)*(-z^3+z^2+z+3) / (8*sqrt(Pi) * n^(3/2)), where z = 1/(2*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35 + 3*sqrt(129))^(1/3)))) - 1/2*sqrt(8/3-1/3*(280-24*sqrt(129))^(1/3) - 2/3*(35+3*sqrt(129))^(1/3) + 8*sqrt(3/(4+(280-24*sqrt(129))^(1/3) + 2*(35 + 3*sqrt(129))^(1/3)))) = 0.225270426... is the root of the equation 1-2*z^2+z^4-4*z=0. - Vaclav Kotesovec, Sep 11 2013
G.f.: 1/G(0) where G(k) = 1 - q/(1 - (q + q^2) / G(k+1) ). - Joerg Arndt, Dec 06 2014
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k-1,k) * binomial(2*n-3*k+1,n-2*k)/(2*n-3*k+1). - Seiichi Manyama, Aug 28 2023
Conjecture: A(x) = 1 + x*exp(Sum_{n >= 1} g(n, x)*x^n/n), where g(n, x) = Sum_{k = 0..n} binomial(n, k)^2*(1 + x)^k. Cf. A105633 and A167638. - Peter Bala, Sep 10 2024

A122446 G.f. satisfies: A(x) = 1 + x*A(x)^2 + 2*x^2*(A(x)^2 - A(x)); equals the base sequence of pendular trinomial triangle A122445.

Original entry on oeis.org

1, 1, 2, 7, 24, 88, 336, 1321, 5316, 21788, 90640, 381750, 1624592, 6975136, 30177056, 131428917, 575765820, 2535433668, 11216757104, 49829385786, 222193501760, 994153952528, 4461915817760, 20082611971226, 90625360612296
Offset: 0

Views

Author

Paul D. Hanna, Sep 07 2006

Keywords

Comments

Functional equation for the g.f. is derived from the recurrence of the pendular triangle A122445. Iterated convolutions of this sequence with the central terms (A122447) generates all diagonals of A122445. For example: A122448 = A122446 * A122447; A122449 = A122446^2 * A122447.
Diagonal sums of triangle T with T(n,k) = 2^k*A133336(n,k). - Philippe Deléham, Nov 10 2009

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 30);
    Coefficients(R!( (1+2*x^2 -Sqrt(1-4*x-4*x^2+4*x^4))/(2*x*(1+2*x)) )); // G. C. Greubel, Mar 16 2021
  • Maple
    m:=30; S:=series( (1+2*x^2 -sqrt(1-4*x-4*x^2+4*x^4))/(2*x*(1+2*x)), x, m+1):
    seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Mar 16 2021
  • Mathematica
    CoefficientList[Series[(1+2*x^2-Sqrt[1-4*x-4*x^2+4*x^4])/(2*x*(1+2*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 17 2013 *)
  • PARI
    {a(n)=polcoeff(2/(1+2*x^2+sqrt(1-4*x-4*x^2+4*x^4+x*O(x^n))),n)}
    
  • Sage
    def A122446_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( (1+2*x^2 -sqrt(1-4*x-4*x^2+4*x^4))/(2*x*(1+2*x)) ).list()
    A122446_list(30) # G. C. Greubel, Mar 16 2021
    

Formula

G.f.: A(x) = (1 + 2*x^2 - sqrt(1 -4*x -4*x^2 +4*x^4))/(2*x*(1+2*x)).
Recurrence: (n+1)*a(n) = 2*(n-2)*a(n-1) + 12*(n-1)*a(n-2) + 8*(n-2)*a(n-3) - 4*(n-5)*a(n-4) - 8*(n-5)*a(n-5). - Vaclav Kotesovec, Sep 17 2013
a(n) ~ c*d^n/(sqrt(Pi)*n^(3/2)), where d = (1/6)*(6+sqrt(6*(10 + 2^(2/3)*(43-3*sqrt(177))^(1/3) + 2^(2/3)*(43+3*sqrt(177))^(1/3))) + sqrt(6*(20-2^(2/3)*(43-3*sqrt(177))^(1/3) - 2^(2/3)*(43+3*sqrt(177))^(1/3) + 24*sqrt(6/(10+2^(2/3)*(43-3*sqrt(177))^(1/3) + 2^(2/3)*(43+3*sqrt(177))^(1/3)))))) = 4.797536514160165558... is the root of the equation 4 - 4*d^2 - 4*d^3 + d^4 = 0 and c = 0.908214882020417619380249683... is the positive root of the equation -59 - 944*c^2 - 2032*c^4 - 320*c^6 + 5184*c^8 = 0. - Vaclav Kotesovec, Sep 17 2013, updated Mar 18 2024

