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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A239101 Riordan array read by rows, corresponding to array in A180562.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 10, 5, 2, 1, 26, 13, 6, 2, 1, 70, 35, 16, 7, 2, 1, 192, 96, 45, 19, 8, 2, 1, 534, 267, 126, 56, 22, 9, 2, 1, 1500, 750, 357, 160, 68, 25, 10, 2, 1, 4246, 2123, 1016, 463, 198, 81, 28, 11, 2, 1, 12092, 6046, 2907, 1337, 586, 240, 95, 31
Offset: 0

Views

Author

N. J. A. Sloane, Mar 25 2014

Keywords

Comments

Take lower triangle of square array in A180562, read from right to left.
Row sums are in A225034. - Philippe Deléham, Mar 25 2014
Riordan array (f(x), (f(x)-1)/(2*f(x))) where f(x) = sqrt((1+x)/(1-3*x)). - Philippe Deléham, Mar 25 2014

Examples

			Triangle begins:
1
2 1
4 2 1
10 5 2 1
26 13 6 2 1
70 35 16 7 2 1
192 96 45 19 8 2 1
...
192 = 2*96, 96 = 70 - 35 + 16 + 45, 45 = 35 - 16 + 7 + 19, etc. - _Philippe Deléham_, Mar 25 2014
Production matrix is:
2, 1
0, 0, 1
2, 1, 0, 1
2, 1, 1, 0, 1
2, 1, 1, 1, 0, 1
2, 1, 1, 1, 1, 0, 1
2, 1, 1, 1, 1, 1, 0, 1
2, 1, 1, 1, 1, 1, 1, 0, 1
... _Philippe Deléham_, Sep 15 2014
		

Crossrefs

Cf. A180562.
Cf. T(n,0) = A025565(n+1), T(n+1,1) = A005773(n+1), T(n+2,2) = A005717(n+1), A225034 (Row sums). - Philippe Deléham, Mar 25 2014

Formula

T(0,0) = 1, T(n,0) = 2*T(n,1) for n>0, T(n,k) = T(n-1,k-1) - T(n-1,k) + T(n-1,k+1) + T(n,k+1) for k>0, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 25 2014

Extensions

More terms from Philippe Deléham, Mar 25 2014

A273020 a(n) = Sum_{k=0..n} C(n,k)*((-1)^n*(C(k,n-k)-C(k,n-k-1))+C(n-k,k+1)).

Original entry on oeis.org

1, 1, 3, 5, 19, 39, 141, 321, 1107, 2675, 8953, 22483, 73789, 190345, 616227, 1621413, 5196627, 13882947, 44152809, 119385663, 377379369, 1030434069, 3241135527, 8921880135, 27948336381, 77459553549, 241813226151, 674100041501, 2098240353907, 5878674505303, 18252025766941
Offset: 0

Views

Author

Peter Luschny, May 13 2016

Keywords

Crossrefs

Programs

  • Maple
    seq(simplify(hypergeom([-n,1/2],[2],4) + n*hypergeom([-n/2+1,-n/2+1/2],[2],4)), n=0..30);
  • Mathematica
    Table[ JacobiP[n, 1, -n-3/2, -7]/(n+1) + GegenbauerC[n-1,-n,-1/2], {n,0,30} ]
  • Sage
    def A():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        while True:
            yield d + b*(1-(-1)^n)
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)//((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)//n
    A273020 = A()
    print([next(A273020) for _ in range(31)])

Formula

a(n) = JacobiP(n, 1, -n-3/2, -7)/(n+1) + GegenbauerC(n-1, -n, -1/2), with a(0) = 1.
a(n) = hypergeom([-n,1/2], [2], 4) + n*hypergeom([-n/2+1,-n/2+1/2], [2], 4).
a(n) = (-1)^n*A005043(n) + A005717(n).
a(2*n) = A082758(n).
a(2*n+1) = A273019(n).

A292628 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of e.g.f. exp(k*x)*BesselI(1,2*x).

