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.

Previous Showing 31-40 of 51 results. Next

A213684 Logarithmic derivative of A001002.

Original entry on oeis.org

1, 5, 22, 105, 511, 2534, 12720, 64449, 328900, 1688115, 8705060, 45064110, 234054198, 1219053680, 6364813192, 33302104593, 174570695175, 916628799380, 4820160541350, 25381091113455, 133808636072595, 706211862466500, 3730964595817680, 19729042153581150
Offset: 1

Views

Author

Paul D. Hanna, Jun 22 2012

Keywords

Comments

A001002(n) is the number of dissections of a convex (n+2)-gon into triangles and quadrilaterals by nonintersecting diagonals.
The g.f. of A001002 satisfies: G(x) = 1 + x*G(x)^2 + x^2*G(x)^3.
Central terms in A155161: a(n) = A155161(2*n,n). - Reinhard Zumkeller, Apr 17 2013
a(n) is the 2n-th term of the n-fold self-convolution of the Fibonacci numbers. - Alois P. Heinz, Feb 07 2021

Examples

			L.g.f.: L(x) = x + 5*x^2/2 + 22*x^3/3 + 105*x^4/4 + 511*x^5/5 +...
such that
L(x) = x*(1+x) + d/dx x^3*(1+x)^2/2! + d^2/dx^2 x^5*(1+x)^3/3! + d^3/dx^3 x^7*(1+x)^4/4! +...
The g.f. of A001002 begins:
exp(L(x)) = 1 + x + 3*x^2 + 10*x^3 + 38*x^4 + 154*x^5 + 654*x^6 +...
		

Crossrefs

Programs

  • Haskell
    a213684 n = a155161 (2*n) n  -- Reinhard Zumkeller, Apr 17 2013
  • Maple
    with(orthopoly): seq(add(i, i in [seq((-1)^iquo(n-k,2)*coeff(G(n,n,x/2), x, k), k=0..n)]), n=1..24); # Peter Luschny, Jan 26 2018
  • Mathematica
    Table[n*Sum[Binomial[k+n-1,n]*Binomial[k,n-k]/k,{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
  • PARI
    {a(n)=n*sum(r=1,n,binomial(r+n-1,n)*binomial(r,n-r)/r)}
    for(n=1, 30, print1(a(n), ", "))
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
    {a(n)=local(A=1); A=(sum(m=1, n+1, Dx(m-1, x^(2*m-1)*(1+x)^m/m!)+x*O(x^n))); n*polcoeff(A, n)}
    for(n=1, 30, print1(a(n), ", "))
    

Formula

a(n) = n * Sum_{r=1..n} binomial(r+n-1,n) * binomial(r,n-r) / r.
L.g.f.: Sum_{n>=1} d^(n-1)/dx^(n-1) x^(2*n-1)*(1+x)^n/n! = Sum_{n>=1} a(n)*x^n/n.
Recurrence: 75*(n-1)*n*a(n) = 5*(n-1)*(59*n-12)*a(n-1) + (559*n^2-1503*n+1100)* a(n-2) + 21*(3*n-8)*(3*n-7)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 3^(3*n)/(2*5^(n-1/2)*sqrt(6*Pi*n)). - Vaclav Kotesovec, Oct 20 2012
a(n) = A037027(2*n-1,n-1). - Vladimir Kruchinin, Feb 28 2013
a(n) = Sum_{k=0..n} (-1)^floor((n-k)/2) [x^k] G(n,n,x/2), where G(n,a,x) denotes the n-th Gegenbauer polynomial; row sums of A298610. - Peter Luschny, Jan 26 2018
a(n) = [x^n] (1/(1-x-x^2))^n. - Alois P. Heinz, Feb 07 2021
From Peter Bala, Mar 11 2025: (Start)
a(n) = Sum_{k = 0..n} n/(2*n-k) * binomial(2*n-k, k)*binomial(2*n-2*k, n).
a(n) = (1/2)*binomial(2*n, n)*hypergeom([-n/2, (-n+1)/2], [-2*n+1], -4). Cf. A350383.
Second-order recurrence: 5*n*(n-1)*(8*n-13)*a(n) = 2*(n-1)*(88*n^2-187*n+75)*a(n-1) + 3*(8*n-5)*(3*n-4)*(3*n-5)*a(n-2) with a(1) = 1 and a(2) = 5. (End)

