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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A201204 Half-convolution of Catalan sequence A000108 with itself.

Original entry on oeis.org

1, 1, 3, 7, 23, 66, 227, 715, 2529, 8398, 30275, 104006, 380162, 1337220, 4939443, 17678835, 65844845, 238819350, 895451117, 3282060210, 12374186318, 45741281820, 173257703723, 644952073662, 2452607696798, 9183676536076, 35042725663002, 131873975875180, 504697422982484, 1907493251046152
Offset: 0

Views

Author

Wolfdieter Lang, Jan 02 2012

Keywords

Comments

In general the half-convolution of a sequence {b(n)}_0^infty with itself is defined by chat(n):=sum(b(k)*b(n-k), k=0..floor(n/2)), n>=0. The o.g.f. of the sequence {chat(n)} is obtained from the bisection 2*chat(2*k) - b(k)^2 = c(2*k), k>=0, with the ordinary convolution c(n):=sum(b(k)*b(n-k),k=0..n), n>=0, and 2*chat(2*k+1) = c(2*k+1), k>=0. This leads to the o.g.f.s for the corresponding even (e) and odd (o) parts:
2*Chate(x) - B2(x) = Ce(x) and 2*Chato(x) = Co(x), where Chate(x):= sum(chat(2*k)*x^k,k=0..infty), Chato(x):= sum(chat(2*k+1)*x^k,k=0..infty), B2(x) := sum(b(k)^2*x^k, k=0..infty), Ce(x) := sum(c(2*k)*x^k, k=0..infty) and Co(x) := sum(c(2*k+1)*x^k, k=0..infty). Thus Chate(x)=(Ce(x) + B2(x))/2 and Chato(x)=Co(x)/2. Expressing this in terms of C(x), the o.g.f. of {c(n)}, and B2(x) leads to the result: Chat(x)= (C(x) + B2(x^2))/2.
In the Catalan case b(n)=A000108(n), c(n)=b(n+1), C(x)= (cata(x)+1)/x, with the o.g.f. of A000108 cata(x)=(1-sqrt(1-4*x))/(2*x), and B2(x) is found under A001246 to be (-1 + hypergeom([-1/2,-1/2],[1],16*x))/(4*x). This produces the o.g.f. given in the formula section.
This computation was motivated by a question about the o.g.f. of A000992 ("half-Catalan numbers"). Note, however, that this sequence is not the half-convolution of the Catalan numbers presented here.
Apparently the number of hills to the left of or at the midpoint in all Dyck paths of semilength n+1. [David Scambler, Apr 30 2013]

Crossrefs

A000108, bisection: A201205 and A065097.

Programs

  • Maple
    C:= n -> binomial(2*n,n)/(n+1):
    A:= n -> add(C(k)*C(n-k),k=0..floor(n/2));
    seq(A(i),i=1..100); # Robert Israel, Jun 06 2014
  • Mathematica
    Table[Sum[CatalanNumber[k]CatalanNumber[n-k],{k,0,Floor[n/2]}],{n,0,30}] (* Harvey P. Dale, Jun 12 2012 *)
    Table[CatalanNumber[n + 1]/2 + 2^(2 n + 1) Binomial[1/2, n/2 + 1]^2, {n, 0, 30}] (* Vladimir Reshetnikov, Oct 03 2016 *)

Formula

a(n) = sum(Catalan(k)*Catalan(n-k),k=0..floor(n/2)), n>=0, with Catalan(n)=A000108(n).
O.g.f.: G(x)=(catalan(x)-1)/(2*x)+(-1+hypergeom([-1/2,-1/2],[1],16*x^2))/(8*x^2), with the o.g.f. catalan(x) of the Catalan numbers (see also the comment section).
a(n) ~ 2^(2*n+1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Oct 15 2014
a(n) = A000108(n+1)/2 + 2^(2*n+1) * binomial(1/2, n/2+1)^2. - Vladimir Reshetnikov, Oct 03 2016
D-finite with recurrence: (n+1)*(n+2)^2*a(n) +6*(n-2)*(n+1)^2*a(n-1) +4*(-16*n^3+25*n^2+4*n-4)*a(n-2) +16*(-4*n^3+25*n^2-56*n+41)*a(n-3) +192*(4*n-7)*(n-3)^2*a(n-4) -256*(2*n-7)*(n-4)^2*a(n-5)=0. - R. J. Mathar, Feb 21 2020

A358588 Number of n-node ordered rooted trees of height equal to the number of internal (non-leaf) nodes.