A131198 Triangle T(n,k), 0 <= k <= n, read by rows, given by [1,0,1,0,1,0,1,0,...] DELTA [0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 6, 1, 0, 1, 10, 20, 10, 1, 0, 1, 15, 50, 50, 15, 1, 0, 1, 21, 105, 175, 105, 21, 1, 0, 1, 28, 196, 490, 490, 196, 28, 1, 0, 1, 36, 336, 1176, 1764, 1176, 336, 36, 1, 0, 1, 45, 540, 2520, 5292, 5292, 2520, 540, 45, 1, 0
Offset: 0

Views

Author

Philippe Deléham, Oct 20 2007

Keywords

Comments

Mirror image of triangle A090181, another version of triangle of Narayana (A001263).
Equals A133336*A130595 as infinite lower triangular matrices. - Philippe Deléham, Oct 23 2007

Examples

			Triangle begins:
  1;
  1,  0;
  1,  1,   0;
  1,  3,   1,   0;
  1,  6,   6,   1,   0;
  1, 10,  20,  10,   1,   0;
  1, 15,  50,  50,  15,   1,  0;
  1, 21, 105, 175, 105,  21,  1, 0;
  1, 28, 196, 490, 490, 196, 28, 1, 0; ...
		

Crossrefs

Programs

  • Magma
    [[n le 0 select 1 else (n-k)*Binomial(n,k)^2/(n*(k+1)): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018
  • Maple
    T := (n,k) -> `if`(n=0, 0^n, binomial(n,k)^2*(n-k)/(n*(k+1)));
    seq(print(seq(T(n,k), k=0..n)), n=0..5); # Peter Luschny, Jun 08 2014
    R := n -> simplify(hypergeom([1 - n, -n], [2], x)):
    Trow := n -> seq(coeff(R(n, x), x, k), k = 0..n):
    seq(print(Trow(n)), n = 0..9); # Peter Luschny, Apr 26 2022
  • Mathematica
    Table[If[n == 0, 1, (n-k)*Binomial[n,k]^2/(n*(k+1))], {n,0,10}, {k,0,n}] //Flatten (* G. C. Greubel, Feb 06 2018 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(if(n==0,1, (n-k)*binomial(n,k)^2/(n* (k+1))), ", "))) \\ G. C. Greubel, Feb 06 2018
    

Formula

Sum_{k=0..n} T(n,k)*x^k = A000012(n), A000108(n), A001003(n), A007564(n), A059231(n), A078009(n), A078018(n), A081178(n), A082147(n), A082181(n), A082148(n), A082173(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000007(n), A000108(n), A006318(n), A047891(n+1), A082298(n), A082301(n), A082302(n), A082305(n), A082366(n), A082367(n), for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Oct 23 2007
Sum_{k=0..floor(n/2)} T(n-k,k) = A004148(n). - Philippe Deléham, Nov 06 2007
T(2*n,n) = A125558(n). - Philippe Deléham, Nov 16 2011
T(n, k) = [x^k] hypergeom([1 - n, -n], [2], x). - Peter Luschny, Apr 26 2022

A154825 Reversion of x*(1-2*x)/(1-3*x).