A122075 Coefficients of a generalized Pell-Lucas polynomial read by rows.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 5, 7, 4, 1, 8, 15, 12, 5, 1, 13, 30, 31, 18, 6, 1, 21, 58, 73, 54, 25, 7, 1, 34, 109, 162, 145, 85, 33, 8, 1, 55, 201, 344, 361, 255, 125, 42, 9, 1, 89, 365, 707, 850, 701, 413, 175, 52, 10, 1, 144, 655, 1416, 1918, 1806, 1239, 630, 236, 63, 11, 1
Offset: 0

Views

Author

R. J. Mathar, Oct 16 2006

Keywords

Comments

A122075 is jointly generated with A037027 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+(x+1)*v(n-1)x and v(n,x)=u(n-1,x)+x*v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 05 2012
Subtriangle of the triangle T(n,k) given by (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 11 2012

Examples

			Triangle begins:
  1
  2 1
  3 3 1
  5 7 4 1
  8 15 12 5 1
  13 30 31 18 6 1
A055830 = (1, 1, -1, 0, 0, 0, ...) DELTA (0, 1, 0, 0, 0, 0, ...) begins:
  1
  1, 0
  2, 1, 0
  3, 3, 1, 0
  5, 7, 4, 1, 0
  8, 15, 12, 5, 1, 0
  13, 30, 31, 18, 6, 1, 0
		

Crossrefs

See A055830 for another version.

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x];
    v[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]    (* A122075 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A037027 *)
    (* Clark Kimberling, Mar 05 2012 *)
    CoefficientList[CoefficientList[Series[-(1 + x)/(-1 + x*y + x + x^2), {x, 0, 10}, {y, 0, 10}], x], y] // Flatten (* G. C. Greubel, Dec 24 2017 *)
  • PARI
    T(n,k)={ sum(j=0,n-k+1, binomial(n-k-j+1,j)*binomial(n-j,k)) ; } { nmax=10 ; for(n=0,nmax, for(k=0,n, print1(T(n,k),",") ; ); ); }

Formula

T(n,k) = Sum_{j=0..n-k+1} binomial(n-k-j+1,j)*binomial(n-j,k).
Sum_{k>=0} T(n-k,k) = 2^n.
Sum_{k>=0} (-1)^k*T(n-k,k) = 2-delta(0,n).
G.f.: -(1+x)/(-1+x*y+x+x^2). - R. J. Mathar, Aug 11 2015

A057280 Coefficient triangle of polynomials (rising powers) related to Fibonacci convolutions. Companion triangle to A057995.

Original entry on oeis.org

2, 17, 5, 225, 120, 15, 4080, 3050, 700, 50, 94440, 89225, 28625, 3775, 175, 2666880, 3006000, 1208975, 223175, 19225, 625, 89016480, 115299900, 54824650, 12689800, 1537100, 93500, 2250, 3430929600, 4973077800, 2695596850, 737744125
Offset: 0

Views

Author

Wolfdieter Lang, Sep 13 2000

Keywords

Comments

The row polynomials are q(k,x) := sum(a(k,m)*x^m,m=0..k), k=0,1,2,...
The k-th convolution of F0(n) := A000045(n+1), n >= 0, (Fibonacci numbers starting with F0(0)=1) with itself is Fk(n) := A037027(n+k,k) =( p(k-1,n)*(n+1)*F0(n+1) + q(k-1,n)*(n+2)*F0(n))/(k!*5^k), k=1,2,..., where the companion polynomials p(k,n) := sum(b(k,m)*n^m,m=0..k), k >= 0, are the row polynomials of triangle b(k,m)= A057995(k,m).

Examples

			k=2: F2(n)=((16+5*n)*(n+1)*F0(n+1)+(17+5*n)*(n+2)*F0(n))/50, cf. A001628.
		