Original entry on oeis.org

0, 0, 0, 0, 1, 8, 41, 171, 633, 2171, 7070, 22195, 67830, 203130, 598806, 1743258, 5023711, 14356226, 40737383, 114904941, 322432215, 900707165, 2506181060, 6948996085, 19207795836, 52944197508, 145567226556, 399314965956, 1093107693133, 2986640695436
Offset: 1

Views

Author

Gus Wiseman, Nov 25 2022

Keywords

Examples

			The a(5) = 1 and a(6) = 8 ordered trees:
  ((o)(o))  ((o)(o)o)
            ((o)(oo))
            ((o)o(o))
            ((oo)(o))
            (o(o)(o))
            (((o))(o))
            (((o)(o)))
            ((o)((o)))
		

Crossrefs

For leaves instead of height we have A000891, unordered A185650 aerated.
The unordered version is A358587, ranked by A358576.
For leaves instead of internal nodes we have A358590, unordered A358589.
A000108 counts ordered rooted trees, unordered A000081.
A001263 counts ordered rooted trees by nodes and leaves, unordered A055277.
A080936 counts ordered rooted trees by nodes and height, unordered A034781.
A090181 counts ordered rooted trees by nodes and internals, unord. A358575.

Programs

  • Mathematica
    aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[aot[n],Count[#,[_],{0,Infinity}]==Depth[#]-1&]],{n,1,10}]
  • PARI
    \\ Needs R(n,f) defined in A358590.
    seq(n) = {Vec(R(n, (h,p)->polcoef(subst(p, x, x/y), -h, y)), -n)} \\ Andrew Howroyd, Jan 01 2023

Formula

Conjectures from Chai Wah Wu, Apr 14 2024: (Start)
a(n) = 9*a(n-1) - 32*a(n-2) + 58*a(n-3) - 58*a(n-4) + 32*a(n-5) - 9*a(n-6) + a(n-7) for n > 7.
G.f.: x^5*(-x^2 + x - 1)/((x - 1)^3*(x^2 - 3*x + 1)^2). (End)

Extensions

Terms a(16) and beyond from Andrew Howroyd, Jan 01 2023

A358584 Number of rooted trees with n nodes, at most half of which are leaves.

Original entry on oeis.org

0, 1, 1, 3, 5, 15, 28, 87, 176, 550, 1179, 3688, 8269, 25804, 59832, 186190, 443407, 1375388, 3346702, 10348509, 25632265, 79020511, 198670299, 610740694, 1555187172, 4768244803, 12276230777, 37546795678, 97601239282, 297831479850, 780790439063, 2377538260547
Offset: 1

Views

Author

Gus Wiseman, Nov 23 2022

Keywords

Examples

			The a(2) = 1 through a(6) = 15 trees:
  (o)  ((o))  ((oo))   (((oo)))   (((ooo)))
              (o(o))   ((o)(o))   ((o)(oo))
              (((o)))  ((o(o)))   ((o(oo)))
                       (o((o)))   ((oo(o)))
                       ((((o))))  (o((oo)))
                                  (o(o)(o))
                                  (o(o(o)))
                                  (oo((o)))
                                  ((((oo))))
                                  (((o)(o)))
                                  (((o(o))))
                                  ((o)((o)))
                                  ((o((o))))
                                  (o(((o))))
                                  (((((o)))))
		

Crossrefs

For equality we have A185650 aerated, ranked by A358578.
The complement is A358581.
The strict case is A358582.
The opposite version is A358583.
A000081 counts rooted trees, ordered A000108.
A055277 counts rooted trees by nodes and leaves, ordered A001263.
A358575 counts rooted trees by nodes and internal nodes, ordered A090181.
A358589 counts square trees, ranked by A358577, ordered A358590.