Original entry on oeis.org

1, -1, -1, 1, 5, 3, -21, -51, 41, 391, 407, -1927, -6227, 2507, 49347, 71109, -236079, -966129, 9519, 7408497, 13685205, -32079981, -167077221, -60639939, 1209248505, 2761755543, -4457338681, -30629783831, -22124857219, 206064020315, 572040039283, -590258340811
Offset: 0

Views

Author

Paul Barry, Jan 15 2009

Keywords

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( (1+3*x-Sqrt(1-2*x+9*x^2))/(4*x) )); // G. C. Greubel, May 24 2022
    
  • Maple
    A154825_list := proc(n) local j, a, w; a := array(0..n); a[0] := 1;
    for w from 1 to n do a[w] := -a[w-1]+2*add(a[j]*a[w-j-1],j=1..w-1)od;
    convert(a, list) end: A154825_list(28); # Peter Luschny, May 19 2011
  • Mathematica
    CoefficientList[Series[(1+3*x-Sqrt[1-2*x+9*x^2])/(4*x), {x, 0, 40}], x] (* Vaclav Kotesovec, Feb 08 2014 *)
  • SageMath
    [sum(binomial(n+k,n-k)*catalan_number(k)*2^k*(-3)^(n-k) for k in (0..n)) for n in (0..40)] # G. C. Greubel, May 24 2022

Formula

G.f.: (1+3*x-sqrt(1-2*x+9*x^2))/(4*x). - corrected by Vaclav Kotesovec, Feb 08 2014
G.f.: 1/(1+x/(1-2x/(1+x/(1-2x/(1+x/(1-2x/(1+.... (continued fraction).
a(n) = Sum_{k=0..n} binomial(n+k, 2k)*A000108(k)*2^k*(-3)^(n-k).
From Philippe Deléham, Jan 17 2009: (Start)
a(n) = Sum_{k=0..n} A131198(n,k)*(-1)^(n-k)*2^k.
a(n) = Sum_{k=0..n} A090181(n,k)*(-1)^k*2^(n-k).
a(n) = Sum_{k=0..n} A060693(n,k)*2^(n-k)*(-3)^k.
a(n) = Sum_{k=0..n} A088617(n,k)*2^k*(-3)^(n-k).
a(n) = Sum_{k=0..n} A086810(n,k)*(-1)^k*3^(n-k).
a(n) = Sum_{k=0..n} A133336(n,k)*3^k*(-1)^(n-k). (End)
D-finite with recurrence (n+1)*a(n) = (2*n-1)*a(n-1) - 9*(n-2)*a(n-2). - R. J. Mathar, Nov 15 2012
a(n) = (-3)^n*Hypergeometric2F1([-n, n+1], [2]; 2/3). - G. C. Greubel, May 24 2022

A152601 a(n) = Sum_{k=0..n} C(n+k,2k)*A000108(k)*3^k*2^(n-k).

Original entry on oeis.org

1, 5, 40, 395, 4360, 51530, 637840, 8163095, 107140360, 1434252230, 19507077040, 268796321870, 3744480010960, 52647783144980, 746145741252640, 10648007952942095, 152877753577617160, 2206713692628578030
Offset: 0

Views

Author

Paul Barry, Dec 09 2008

Keywords

Comments

Hankel transform is 15^C(n+1,2).

Crossrefs

Formula

a(n) = A152600(n+1)/2.
a(n) = Sum_{k=0..n} A088617(n,k)*3^k*2^(n-k) = Sum_{k=0..n} A060693(n,k)*2^k*3^(n-k). - Philippe Deléham, Dec 10 2008
a(n) = Sum_{k=0..n} A090181(n,k)*5^k*3^(n-k). - Philippe Deléham, Dec 10 2008
a(n) = Sum_{k=0..n} A131198(n,k)*3^k*5^(n-k). - Philippe Deléham, Dec 10 2008
a(n) = Sum_{k=0..n} A133336(n,k)*(-2)^k*5^(n-k) = Sum_{k=0..n} A086810(n,k)*5^k*(-2)^(n-k). - Philippe Deléham, Dec 10 2008
G.f.: 1/(1-5x/(1-3x/(1-5x/(1-3x/(1-5x/(1-3x/(1-5x/(1-... (continued fraction). - Philippe Deléham, Nov 28 2011
Conjecture: (n+1)*a(n) +8*(-2*n+1)*a(n-1) +4*(n-2)*a(n-2)=0. - R. J. Mathar, Nov 24 2012
G.f.: 1/G(x), with G(x) = 1-2*x-(3*x)/G(x) (continued fraction). - Nikolaos Pantelidis, Jan 09 2023
Showing 1-6 of 6 results.