Crossrefs

A057995 Coefficient triangle of polynomials (rising powers) related to Fibonacci convolutions. Companion triangle to A057280.

Original entry on oeis.org

1, 16, 5, 300, 160, 20, 6840, 4850, 1075, 75, 186120, 159650, 48175, 6100, 275, 5916240, 5846700, 2168650, 379700, 31550, 1000, 215717040, 238437900, 103057800, 22426825, 2605175, 153875, 3625, 8888140800, 10772348400
Offset: 0

Views

Author

Wolfdieter Lang, Sep 13 2000

Keywords

Comments

The row polynomials are p(k,x) := sum(a(k,m)*x^m,m=0..k), k=0,1,2,...
The k-th convolution of F0(n) := A000045(n+1) n >= 0, (Fibonacci starting with F0(0)=1) with itself is Fk(n) := A037027(n+k,k) = (p(k-1,n)*(n+1)*F0(n+1) + q(k-1,n)*(n+2)*F0(n))/(k!*5^k)), k=1,2,..., where the companion polynomials q(k,n) := sum(b(k,m)*n^m,m=0..k), k >= 0, are the row polynomials of triangle b(k,m)= A057280).

Examples

			k=2: F2(n)=((16+5*n)*(n+1)*F0(n+1)+(17+5*n)*(n+2)*F0(n))/50, cf. A001628.
		

Crossrefs

A178819 Pascal's prism (3-dimensional array) read by folded antidiagonal cross-sections: (h+i; h, i-j, j), h >= 0, i >= 0, 0 <= j <= i.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 3, 3, 1, 3, 6, 3, 3, 3, 1, 1, 4, 4, 6, 12, 6, 4, 12, 12, 4, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 5, 20, 30, 20, 5, 10, 30, 30, 10, 10, 20, 10, 5, 5, 1, 1, 6, 6, 15, 30, 15, 20, 60, 60, 20, 15, 60, 90, 60, 15, 6, 30, 60, 60, 30, 6, 1, 6, 15, 20, 15, 6, 1
Offset: 0

Views

Author

Harlan J. Brothers, Jun 16 2010

Keywords

Comments

P_h = level h of Pascal's prism where P_1 = Pascal's triangle (A007318) and P_2 = denominators of Leibniz harmonic triangle (A003506). A sequence of length k through P is defined by P for n = {1, 2, 3, ..., k}.

Examples

			Prism begins (levels 1-4):
1
1 1
1 2 1
1 3 3 1
1
2 2
3 6 3
4 12 12 4
1
3 3
6 12 6
10 30 30 10
1
4 4
10 20 10
20 60 60 20
		

Crossrefs

Level 1 = A007318.
Level 2 = A003506.
Level 3 = A094305.
Level 4 = A178820.
Level 5 = A178821.
Level 6 = A178822.
Sums of shallow diagonals for each level correspond to rows of square A037027.
Contains A109649 and A046816.
P = A000984.
P = A006480.
P = A000897.
P<3n-2, 3n-2, n> = A113424.

Programs

  • Mathematica
    end = 5; Column/@Table[Multinomial[h, i-j, j], {h, 0, end}, {i, 0, end}, {j, 0, i}]

Formula

a_(h, i, j) = (h+i-2; h-1, i-j, j-1), h >= 1, i >= 1, 1 <= j <= i.
Recurrence:
For P_h, element a is given by: a_(1, 1) = 1; a_(i, j) = ((i+h-2)/(i-1)) (a_(i-1, j) + a_(i-1, j-1)).

