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 1-10 of 12 results. Next

A089627 T(n,k) = binomial(n,2*k)*binomial(2*k,k) for 0 <= k <= n, triangle read by rows.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 6, 0, 0, 1, 12, 6, 0, 0, 1, 20, 30, 0, 0, 0, 1, 30, 90, 20, 0, 0, 0, 1, 42, 210, 140, 0, 0, 0, 0, 1, 56, 420, 560, 70, 0, 0, 0, 0, 1, 72, 756, 1680, 630, 0, 0, 0, 0, 0, 1, 90, 1260, 4200, 3150, 252, 0, 0, 0, 0, 0, 1, 110, 1980, 9240, 11550, 2772, 0, 0, 0, 0, 0, 0, 1, 132, 2970, 18480, 34650, 16632, 924, 0, 0, 0, 0, 0, 0, 1, 156, 4290, 34320, 90090, 72072, 12012, 0, 0, 0, 0, 0, 0, 0, 1, 182, 6006, 60060, 210210, 252252, 84084, 3432, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Dec 31 2003

Keywords

Comments

The rows of this triangle are the gamma vectors of the n-dimensional type B associahedra (Postnikov et al., p.38 ). Cf. A055151 and A101280. - Peter Bala, Oct 28 2008
T(n,k) is the number of Grand Motzkin paths of length n having exactly k upsteps (1,1). Cf. A109189, A055151. - Geoffrey Critzer, Feb 05 2014
The result Sum_{k = 0..floor(n/2)} C(n,2*k)*C(2*k,k)*x^k = (sqrt(1 - 4*x))^n* P(n,1/sqrt(1 - 4*x)) expressing the row polynomials of this triangle in terms of the Legendre polynomials P(n,x) is due to Catalan. See Laden, equation 7.10, p. 56. - Peter Bala, Mar 18 2018

Examples

			Triangle begins:
  1
  1,   0
  1,   2,    0
  1,   6,    0,     0
  1,  12,    6,     0,     0
  1,  20,   30,     0,     0,     0
  1,  30,   90,    20,     0,     0,   0
  1,  42,  210,   140,     0,     0,   0, 0
  1,  56,  420,   560,    70,     0,   0, 0, 0
  1,  72,  756,  1680,   630,     0,   0, 0, 0, 0
  1,  90, 1260,  4200,  3150,   252,   0, 0, 0, 0, 0
  1, 110, 1980,  9240, 11550,  2772,   0, 0, 0, 0, 0, 0
  1, 132, 2970, 18480, 34650, 16632, 924, 0, 0, 0, 0, 0, 0
Relocating the zeros to be evenly distributed and interpreting the triangle as the coefficients of polynomials
                     1
                     1
                 1 + 2 q^2
                 1 + 6 q^2
            1 + 12 q^2 +  6 q^4
            1 + 20 q^2 + 30 q^4
       1 + 30 q^2 +  90 q^4 +  20 q^6
       1 + 42 q^2 + 210 q^4 + 140 q^6
  1 + 56 q^2 + 420 q^4 + 560 q^6 + 70 q^8
then the substitution q^k -> 1/(floor(k/2)+1) gives the Motzkin numbers A001006.
- _Peter Luschny_, Aug 29 2011
		

Crossrefs

Row sums A002426. Antidiagonal sums A098479.

Programs

  • Maple
    for i from 0 to 12 do seq(binomial(i, j)*binomial(i-j, j), j=0..i) od; # Zerinvary Lajos, Jun 07 2006
    # Alternatively:
    R := (n, x) -> simplify(hypergeom([1/2 - n/2, -n/2], [1], 4*x)):
    Trow := n -> seq(coeff(R(n,x), x, j), j=0..n):
    seq(print(Trow(n)), n=0..9); # Peter Luschny, Mar 18 2018
  • Mathematica
    nn=15;mxy=(1-x-(1-2x+x^2-4x^2y)^(1/2))/(2x^2 y);Map[Select[#,#>0&]&, CoefficientList[Series[1/(1-x-2y x^2mxy),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Feb 05 2014 *)
  • PARI
    T(n,k) = binomial(n,2*k)*binomial(2*k,k);
    concat(vector(15, n, vector(n, k, T(n-1, k-1)))) \\ Gheorghe Coserea, Sep 01 2018