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 2, 3, 0, 1, 4, 6, 0, 0, 1, 6, 15, 16, 10, 0, 1, 8, 30, 56, 45, 0, 0, 1, 10, 51, 144, 210, 126, 35, 0, 1, 12, 78, 304, 685, 792, 357, 0, 0, 1, 14, 111, 560, 1770, 3258, 3003, 1016, 126, 0, 1, 16, 150, 936, 3885, 10224, 15533, 11440, 2907, 0, 0, 1, 18, 195, 1456, 7570, 26550, 58947, 74280, 43758, 8350, 462
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 20 2017

Keywords

Comments

A(n,k) is the k-th binomial transform of A138364 evaluated at n.

Examples

			E.g.f. of column k: A_k(x) = x/1! + 2*k*x^2/2! + 3*(k^2 + 1)*x^3/3! + 4*k*(k^2 + 3)*x^4/4! + 5*(k^4 + 6*k^2 + 2)*x^5/5! + ...
Square array begins:
   0,   0,    0,    0,     0,     0,  ...
   1,   1,    1,    1,     1,     1,  ...
   0,   2,    4,    6,     8,    10,  ...
   3,   6,   15,   30,    51,    78,  ...
   0,  16,   56,  144,   304,   560,  ...
  10,  45,  210,  685,  1770,  3885,  ...
		

Crossrefs

Columns k=0..3 give A138364, A005717, A001791, A026376.
Main diagonal gives A292629.
Cf. A292627.

Programs

  • Mathematica
    Table[Function[k, n! SeriesCoefficient[Exp[k x] BesselI[1, 2 x], {x, 0, n}]][j - n], {j, 0, 11}, {n, 0, j}] // Flatten

Formula

E.g.f. of column k: exp(k*x)*BesselI(1,2*x).

A383583 a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k-1,k) * binomial(k,n-3*k).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 2, 3, 0, 3, 12, 10, 4, 30, 60, 40, 60, 210, 286, 231, 560, 1267, 1428, 1722, 4208, 7182, 8064, 13275, 28080, 40656, 51754, 97020, 176088, 240251, 355872, 667810, 1081092, 1506648, 2475616, 4401696, 6693492, 9904752, 16950662, 28359201
Offset: 0

Views

Author

Seiichi Manyama, May 01 2025

Keywords

Crossrefs

Programs

  • Magma
    [&+[Binomial(n-2*k-1, k) * Binomial(k, n-3*k): k in [0..Floor(n div 3)]]: n in [0..45]]; // Vincenzo Librandi, May 03 2025
  • Mathematica
    Table[Sum[Binomial[n-2*k-1,k]* Binomial[k,n-3*k],{k,0,Floor[n/3]}],{n,0,45}] (* Vincenzo Librandi, May 03 2025 *)
  • PARI
    a(n) = sum(k=0, n\3, binomial(n-2*k-1, k)*binomial(k, n-3*k));
    

Formula

G.f.: (1/2) * ( 1 + 1/sqrt(1 - 4*x^4/(1-x^3)^2) ).
a(n) ~ phi^(n-1) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, May 01 2025

A383584 a(n) = Sum_{k=0..floor(n/4)} binomial(n-3*k-1,k) * binomial(k,n-4*k).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 0, 0, 0, 2, 3, 0, 0, 3, 12, 10, 0, 4, 30, 60, 35, 5, 60, 210, 280, 132, 105, 560, 1260, 1267, 630, 1260, 4200, 6938, 5796, 4236, 11550, 27729, 36396, 28644, 34155, 90100, 168663, 188100, 163020, 276573, 631290, 973830, 995280, 1068222, 2111252, 4104100
Offset: 0

Views

Author

Seiichi Manyama, May 01 2025

Keywords

Crossrefs

Programs

  • Magma
    [&+[Binomial(n-3*k-1,k) * Binomial(k,n-4*k): k in [0..Floor(n div 4)]]: n in [0..45]]; // Vincenzo Librandi, May 02 2025
  • Mathematica
    Table[Sum[Binomial[n-3*k-1,k]* Binomial[k,n-4*k],{k,0,Floor[n/4]}],{n,0,40}] (* Vincenzo Librandi, May 02 2025 *)
  • PARI
    a(n) = sum(k=0, n\4, binomial(n-3*k-1, k)*binomial(k, n-4*k));
    

Formula

G.f.: (1/2) * ( 1 + 1/sqrt(1 - 4*x^5/(1-x^4)^2) ).