Extensions

Keyword tabf by Michel Marcus, Oct 22 2017

A214178 Triangle T(n,k) by rows: the k-th derivative of the Fibonacci Polynomial F_n(x) evaluated at x=1.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 2, 2, 2, 0, 3, 5, 6, 6, 0, 5, 10, 18, 24, 24, 0, 8, 20, 44, 84, 120, 120, 0, 13, 38, 102, 240, 480, 720, 720, 0, 21, 71, 222, 630, 1560, 3240, 5040, 5040, 0, 34, 130, 466, 1536, 4560, 11760, 25200, 40320, 40320, 0, 55, 235, 948, 3564, 12264
Offset: 0

Views

Author

Keywords

Comments

T(n,0) = A000045(n), Fibonacci numbers;
T(n,1) = A001629(n) for n > 0;
T(n,n-3) = A038720(n-2) for n > 2;
T(n,n-2) = A000142(n-1) for n > 1;
T(n,n-1) = A000142(n-1) for n > 0;
T(n,n) = 0.

Examples

			The triangle begins:
.   0: [0]
.   1: [1, 0]
.   2: [1, 1, 0]
.   3: [2, 2, 2, 0]
.   4: [3, 5, 6, 6, 0]
.   5: [5, 10, 18, 24, 24, 0]
.   6: [8, 20, 44, 84, 120, 120, 0]
.   7: [13, 38, 102, 240, 480, 720, 720, 0]
.   8: [21, 71, 222, 630, 1560, 3240, 5040, 5040, 0]
.   9: [34, 130, 466, 1536, 4560, 11760, 25200, 40320, 40320, 0]
.  10: [55, 235, 948, 3564, 12264, 37800, 100800, 221760, 362880, 362880, 0]
       ...
		

Crossrefs

Programs

  • Haskell
    a214178 n k = a214178_tabl !! n !! k
    a214178_row n = a214178_tabl !! n
    a214178_tabl = [0] : map f a037027_tabl where
       f row = (zipWith (*) a000142_list row) ++ [0]
  • Mathematica
    T[n_, k_] := D[Fibonacci[n, x], {x, k}] /. x -> 1;
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 20 2021 *)

Formula

T(n,k) = A037027(n,k)*k!, 0 <= k < n; T(n,n) = 0.

A057282 Coefficient triangle of polynomials (falling powers) related to Fibonacci convolutions. Companion triangle to A057281.

Original entry on oeis.org

2, 5, 17, 15, 120, 225, 50, 700, 3050, 4080, 175, 3775, 28625, 89225, 94440, 625, 19225, 223175, 1208975, 3006000, 2666880, 2250, 93500, 1537100, 12689800, 54824650, 115299900, 89016480, 8125, 438250, 9670750, 112454500, 737744125
Offset: 1

Views

Author

Wolfdieter Lang, Sep 13 2000

Keywords

Comments

The row polynomials are q(k,x) := sum(a(k,m)*x^(k-m),m=0..k), k=0,1,2,..
The k-th convolution of F0(n) := A000045(n+1), n >= 0, (Fibonacci numbers starting with F0(0)=1) with itself is Fk(n) := A037027(n+k,k) = (p(k-1,n)*(n+1)*F0(n+1) + q(k-1,n)*(n+2)*F0(n))/(k!*5^k), k=1,2,..., where the companion polynomials p(k,n) := sum(b(k,m)*n^(k-m),m=0..k) are the row polynomials of triangle b(k,m)= A057281(k,m).
a(k,0)= A020876(k), k >= 0.

Examples

			k=2: F2(n)=((5*n^2+21*n+16)*F(n+2)+(5*n^2+27*n+34)*F(n+1))/50, F(n) := A000045(n); see A001628.
2; 5,17; 15,120,225; 50,700,3050,4080; 175,3775,28625,89225,94440; ...
		

Crossrefs

A092565 Triangle of coefficients T(n,k) (n>=0, 0<=k<=2*n), read by rows, where the n-th row polynomial equals the numerator of the n-th convergent of the continued fraction [1+x+x^2;1+x+x^2,1+x+x^2,...] for n>0, with the zeroth row defined as T(0,0)=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 2, 1, 3, 5, 8, 7, 6, 3, 1, 5, 10, 19, 22, 22, 16, 10, 4, 1, 8, 20, 42, 58, 69, 63, 49, 30, 15, 5, 1, 13, 38, 89, 142, 191, 206, 191, 146, 95, 50, 21, 6, 1, 21, 71, 182, 327, 491, 602, 637, 573, 447, 296, 167, 77, 28, 7, 1, 34, 130, 363, 722, 1191, 1626
Offset: 0

Views

Author

Paul D. Hanna, Feb 28 2004

Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0),(2,0),(1,1),(1,2). - Joerg Arndt, Jul 01 2011
Diagonal forms A092566, row sums form A006190. Column T(n,0) forms Fibonacci numbers A000045, T(n,1) forms A001629.