Programs

  • Mathematica
    art[n_]:=If[n==1,{{}},Join@@Table[Select[Tuples[art/@c],OrderedQ],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[art[n],Count[#,{},{0,Infinity}]<=Count[#,[_],{0,Infinity}]&]],{n,0,10}]
  • PARI
    R(n) = {my(A = O(x)); for(j=1, n, A = x*(y - 1  + exp( sum(i=1, j, 1/i * subst( subst( A + O(x*x^(j\i)), x, x^i), y, y^i) ) ))); Vec(A)};
    seq(n) = {my(A=R(n)); vector(n, n, vecsum(Vecrev(A[n]/y)[1..n\2]))} \\ Andrew Howroyd, Dec 30 2022

Formula

A358581(n) + A358584(n) = A000081(n).
A358582(n) + A358583(n) = A000081(n).
a(n) = Sum_{k=1..floor(n/2)} A055277(n, k). - Andrew Howroyd, Dec 30 2022

Extensions

Terms a(19) and beyond from Andrew Howroyd, Dec 30 2022

A358591 Number of 2n-node rooted trees whose height, number of leaves, and number of internal (non-leaf) nodes are all equal.

Original entry on oeis.org

0, 0, 2, 17, 94, 464, 2162, 9743, 42962, 186584, 801316, 3412034, 14430740, 60700548, 254180426, 1060361147, 4409342954, 18285098288, 75645143516, 312286595342, 1286827096964, 5293833371408, 21745951533236, 89208948855542, 365523293690804, 1496048600896784
Offset: 1

Views

Author

Gus Wiseman, Nov 23 2022

Keywords

Examples

			The a(3) = 2 and a(4) = 17 trees:
  ((o)(oo))  (((o))(ooo))
  (o(o)(o))  (((o)(ooo)))
             (((oo))(oo))
             (((oo)(oo)))
             ((o)((ooo)))
             ((o)(o(oo)))
             ((o)(oo(o)))
             ((o(o)(oo)))
             ((oo)(o(o)))
             ((oo(o)(o)))
             (o((o))(oo))
             (o((o)(oo)))
             (o(o)((oo)))
             (o(o)(o(o)))
             (o(o(o)(o)))
             (oo((o)(o)))
             (oo(o)((o)))
		

Crossrefs

For leaves = internals we have A185650 aerated, ranked by A358578.
For height = internals we have A358587, ranked by A358576, ordered A358588.
For height = leaves we have A358589, ranked by A358577, ordered A358590.
These trees are ranked by A358592.
A000081 counts rooted trees, ordered A000108.
A034781 counts rooted trees by nodes and height, ordered A080936.
A055277 counts rooted trees by nodes and leaves, ordered A001263.
A358575 counts rooted trees by nodes and internal nodes, ordered A090181.

Programs

  • Mathematica
    art[n_]:=If[n==1,{{}},Join@@Table[Select[Tuples[art/@c],OrderedQ],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[art[n],Count[#,[_],{0,Infinity}]==Count[#,{},{0,Infinity}]==Depth[#]-1&]],{n,2,15,2}]
  • PARI
    \\ Needs R(n,f) defined in A358589.
    seq(n) = {Vecrev(R(2*n, (h,p)->if(h<=n, x^h*polcoef(polcoef(p, 2*h, x), h, y))), -n)} \\ Andrew Howroyd, Jan 01 2023

Extensions

Terms a(10) and beyond from Andrew Howroyd, Jan 01 2023

A358582 Number of rooted trees with n nodes, most of which are not leaves.

Original entry on oeis.org

0, 0, 1, 1, 5, 7, 28, 48, 176, 336, 1179, 2420, 8269, 17855, 59832, 134289, 443407, 1025685, 3346702, 7933161, 25632265, 62000170, 198670299, 488801159, 1555187172, 3882403641, 12276230777, 31034921462, 97601239282, 249471619165, 780790439063, 2015194486878
Offset: 1

Views

Author

Gus Wiseman, Nov 23 2022

Keywords

Examples

			The a(3) = 1 through a(6) = 7 trees:
  ((o))  (((o)))  (((oo)))   ((((oo))))
                  ((o)(o))   (((o)(o)))
                  ((o(o)))   (((o(o))))
                  (o((o)))   ((o)((o)))
                  ((((o))))  ((o((o))))
                             (o(((o))))
                             (((((o)))))
		

Crossrefs

For equality we have A185650 aerated, ranked by A358578.
The opposite version is A358581, non-strict A358583.
The non-strict version is A358584.
The ordered version is A358585, odd-indexed terms A065097.
A000081 counts rooted trees, ordered A000108.
A055277 counts rooted trees by nodes and leaves, ordered A001263.
A358575 counts rooted trees by nodes and internal nodes, ordered A090181.
A358589 counts square trees, ranked by A358577, ordered A358590.