A102004 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and having k branches of even length (n>=0, 0<=k<=floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 6, 7, 1, 16, 20, 6, 40, 64, 26, 2, 109, 196, 108, 16, 297, 619, 414, 96, 4, 836, 1940, 1557, 484, 45, 2377, 6142, 5690, 2247, 331, 9, 6869, 19454, 20535, 9792, 2010, 126, 20042, 61893, 73123, 40997, 10820, 1116, 21, 59071, 197280, 258220
Offset: 0

Views

Author

Emeric Deutsch, Dec 25 2004

Keywords

Comments

Row n has 1+floor(n/2) terms.
Row sums are the Catalan numbers (A000108).
T(2n,n) = A001006(n-1) for n>=1 (the Motzkin numbers).
T(2n+1,n) = A005717(n+1) for n>=0.

Examples

			T(3,0)=3 because we have: (i) tree with 3 edges hanging from the root, (ii) tree with one edge hanging from the root, at the end of which 2 edges are hanging and (iii) tree with a path of length 3 hanging from the root.
Triangle starts:
1;
1;
1,   1;
3,   2;
6,   7, 1;
16, 20, 6;
		

Crossrefs

Programs

  • Maple
    G:=1/2/(t*z^2+z)*(-z^2+z+1+t*z^2-sqrt(-5*z^2-6*t*z^3-2*z+2*z^3-3*t^2*z^4-2*t*z^2+2*t*z^4+1+z^4)): Gserz:=simplify(series(G,z=0,16)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(expand(coeff(Gserz,z^n))) od:for n from 0 to 14 do seq(coeff(t*P[n], t^k),k=1..1+floor(n/2)) od;

Formula

G.f. G = G(t,z) satisfies z(1+tz)G^2-(1+z-z^2+tz^2)G+1+z-z^2+tz^2=0.

A127905 Construct triangle in which n-th row is obtained by expanding (1+x+x^3)^n and take the next-to-central column.

Original entry on oeis.org

0, 1, 2, 3, 8, 25, 66, 168, 456, 1269, 3490, 9581, 26544, 73944, 206220, 576045, 1613264, 4527661, 12725946, 35818135, 100950440, 284869263, 804726934, 2275500998, 6440230392, 18242735800, 51714552656
Offset: 0

Views

Author

Paul Barry, Feb 05 2007

Keywords

Crossrefs

Cf. A005717.

Programs

  • Magma
    [0] cat [n*(&+[Binomial(n-1,3*k)*Binomial(3*k,k)/(2*k+1): k in [0..Floor((n-1)/3)]]): n in [1..30]]; // G. C. Greubel, Apr 30 2018
  • Maple
    A127905 := proc(n)
        n*add(binomial(n-1,3*k)*binomial(3*k,k)/(2*k+1),k=0..floor((n-1)/3)) ;
    end proc: # R. J. Mathar, Feb 23 2015
  • Mathematica
    Table[n*Sum[Binomial[n-1,3*k]*Binomial[3*k,k]/(2*k+1), {k, 0, Floor[(n -1)/3]}], {n, 0, 50}] (* G. C. Greubel, Apr 30 2018 *)
  • PARI
    a(n)=if(n<0, 0, polcoeff((1+x+x^3)^n, n-1));
    
  • PARI
    a(n)=if(n<0, 0, n++; n*polcoeff(serreverse(x/(1+x+x^3)+x*O(x^n)), n))
    