Formula

T(n,k) = n!/((n-2*k)!*k!*k!).
E.g.f.: exp(x)*BesselI(0, 2*x*sqrt(y)). - Vladeta Jovovic, Apr 07 2005
O.g.f.: ( 1 - x - sqrt(1 - 2*x + x^2 - 4*x^2*y))/(2*x^2*y). - Geoffrey Critzer, Feb 05 2014
R(n, x) = hypergeom([1/2 - n/2, -n/2], [1], 4*x) are the row polynomials. - Peter Luschny, Mar 18 2018
From Peter Bala, Jun 23 2023: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^i*binomial(n, i)*binomial(n-i, k-i)^2. Cf. A063007(n,k) = Sum_{i = 0..k} binomial(n, i)^2*binomial(n-i, k-i).
T(n,k) = A063007(n-k,k); that is, the diagonals of this table are the rows of A063007. (End)

A098551 Inverse of A098550.

Original entry on oeis.org

1, 2, 3, 4, 9, 10, 15, 6, 5, 16, 22, 12, 23, 8, 7, 14, 30, 31, 43, 18, 17, 20, 51, 33, 11, 25, 19, 27, 61, 39, 62, 29, 24, 35, 13, 37, 79, 41, 21, 48, 87, 44, 88, 46, 26, 56, 101, 52, 40, 50, 28, 54, 114, 69, 34, 58, 47, 63, 127, 71, 132, 60, 42, 65, 36, 73, 142, 67, 49, 80, 153
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 14 2004

Keywords

Comments

Now known to be a permutation of the natural numbers: see the 2015 article by Applegate, Havermann, Selcoe, Shevelev, Sloane, and Zumkeller.

Crossrefs

Cf. A249943 (partial maxima).

Programs

  • Haskell
    import Data.List (elemIndex); import Data.Maybe (fromJust)
    a098551 = (+ 1) . fromJust . (`elemIndex` a098550_list)
    -- Reinhard Zumkeller, Nov 21 2014
  • Mathematica
    f[lst_List] := Block[{k = 4}, While[ GCD[ lst[[-2]], k] == 1 || GCD[ lst[[-1]], k] > 1 || MemberQ[lst, k], k++]; Append[lst, k]]; Table[ Position[ Nest[ f, {1, 2, 3}, 120], n], {n, 71}] // Flatten (* Robert G. Wilson v, Nov 21 2014 *)

Formula

A098553(n) = a(a(n)).

A376791 Expansion of 1/sqrt((1 - x^3)^2 - 4*x).

Original entry on oeis.org

1, 2, 6, 21, 76, 282, 1065, 4074, 15732, 61193, 239406, 941064, 3713701, 14703896, 58383138, 232383841, 926943678, 3704410890, 14828984641, 59450138412, 238659074286, 959247218253, 3859777477944, 15546444564846, 62675854384977, 252893414725842, 1021208266423260
Offset: 0

Views

Author

Seiichi Manyama, Oct 04 2024

Keywords

Comments

From Seiichi Manyama, Apr 30 2025: (Start)
Number of lattice paths from (0,0) to (n,n) using steps (1,0),(0,1),(3,3).
Diagonal of the rational function 1 / (1 - x - y - x^3*y^3). (End)

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/sqrt((1-x^3)^2-4*x))
    
  • PARI
    a(n) = sum(k=0, n\3, binomial(2*n-5*k, k)*binomial(2*n-6*k, n-3*k));

Formula

a(n) = Sum_{k=0..floor(n/3)} binomial(2*n-5*k,k) * binomial(2*n-6*k,n-3*k).

A098482 Expansion of 1/sqrt((1-x)^2 - 4*x^4).