Programs

  • Mathematica
    art[n_]:=If[n==1,{{}},Join@@Table[Select[Tuples[art/@c],OrderedQ],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[art[n],Count[#,{},{0,Infinity}][_],{0,Infinity}]&]],{n,0,10}]
  • PARI
    \\ See A358584 for R(n).
    seq(n) = {my(A=R(n)); vector(n, n, vecsum(Vecrev(A[n]/y)[1..(n-1)\2]))} \\ Andrew Howroyd, Dec 30 2022

Formula

A358581(n) + A358584(n) = A000081(n).
A358582(n) + A358583(n) = A000081(n).
a(n) = Sum_{k=1..floor((n-1)/2)} A055277(n, k). - Andrew Howroyd, Dec 30 2022

Extensions

Terms a(19) and beyond from Andrew Howroyd, Dec 30 2022

A358585 Number of ordered rooted trees with n nodes, most of which are leaves.

Original entry on oeis.org

1, 0, 1, 1, 7, 11, 66, 127, 715, 1549, 8398, 19691, 104006, 258194, 1337220, 3467115, 17678835, 47440745, 238819350, 659060677, 3282060210, 9271024542, 45741281820, 131788178171, 644952073662, 1890110798926, 9183676536076, 27316119923002, 131873975875180, 397407983278484
Offset: 1

Views

Author

Gus Wiseman, Nov 24 2022

Keywords

Examples

			The a(1) = 1 through a(6) = 11 ordered trees:
  o  .  (oo)  (ooo)  (oooo)   (ooooo)
                     ((o)oo)  ((o)ooo)
                     ((oo)o)  ((oo)oo)
                     ((ooo))  ((ooo)o)
                     (o(o)o)  ((oooo))
                     (o(oo))  (o(o)oo)
                     (oo(o))  (o(oo)o)
                              (o(ooo))
                              (oo(o)o)
                              (oo(oo))
                              (ooo(o))
		

Crossrefs

For equality we have A000891, unordered A185650.
Odd-indexed terms are A065097.
The unordered version is A358581.
The opposite is the same, unordered A358582.
The non-strict version is A358586, unordered A358583.
A000108 counts ordered rooted trees, unordered A000081.
A001263 counts ordered rooted trees by nodes and leaves, unordered A055277.
A080936 counts ordered rooted trees by nodes and height, unordered A034781.
A090181 counts ordered rooted trees by nodes and internals, unord. A358575.
A358590 counts square ordered trees, unordered A358589 (ranked by A358577).

Programs

  • Mathematica
    aot[n_]:=If[n==1,{{}},Join@@Table[Tuples[aot/@c],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[aot[n],Count[#,{},{0,Infinity}]>Count[#,[_],{0,Infinity}]&]],{n,10}]
  • PARI
    a(n) = if(n==1, 1, n--; (binomial(2*n,n)/(n+1) - if(n%2, binomial(n, (n-1)/2)^2 / n))/2) \\ Andrew Howroyd, Jan 13 2024

Formula

From Andrew Howroyd, Jan 13 2024: (Start)
a(n) = Sum_{k=1..floor((n-1)/2)} A001263(n-1, k) for n >= 2.
a(2*n) = (A000108(2*n-1) - A000891(n-1))/2 for n >= 1;
a(2*n+1) = A000108(2*n)/2 for n >= 1. (End)

Extensions

a(16) onwards from Andrew Howroyd, Jan 13 2024

A358583 Number of rooted trees with n nodes, at least half of which are leaves.

Original entry on oeis.org

1, 1, 1, 3, 4, 13, 20, 67, 110, 383, 663, 2346, 4217, 15118, 27979, 101092, 191440, 695474, 1341974, 4893067, 9589567, 35055011, 69612556, 254923825, 511987473, 1877232869, 3807503552, 13972144807, 28585315026, 104955228432, 216381073935, 794739865822
Offset: 1

Views

Author

Gus Wiseman, Nov 23 2022

Keywords

Examples

			The a(1) = 1 through a(6) = 13 trees:
  o  (o)  (oo)  (ooo)   (oooo)   (ooooo)
                ((oo))  ((ooo))  ((oooo))
                (o(o))  (o(oo))  (o(ooo))
                        (oo(o))  (oo(oo))
                                 (ooo(o))
                                 (((ooo)))
                                 ((o)(oo))
                                 ((o(oo)))
                                 ((oo(o)))
                                 (o((oo)))
                                 (o(o)(o))
                                 (o(o(o)))
                                 (oo((o)))
		

Crossrefs

For equality we have A185650 aerated, ranked by A358578.
The strict case is A358581.
The opposite version is A358584, strict A358582.
The ordered version is A358586, strict A358585.
A000081 counts rooted trees, ordered A000108.
A055277 counts rooted trees by nodes and leaves, ordered A001263.
A358575 counts rooted trees by nodes and internal nodes, ordered A090181.
A358589 counts square rooted trees, ranked by A358577, ordered A358590.