Examples

			Ratio of row polynomials R(3)/R(2) = (3+5*x+8*x^2+7*x^3+6*x^4+3*x^5+x^6)/(2+2*x+3*x^2+2*x^3+x^4) = [1+x+x^2;1+x+x^2,1+x+x^2].
Rows begin:
  1;
  1, 1, 1;
  2, 2, 3, 2, 1;
  3, 5, 8, 7, 6, 3, 1;
  5, 10, 19, 22, 22, 16, 10, 4, 1;
  8, 20, 42, 58, 69, 63, 49, 30, 15, 5, 1;
  13, 38, 89, 142, 191, 206, 191, 146, 95, 50, 21, 6, 1;
  21, 71, 182, 327, 491, 602, 637, 573, 447, 296, 167, 77, 28, 7, 1;
  34, 130, 363, 722, 1191, 1626, 1921, 1958, 1752, 1366, 931, 546, 273, 112, 36, 8, 1;
  ...
		

Crossrefs

Programs

  • Maple
    T:= proc(x, y) option remember; `if`(y<0 or y>2*x, 0, `if`(x=0, 1,
          add(T(x-l[1], y-l[2]), l=[[1, 0], [2, 0], [1, 1], [1, 2]])))
        end:
    seq(seq(T(n,k), k=0..2*n), n=0..10); # Alois P. Heinz, Apr 16 2013
  • Mathematica
    A037027[n_, k_] := Sum[Binomial[k+j, k]*Binomial[j, n-j-k], {j, 0, n-k}]; A037027[n_, 0] = Fibonacci[n + 1]; row[n_] := CoefficientList[ Sum[A037027[n, k] x^k (1+x)^k, {k, 0, n}], x]; Flatten[Table[row[n], {n, 0, 8}]][[1 ;; 70]] (* Jean-François Alcover, Jul 18 2011 *)
  • PARI
    T(n,k)=if(2*n
    				
  • PARI
    /* same as in A092566, but last line (output) replaced by the following */
    /* show as triangle (0<=k<=2*n): */
    {for (n=1,N, for (k=1,2*n-1, print1(M[n,k],", "); ); print(); );}
    /* Joerg Arndt, Jul 01 2011 */

Formula

n-th row polynomial R(n) = Sum_{k=0..n} A037027(n, k)*x^k*(1+x)^k; R(n+1)/R(n) = [1+x+x^2;1+x+x^2, ...(n+1)times..., 1+x+x^2] for n>=0; R(0)=1.

A152440 Riordan matrix (1/(1-x-x^2),x/(1-x-x^2)^2).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 3, 9, 5, 1, 5, 22, 20, 7, 1, 8, 51, 65, 35, 9, 1, 13, 111, 190, 140, 54, 11, 1, 21, 233, 511, 490, 255, 77, 13, 1, 34, 474, 1295, 1554, 1035, 418, 104, 15, 1, 55, 942, 3130, 4578, 3762, 1925, 637, 135, 17, 1, 89, 1836, 7285, 12720, 12573, 7865, 3276
Offset: 0

Views

Author

Emanuele Munarini, Dec 04 2008, Dec 05 2008

Keywords

Comments

From Philippe Deléham, Feb 20 2014: (Start)
T(n,0) = A000045(n+1);
T(n+1,1) = A001628(n);
T(n+2,2) = A001873(n);
T(n+3,3) = A001875(n).
Row sums are A238236(n). (End)

Examples

			Triangle begins:
1;
1, 1;
2, 3, 1;
3, 9, 5, 1;
5, 22, 20, 7, 1;
8, 51, 65, 35, 9, 1;
13, 111, 190, 140, 54, 11, 1;
21, 233, 511, 490, 255, 77, 13, 1, etc.
- _Philippe Deléham_, Feb 20 2014
		

Crossrefs

The first row is given by A000045.