Original entry on oeis.org

1, 1, 1, 1, 3, 7, 13, 21, 37, 73, 147, 283, 531, 1007, 1953, 3817, 7423, 14371, 27877, 54333, 106189, 207585, 405743, 793719, 1554889, 3049525, 5984803, 11751067, 23086695, 45388291, 89289765, 175746797, 346077153, 681795925, 1343790319, 2649687079, 5226711507
Offset: 0

Views

Author

Paul Barry, Sep 10 2004

Keywords

Comments

From Joerg Arndt, Jul 01 2011: (Start)
Empirical: Number of lattice paths from (0,0) to (n,n) using steps (4,0), (0,4), (1,1).
It appears that 1/sqrt((1-x)^2-4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1).
Empirical: Number of lattice paths from (0,0) to (n,n) using steps (3,1), (1,3), (1,1). (End)
1/sqrt((1-x)^2-4*r*x^4) expands to sum(k=0..floor(n/2), binomial(n-2*k,k)*binomial(n-3*k,k)*r^k ).
Diagonal of the rational function 1 / ((1-x)*(1-y) - x^3*y^4). - Seiichi Manyama, Apr 29 2025
a(n) is the number of ways to tile a strip of length n with 1 X 1 squares and 1 X 2 red dominos and 1 X 2 blue dominos, with an equal number of red and blue dominos. - Greg Dresden and Leo Zhang, Jul 08 2025

Examples

			From _Joerg Arndt_, Jul 01 2011: (Start)
The triangle of lattice paths from (0,0) to (n,k) using steps (3,1), (1,3), (1,1) begins
  1;
  0, 1;
  0, 0, 1;
  0, 1, 0, 1;
  0, 0, 2, 0, 3;
  0, 0, 0, 3, 0, 7;
  0, 0, 1, 0, 4, 0, 13;
  0, 0, 0, 3, 0, 8, 0, 21;
  0, 0, 0, 0, 6, 0, 18, 0, 37;
  0, 0, 0, 1, 0, 10, 0, 37, 0, 73;
The triangle of lattice paths from (0,0) to (n,k) using steps (4,0), (0,4), (1,1) begins
  1;
  0, 1;
  0, 0, 1;
  0, 0, 0, 1;
  1, 0, 0, 0, 3;
  0, 2, 0, 0, 0, 7;
  0, 0, 3, 0, 0, 0, 13;
  0, 0, 0, 4, 0, 0, 0, 21;
  1, 0, 0, 0, 8, 0, 0, 0, 37;
  0, 3, 0, 0, 0, 18, 0, 0, 0, 73;
The diagonals of both appear to be this sequence.  (End)
		

Crossrefs

Programs

  • Maple
    seq(add(binomial(n-3*k,k)*binomial(n-2*k,k),k=0..floor(n/3)),n=0..34); # Zerinvary Lajos, Apr 03 2007
  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-x)^2-4*x^4], {x, 0, 20}], x] (* Vaclav Kotesovec, Jun 23 2014 *)
  • PARI
    /* as lattice paths, assuming the first comment is true */
    /* same as in A092566 but use either of */
    steps=[[4,0], [0,4], [1,1]];
    steps=[[3,1], [1,3], [1,1]];
    /* Joerg Arndt, Jul 01 2011 */

Formula

a(n) = Sum_{k=0..floor(n/4)} binomial(n-2*k, k) * binomial(n-3*k, k).
D-finite with recurrence: n*a(n) = (2*n-1)*a(n-1) - (n-1)*a(n-2) + 4*(n-2)*a(n-4). - Vaclav Kotesovec, Jun 23 2014
a(n) ~ 2^(n+1/2) / sqrt(3*Pi*n). - Vaclav Kotesovec, Jun 23 2014
G.f.: 1/(1 - x - 2*x^4/(1 - x - x^4/(1 - x - x^4/(1 - x - x^4/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Nov 19 2021
a(n) = Sum_{k=0..floor(n/4)} binomial(n-2*k, 2*k) * binomial(2*k, k). - Greg Dresden and Leo Zhang, Jul 08 2025

A098480 Expansion of 1/sqrt((1-x)^2-8x^3).