Programs

  • Mathematica
    art[n_]:=If[n==1,{{}},Join@@Table[Select[Tuples[art/@c],OrderedQ],{c,Join@@Permutations/@IntegerPartitions[n-1]}]];
    Table[Length[Select[art[n],Count[#,{},{0,Infinity}]>=Count[#,[_],{0,Infinity}]&]],{n,1,10}]
  • PARI
    \\ See A358584 for R(n).
    seq(n) = {my(A=R(n)); vector(n, n, my(u=Vecrev(A[n]/y)); vecsum(u[(n-1)\2+1..#u]))} \\ Andrew Howroyd, Dec 31 2022

Formula

A358581(n) + A358584(n) = A000081(n).
A358582(n) + A358583(n) = A000081(n).
a(n) = Sum_{k=floor((n-1)/2)+1..n} A055277(n, k). - Andrew Howroyd, Dec 31 2022

Extensions

Terms a(19) and beyond from Andrew Howroyd, Dec 31 2022

A001795 Coefficients of Legendre polynomials.

Original entry on oeis.org

1, 1, 7, 33, 715, 4199, 52003, 334305, 17678835, 119409675, 1641030105, 11435320455, 322476036831, 2295919134019, 32968493968795, 238436656380769, 27767032438524099, 203236010537432691, 2989949596465113373
Offset: 0

Views

Author

Keywords

Comments

Numerators in expansion of sqrt(c(x)), c(x) the g.f. of A000108. - Paul Barry, Jul 12 2005
Coefficient of Legendre_0(x) when x^n is written in term of Legendre polynomials. - Michel Marcus, May 28 2013

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Divisor of A048990 and A065097.
Apparently a bisection of A002596.
Bisection of A099024.

Programs

  • Magma
    A001795:= func< n | Numerator(Catalan(2*n)/4^n) >;
    [A001795(n): n in [0..25]]; // G. C. Greubel, Apr 22 2025
    
  • Mathematica
    Table[Numerator[CatalanNumber[2*n]/4^n], {n,0,30}] (* G. C. Greubel, Apr 22 2025 *)
  • PARI
    my(x='x+O('x^30)); apply(numerator, Vec(((1-sqrt(1-4*x))/(2*x))^(1/2))) \\ Michel Marcus, Feb 04 2022
    
  • PARI
    a(n)=numerator(binomial(2*n-1/2, n)/(2*n+1)) \\ Tani Akinari, Oct 22 2024
    
  • SageMath
    def A001795(n): return numerator(catalan_number(2*n)/4^n)
    print([A001795(n) for n in range(31)]) # G. C. Greubel, Apr 22 2025

Formula

1/(sqrt(1-x) + sqrt(1+x)) = Sum_{n>=0} (a(n)/b(n))*x^(2*n) where b(n) is a power of 2. - Benoit Cloitre, Mar 12 2002
For n >= 1, 2^(n+1)*a(2^(n-1)) = A001791(2^n). - Vladimir Shevelev, Sep 05 2010
a(n) = numerator(binomial(2*n-1/2, n)/(2*n+1)). - Tani Akinari, Oct 22 2024
a(n) = numerator( A000108(2*n)/4^n ). - G. C. Greubel, Apr 22 2025

Extensions

More terms from Benoit Cloitre, Mar 12 2002

A201205 Bisection of half-convolution of Catalan sequence A000108; even part.

Original entry on oeis.org

1, 3, 23, 227, 2529, 30275, 380162, 4939443, 65844845, 895451117, 12374186318, 173257703723, 2452607696798, 35042725663002, 504697422982484, 7319313029400467, 106793147620036005, 1566546633240722681, 23089471526179716182, 341774295456352388245
Offset: 0

Views

Author

Wolfdieter Lang, Jan 02 2012

Keywords

Comments

For the definition of the half-convolution of a sequence with itself see a comment to A201204.
The odd part of this bisection is found under A065097.