Formula

a(n) = n*A071879(n-1).
a(n) = n*Sum_{k=0..floor((n-1)/3)} C(n-1,3*k)*C(3*k,k)/(2*k+1).
a(n) = Sum_{k=0..floor((n-1)/3)} (3*k+1)*C(n,3*k+1)*C(3*k,k)/(2k+1).
a(n) = Sum_{k=0..n-1} Sum_{j=0..floor(k/3)} C(k,3*j)*C(3*j+1,j).
Conjecture: 2*(2*n+1)*(n-1)^2*a(n) -2*n*(6*n^2-12*n+5)*a(n-1) +6*n*(n-1)*(2*n-3)*a(n-2) -31*n*(n-1)*(n-2)*a(n-3)=0. - R. J. Mathar, Feb 23 2015
a(n) ~ (1 + 3/2^(2/3))^(n + 1/2) / sqrt(12*Pi*n). - Vaclav Kotesovec, May 01 2018

Extensions

Edited by Charles R Greathouse IV, Oct 28 2009

A135413 Number of at most 4-way branching ordered (i.e., plane) trees.

Original entry on oeis.org

1, 2, 6, 20, 70, 246, 875, 3144, 11385, 41470, 151778, 557712, 2056210, 7602700, 28180050, 104677280, 389571983, 1452293766, 5422187130, 20271296100, 75878518695, 284339792110, 1066585128810, 4004566131000, 15048213795600
Offset: 1

Views

Author

Andrey Bovykin (indiscernibles(AT)googlemail.com), Mar 01 2008

Keywords

Comments

Obtained by Lagrange inversion of the generating function for at most k-way branching trees.
Solve z = T/(1+T+...T^k) when k = 4. I.e., the n-th term is the coefficient of x^(n-1) in the expansion of (1+x+x^2+x^3+x^4)^n.

Crossrefs

For k=2 this is A005717, for k=3 this is A005726.

Programs

  • Maple
    A135413 := proc(n) local ogf,i ; ogf := 1 ; for i from 1 to n do ogf := taylor(ogf*(1+x+x^2+x^3+x^4),x=0,n) ; od: coeftayl(ogf,x=0,n-1) ; end: seq(A135413(n),n=1..30) ; # R. J. Mathar, Apr 21 2008
  • Mathematica
    Join[{1}, Table[Coefficient[(1 + x + x^2 + x^3 + x^4)^n, x,(n - 1)], {n,2,25}]] (* G. C. Greubel, Oct 13 2016 *)
  • Maxima
    a(n):=sum((-1)^i*binomial(n,i)*binomial(2*n-5*i-2,n-5*i-1),i,0,(n-1)/5); /* Vladimir Kruchinin, Mar 28 2019 */
    
  • PARI
    a(n) = polcoef((1+x+x^2+x^3+x^4)^n, n-1, x); \\ Michel Marcus, Mar 28 2019

Formula

a(n) = [ x^(n-1) ] (1+x+x^2+x^3+x^4)^n.
a(n) = Sum_{i=0..floor((n-1)/5)} (-1)^i * C(n,i) * C(2*n-5*i-2,n-5*i-1). - Vladimir Kruchinin, Mar 28 2019

Extensions

More terms from R. J. Mathar, Apr 21 2008

A335349 a(n) counts anti-chains of size three in "0,1,2" Motzkin trees on n edges.

Original entry on oeis.org

2, 16, 98, 500, 2308, 9920, 40522, 159212, 606790, 2256544, 8224202, 29473012, 104124044, 363374560, 1254711038, 4292365876, 14564351510, 49059814576, 164186524940, 546276316120, 1807990549352, 5955265349696, 19530431537488, 63795464433440, 207623760855106, 673440401953856
Offset: 4

Views

Author

Petros Hadjicostas, Jun 03 2020

Keywords

Comments

"0,1,2" trees are rooted trees where each vertex has outdegree zero, one, or two. They are counted by the Motzkin numbers A001006.
A005717(n+1) is the total number of vertices (= anti-chains of size 1) in all "0,1,2" trees with n edges, while A178834(n) is the total number of anti-chains of size 2 in all "0,1,2" trees on n edges.