Original entry on oeis.org

1, 1, 1, 5, 13, 25, 65, 181, 445, 1113, 2945, 7685, 19821, 51865, 136513, 358229, 942109, 2487385, 6573825, 17387045, 46066253, 122213913, 324512833, 862511605, 2294698109, 6109933657, 16280439937, 43411979845, 115835462445
Offset: 0

Views

Author

Paul Barry, Sep 10 2004

Keywords

Comments

1/sqrt((1-x)^2-4rx^3) expands to sum{k=0..floor(n/2), binomial(n-k,k)binomial(n-2k,k)r^k}

Crossrefs

Programs

  • Mathematica
    Array[Sum[Binomial[# - k, k] Binomial[# - 2 k, k] 2^k, {k, 0, #/2}] &, 29, 0] (* Michael De Vlieger, Jul 16 2019 *)

Formula

a(n)=sum{k=0..floor(n/2), binomial(n-k, k)binomial(n-2k, k)2^k}. D-finite with recurrence: n*a(n) +(-2*n+1)*a(n-1) +(n-1)*a(n-2) +4*(-2*n+3)*a(n-3)=0. - R. J. Mathar, Nov 10 2014

A098481 Expansion of 1/sqrt((1-x)^2 - 12*x^3).

Original entry on oeis.org

1, 1, 1, 7, 19, 37, 115, 361, 937, 2599, 7777, 22195, 62701, 182647, 531829, 1534903, 4461571, 13034917, 38015899, 110994193, 325011151, 952442557, 2792471239, 8198275933, 24093817531, 70852613041, 208516575043, 614145137137
Offset: 0

Views

Author

Paul Barry, Sep 10 2004

Keywords

Comments

1/sqrt((1-x)^2 - 4*r*x^3) expands to Sum_{k=0..floor(n/2)} binomial(n-k, k)*binomial(n-2k, k)*r^k.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-x)^2-12x^3],{x,0,40}],x] (* Harvey P. Dale, Jun 02 2011 *)
  • PARI
    Vec(1/sqrt((1-x)^2 - 12*x^3) + O(x^50)) \\ G. C. Greubel, Jan 30 2017

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*binomial(n-2k, k)*3^k.
D-finite with recurrence: n*a(n) = (2*n-1)*a(n-1) - (n-1)*a(n-2) + 6*(2*n-3)*a(n-3). - Vaclav Kotesovec, Jun 23 2014
a(n) ~ 3^(n+1) / (4*sqrt(Pi*n)). - Vaclav Kotesovec, Jun 23 2014

A120580 Hankel transform of Sum_{k=0..n} C(2k,k).

Original entry on oeis.org

1, 0, -4, -8, 0, 32, 64, 0, -256, -512, 0, 2048, 4096, 0, -16384, -32768, 0, 131072, 262144, 0, -1048576, -2097152, 0, 8388608, 16777216, 0, -67108864, -134217728, 0, 536870912, 1073741824, 0, -4294967296, -8589934592, 0, 34359738368, 68719476736, 0, -274877906944, -549755813888, 0
Offset: 0

Views

Author

Paul Barry, Jun 15 2006

Keywords

Comments

Hankel transform of A006134.
Hankel transform of A098479. - Paul Barry, Sep 19 2008
Hankel transform of A025565. - Paul Barry, Mar 26 2010

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2,-4},{1,0},50] (* Harvey P. Dale, Feb 13 2023 *)

Formula

G.f.: (1-2x)/(1-2x+4x^2).
a(n) = 2^n*(cos(Pi*n/3)-sin(Pi*n/3)/sqrt(3)).
E.g.f.: exp(x)*(cos(sqrt(3)*x) - sin(sqrt(3)*x)/sqrt(3)). - Stefano Spezia, Jul 15 2024

A113179 Expansion of 1/sqrt((1-2x)^2-8x^3).