Crossrefs

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<2, 1+2*n,
          (2*n*(256*n^5-544*n^4+256*n^3+75*n^2-69*n+12)*a(n-1)
           -(8*(4*n-5))*(4*n-3)*(8*n^2+n-1)*(2*n-3)^2*a(n-2))/
          ((2*n+1)*n*(8*n^2-15*n+6)*(n+1)^2))
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Nov 28 2015
  • Mathematica
    Table[(CatalanNumber[2 n + 1] + CatalanNumber[n]^2)/2, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)

Formula

a(n) = sum(Catalan(k)*Catalan(2*n-k),k=0..n), n>=0, with Catalan(n)=A000108(n).
O.g.f: Ge(x)=(catao(x)+cata2(x))/2 with catao(x):= sum(Catalan(2*k+1)*x^k,k=0..infty) = (cata(sqrt(x)) - cata(-sqrt(x)))/(2*x), with the o.g.f. cata(x) of A000108, and cata2(x):=sum(Catalan(n)^2,n=0..infty) given in A001246 as (-1 + hypergeom( [-1/2,-1/2],[1],16*x))/(4*x).
a(n) = A028364(2n,n) = A067323(2n,n). - Alois P. Heinz, Nov 28 2015
a(n) = (A000108(2*n+1) + A000108(n)^2)/2. - Vladimir Reshetnikov, Oct 03 2016

Extensions

Cross-reference corrected by Robert Israel, Jun 06 2014

A265101 a(n) = binomial(6*n + 5, 3*n + 1)/(6*n + 5).

Original entry on oeis.org

1, 30, 1144, 49742, 2340135, 115997970, 5967382200, 315614844558, 17055399281284, 937581428480312, 52267355178398304, 2947837630317717410, 167897169647656366330, 9643503773422181941740, 557939244828083793388560, 32486374828326106197187470
Offset: 0

Views

Author

Peter Bala, Dec 02 2015

Keywords

Comments

Let x = p/q be a positive rational in reduced form with p,q > 0. Define Cat(x) = 1/(2*p + q)*binomial(2*p + q, p). Then Cat(n) = Catalan(n). This sequence is Cat(n + 1/3).
Number of maximal faces of the rational associahedron Ass(3*n + 1, 3*n + 4). Number of lattice paths from (0, 0) to (3*n + 4, 3*n + 1) using steps of the form (1, 0) and (0, 1) and staying above the line y = (3*n + 1)/(3*n + 4)*x. See Armstrong et al.

Crossrefs

Row 3 of A306444.
Cf. A000108, A065097 (Cat(n + 1/2)), A265102 (Cat(n + 1/4)), A265103 (Cat(n + 1/5)).

Programs

  • Magma
    [Binomial(6*n+5, 3*n+1)/(6*n+5): n in [0..15]]; // Vincenzo Librandi, Dec 09 2015
    
  • Maple
    seq(1/(6*n + 5)*binomial(6*n + 5, 3*n + 1), n = 0..15);
  • Mathematica
    Table[1/(6 n + 5) Binomial[6 n + 5, 3 n + 1], {n, 0, 20}] (* Vincenzo Librandi, Dec 09 2015 *)
  • PARI
    a(n) = binomial(6*n + 5, 3*n + 1)/(6*n + 5); \\ Altug Alkan, Dec 07 2015
    
  • Sage
    [binomial(6*n+5, 3*n+1)/(6*n+5) for n in (0..15)] # G. C. Greubel, Feb 16 2019

Formula

a(n) = binomial(6*n + 5, 3*n + 1)/(6*n + 5).
(n + 1)*(3*n - 1)*(3*n + 4)*a(n) = 8*(2*n + 1)*(6*n + 1)*(6*n - 1)*a(n-1) with a(0) = 1.
From Ilya Gutkovskiy, Feb 28 2017: (Start)
O.g.f.: (3F2(-1/6,1/6,1/2; -1/3,4/3; 64*x) - 1)/(2*x).
E.g.f.: 3F3(5/6,7/6,3/2; 2/3,2,7/3; 64*x).
a(n) ~ 4^(3*n+2)/(3*sqrt(3*Pi)*n^(3/2)). (End)
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