Examples

			Out of the A001006(4) = 9 Motzkin rooted trees, there are only two that have anti-chains of size 3 (i.e., 3-sets of pairwise incomparable nodes), and each one has only one such an anti-chain. Thus, a(4) = 1 + 1 = 2.
In the first Motzkin tree below with 4 edges, {E, C, D} is an anti-chain of size 3. In the second one, {G, I, K} is an anti-chain of size 3.
        A                          F
       / \                        / \
      /   \                      /   \
     B     E                    G     H
    / \                              / \
   /   \                            /   \
  C     D                          I     K
		

Crossrefs

Programs

  • PARI
    M(z) = (1 - z - sqrt(1 - 2*z - 3*z^2))/(2*z^2);
    T(z) = 1/sqrt(1 - 2*z - 3*z^2);
    my(z='z+O('z^30)); Vec(2*z^4*T(z)^5*M(z)^3)

Formula

G.f. is A000108(r-1) * z^(2*r-2) * T(z)^(2*r-1) * M(z)^r = 2 * z^4 * T(z)^5 * M(z)^3 (with r = 3), where M(z) = (1 - z - sqrt(1 - 2*z - 3*z^2)) / (2*z^2) is the g.f. of the Motzkin numbers A001006 and T(z) = 1 / sqrt(1 - 2*z - 3*z^2) is the g.f. of the central trinomial numbers A002426.

A335355 a(n) counts anti-chains of size four in "0,1,2" Motzkin trees on n edges.

Original entry on oeis.org

5, 55, 420, 2600, 14175, 70665, 329800, 1462680, 6228945, 25661875, 102847560, 402706500, 1545715325, 5831511195, 21671504880, 79475234200, 288043346370, 1033030388790, 3669961024940, 12927078062500, 45182780785500, 156811313843420, 540722493900480, 1853503409060160
Offset: 6

Views

Author

Petros Hadjicostas, Jun 03 2020

Keywords

Comments

"0,1,2" trees are rooted trees where each vertex has outdegree zero, one, or two. They are counted by the Motzkin numbers A001006.
A005717(n+1) is the total number of vertices (= anti-chains of size 1) in all "0,1,2" trees with n edges, A178834(n) is the total number of anti-chains of size 2 in all "0,1,2" trees on n edges, and A335349(n) is the total number of anti-chains of size 3 in all "0,1,2" trees on n edges.
It would be interesting to examine whether there is an interpretation of this sequence and sequences A178834 and A335349 in terms of Motzkin paths. (Salaam (2008) worked with different families of rooted trees, but not with Motzkin paths.)

Examples

			For n=6, we list below all a(6) = 5 four-element anti-chains in Motzkin rooted trees with 6 edges:
              A               A                    A
             / \             / \                  / \
            /   \           /   \                /   \
           B     C         B     C              B     C
          / \   / \       / \                  / \
         /   \ /   \     /   \                /   \
        D    E F   G    D     E              D     E
        {D, E, F, G}         / \            / \
                            /   \          /   \
                           F     G        F     G
                        {C, D, F, G}         {C, E, F, G}
              A                                A
             / \                              / \
            /   \                            /   \
           B     C                          B     C
                / \                              / \
               /   \                            /   \
              D     E                          D     E
             / \                                    / \
            /   \                                  /   \
           F     G                                F     G
          {B, E, F, G}                        {B, D, F, G}
		

Crossrefs

Programs

  • PARI
    default(seriesprecision, 50);
    M(z) = (1 - z - sqrt(1 - 2*z - 3*z^2))/(2*z^2);
    T(z) = 1/sqrt(1 - 2*z - 3*z^2);
    for(n=0, 30, print1(polcoef(5*z^6*T(z)^7*M(z)^4, n, z), ", "))

Formula

G.f.: A000108(r-1) * z^(2*r-2) * T(z)^(2*r-1) * M(z)^r = 5 * z^6 * T(z)^7 * M(z)^4 (with r = 4), where M(z) = (1 - z - sqrt(1 - 2*z - 3*z^2)) / (2*z^2) is the g.f. of the Motzkin numbers A001006 and T(z) = 1 / sqrt(1 - 2*z - 3*z^2) is the g.f. of the central trinomial numbers A002426.
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