Original entry on oeis.org

1, 2, 4, 12, 40, 128, 408, 1328, 4384, 14560, 48576, 162816, 547936, 1850048, 6263680, 21257856, 72298240, 246345728, 840775424, 2873802240, 9835840512, 33704557568, 115622041600, 397032488960, 1364610270720, 4694145256448
Offset: 0

Views

Author

Paul Barry, Oct 16 2005

Keywords

Comments

In general, 1/sqrt((1-a*x)^2-4*b*x^3) expands to sum{k=0..floor(n/2), C(n-k,k)C(n-2k,k)b^k*a^(n-3k)}.

Crossrefs

Cf. A098479.

Programs

  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-2x)^2-8x^3],{x,0,30}],x] (* Harvey P. Dale, Dec 23 2017 *)

Formula

a(n)=sum{k=0..floor(n/2), C(n-k, k)C(n-2k, k)2^(n-2k)}.
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) +4*(n-1)*a(n-2) +4*(-2*n+3)*a(n-3)=0. [Belbachir]

A383539 a(n) = Sum_{k=0..floor(n/3)} binomial(n-k,k)^2 * binomial(n-2*k,k).

Original entry on oeis.org

1, 1, 1, 5, 19, 49, 137, 481, 1645, 5259, 17309, 59477, 203931, 693865, 2384149, 8277773, 28797631, 100312525, 350891175, 1232122535, 4335809699, 15287669469, 54029225569, 191351513905, 678850904981, 2412164275651, 8584573648693, 30595269827149
Offset: 0

Views

Author

Seiichi Manyama, Apr 29 2025

Keywords

Comments

Diagonal of the rational function 1 / ((1-x)*(1-y)*(1-z) - x^2*y^2*z^3).

Crossrefs

Programs

  • Maple
    f:= proc(n) local k; add(binomial(n-k,k)^2 * binomial(n-2*k,k),k=0..n/3) end proc:
    map(f, [$0..30]); # Robert Israel, May 29 2025
  • PARI
    a(n) = sum(k=0, n\3, binomial(n-k, k)^2*binomial(n-2*k, k));

A182885 Triangle read by rows: T(n,k) is the number of weighted lattice paths in L_n having k (1,0)-steps of weight 2. These are paths that start at (0,0) , end on the horizontal axis and whose steps are of the following four kinds: an (1,0)-step with weight 1; an (1,0)-step with weight 2; a (1,1)-step with weight 2; a (1,-1)-step with weight 1. The weight of a path is the sum of the weights of its steps.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 7, 3, 1, 13, 10, 3, 27, 29, 6, 1, 61, 66, 22, 4, 133, 157, 75, 10, 1, 287, 398, 201, 40, 5, 633, 975, 538, 155, 15, 1, 1407, 2334, 1506, 476, 65, 6, 3121, 5631, 4077, 1414, 280, 21, 1, 6943, 13602, 10695, 4320, 966, 98, 7, 15517, 32623, 27966, 12765, 3150, 462, 28, 1
Offset: 0

Views

Author

Emeric Deutsch, Dec 11 2010

Keywords

Comments

Sum of entries in row n is A051286(n).
T(n,0)=A098479(n).
Sum(k*T(n,k), k=0..n)=A182886(n).

Examples

			T(3,1)=2. Indeed, denoting by h (H) the (1,0)-step of weight 1 (2), and u=(1,1), d=(1,-1), the five paths of weight 3 are ud, du, hH, Hh, and hhh; two of them have exactly one H step.
Triangle starts:
1;
1;
1,1;
3,2;
7,3,1;
13,10,3;
27,29,6,1;
		

References

  • M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306.
  • E. Munarini, N. Zagaglia Salvi, On the rank polynomial of the lattice of order ideals of fences and crowns, Discrete Mathematics 259 (2002), 163-177.

Crossrefs

Programs

  • Maple
    G:=1/sqrt(1-2*z-2*t*z^2+z^2+2*t*z^3+t^2*z^4-4*z^3): Gser:=simplify(series(G,z=0,18)): for n from 0 to 14 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 14 do seq(coeff(P[n],t,k),k=0..floor(n/2)) od; # yields sequence in triangular form

Formula

G.f.: G(t,z) =1/sqrt(1-2z-2tz^2+z^2+2t*z^3+t^2*z^4-4z^3).
Showing 1-10 of 12 results. Next