Formula

a(n,k) = sum( binomial(n-j-k,2k) binomial(n-j-k,j), j=0...(n-k)/2 )
a(n,k) = sum( binomial(i+2k,2k) binomial(n-i+k,i+2k), i=0...(n - k)/2 )
Recurrence: a(n+4,k+1) - 2 a(n+3,k+1) - a(n+3,k) - a(n+2,k+1) + 2 a(n+1,k+1) + a(n,k+1) = 0
GF for columns: 1/(1-x-x^2)(x/(1-x-x^2)^2)^k
GF: (1-x-x^2)/((1-x-x^2)^2-xy)
T(n,k) = A037027(n+k, 2*k). - Philippe Deléham, Feb 20 2014

A184018 Expansion of c(x/(1-x-x^2)) / (1-x-x^2), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 2, 6, 19, 67, 254, 1017, 4236, 18168, 79680, 355635, 1609912, 7373401, 34102976, 159055728, 747211753, 3532452169, 16792693562, 80224098381, 384948157635, 1854469572120, 8965866981294, 43488834409737, 211569299607282
Offset: 0

Views

Author

Paul Barry, Jan 08 2011

Keywords

Comments

Hankel transform is (9,-5) Somos-4 sequence A184019.
The radius of convergence r of the g.f. A(x) satisfies: r = (1-r-r^2)/4 = limit a(n)/a(n+1) = (sqrt(29)-5)/2 = 0.19258240... with A(r) = 1/(2*r) = (sqrt(29)+5)/4 = 2.59629120... - Paul D. Hanna, Sep 06 2011

Programs

  • Maple
    A000108 := proc(n) binomial(2*n,n)/(n+1) ; end proc:
    A037027 := proc(n,m) add( binomial(m+k,m)*binomial(k,n-k-m),k=0..n-m) ; end proc:
    A184018 := proc(n) add( A037027(n,k)*A000108(k),k=0..n) ; end proc:
    seq(A184018(n),n=0..10) ; # R. J. Mathar, Jan 11 2011
  • Mathematica
    CoefficientList[Series[(1 - x - x^2 - Sqrt[1 - 6 x + 3 x^2 + 6 x^3 + x^4])/(2 x (1 - x - x^2)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 04 2014 *)
  • PARI
    {a(n)=polcoeff((1-sqrt(1-4*x/(1-x-x^2 +O(x^(n+2)))))/(2*x),n)} /* Paul D. Hanna, Sep 06 2011 */

Formula

G.f.: ( 1 - x - x^2 - sqrt((1 - x - x^2)*(1 - 5*x - x^2)) )/( 2*x*(1 - x - x^2) ).
G.f.: 1/(1 - x - x^2 - x/(1-x/(1 - x - x^2 - x/(1-x/(1 - x - x^2 - x/(1-x/(1 - x - x^2 - x/(1-x/(1-... (continued fraction).
a(n) = Sum_{k=0..n} (Sum_{j=0..n-k} binomial(k+j,k)*binomial(j,n-k-j))*A000108(k) = Sum_{k=0..n} A037027(n,k)*A000108(k).
G.f. satisfies A(x) = 1/(1-x-x^2) + x*A(x)^2. - Paul D. Hanna, Sep 06 2011
Conjecture: (n+1)*a(n) + 2*(1-3*n)*a(n-1) + 3*(n-1)*a(n-2) + 2*(3*n-5)*a(n-3) + (n-3)*a(n-4) = 0. - R. J. Mathar, Nov 15 2011
a(n) ~ (27+5*sqrt(29)) * sqrt(54*sqrt(29)-290) * (5+sqrt(29))^n / (sqrt(Pi) * n^(3/2) * 2^(n+5)). - Vaclav Kotesovec, Feb 04 2014
Previous Showing 31-40 of 51 results. Next