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 1-9 of 9 results.

A060941 Duchon's numbers: the number of paths of length 5*n from the origin to the line y = 2*x/3 with unit East and North steps that stay below the line or touch it.

Original entry on oeis.org

1, 2, 23, 377, 7229, 151491, 3361598, 77635093, 1846620581, 44930294909, 1113015378438, 27976770344941, 711771461238122, 18293652115906958, 474274581883631615, 12388371266483017545, 325714829431573496525, 8613086428709348334675, 228925936056388155632081
Offset: 0

Views

Author

Philippe Flajolet, May 12 2001

Keywords

Comments

A generalization of the ballot numbers.

Crossrefs

See A293946 for a closely related sequence, also from the Bizley paper.

Programs

  • Magma
    [&+[1/(5*n+i+1)*Binomial(5*n+1, n-i)*Binomial(5*n+2*i, i): i in [0..n]]: n in [0..30]]; // Vincenzo Librandi, Feb 12 2016
  • Maple
    A060941 := n -> hypergeom([-n,5*n/2+1/2,5*n/2+1],[4*n+2,5*n+2],-4)* binomial(5*n,n)/(4*n+1); seq(simplify(A060941(n)),n=0..18); # Peter Luschny, Oct 05 2014
  • Mathematica
    a[n_] := ((5n)!*(5n + 1)!*HypergeometricPFQRegularized[{-n, 5n/2 + 1/2, 5n/2 + 1}, {4n + 2, 5n + 2}, -4])/n!; a /@ Range[0, 16]
    (* Jean-François Alcover, Jun 30 2011, after given formula *)
  • Sage
    A060941 = lambda n : hypergeometric([-n,5*n/2+1/2,5*n/2+1],[4*n+2,5*n+2],-4)*gamma(1+5*n)/(gamma(1+n)*gamma(2+4*n))
    [A060941(n).simplify() for n in range(19)] # Peter Luschny, Oct 05 2014
    

Formula

a(n) = Sum_{i=0..n} 1/(5*n+i+1) * C(5*n+1, n-i) * C(5*n+2*i, i).
a(n) = Sum_{i=0..2*n} (-1)^i/(5*i+1) * C((5*i+1)/2, i) * 1/(1+5*(2*n-i)) * C((1+5*(2*n-i))/2, 2*n-i).
G.f. A(z) satisfies: A(z) = 1+2*z*A^5-z*A^6+z*A^7+z^2*A^10. [Corrected by Bryan T. Ek, Oct 30 2017]
G.f.: A(z) = exp(C(5,2)*z/5 + C(10,4)*z^2/10 + C(15,6)*z^3/15 + ...). - Don Knuth, Oct 05 2014
Recurrence: 216*(n-1)*n*(2*n-1)*(3*n-4)*(3*n-2)*(3*n-1)*(3*n+1)*(6*n-1)*(6*n+1)*(5625*n^4 - 38550*n^3 + 97425*n^2 - 107784*n + 44044)*a(n) = 540*(n-1)*(3*n-4)*(3*n-2)*(126562500*n^10 - 1373625000*n^9 + 6557484375*n^8 - 18192221250*n^7 + 32549973750*n^6 - 39248008800*n^5 + 32203028675*n^4 - 17641491134*n^3 + 6113558828*n^2 - 1191132600*n + 96112128)*a(n-1) - 450*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(63281250*n^9 - 718453125*n^8 + 3556125000*n^7 - 10046426250*n^6 + 17765816250*n^5 - 20240090325*n^4 + 14698993900*n^3 - 6468702396*n^2 + 1533535184*n - 142988160)*a(n-2) + 78125*(n-2)*(5*n-14)*(5*n-13)*(5*n-12)*(5*n-11)*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(5625*n^4 - 16050*n^3 + 15525*n^2 - 6084*n + 760)*a(n-3). - Vaclav Kotesovec, Oct 05 2014
Asymptotics (Duchon, 2000): a(n) ~ c * (3125/108)^n / n^(3/2), where c = 0.0876612192439026461763141944768209255550234422281635788... (constant corrected, in the reference "On the enumeration and generation of generalized Dyck words", p.132 is a wrong value 0.0887). - Vaclav Kotesovec, Oct 05 2014, c = sqrt(5*(10^(2/3) - 5^(1/3)/2^(2/3) - 2))/(18*sqrt(Pi)). - Vaclav Kotesovec, Sep 16 2021
a(n) = Gamma(n+4/5)*Gamma(n+3/5)*Gamma(n+2/5)*3125^n*hypergeom([-n, (5/2)*n+1, (5/2)*n+1/2], [5*n+2, 4*n+2], -4)*Gamma(n+1/5)/ (Pi^2*csc((2/5)*Pi)*csc((1/5)*Pi)*Gamma(4*n+2)). - Robert Israel, Oct 05 2014
a(n) = A002294(n)*hypergeom([-n,5*n/2+1/2,5*n/2+1],[4*n+2,5*n+2],-4). - Peter Luschny, Oct 05 2014
O.g.f. A(x) satisfies: A(x)^5 = 1/x*series reversion( x/((1+x)*C(x))^5 ), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108. See A001450. - Peter Bala, Oct 05 2015
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 2, 50, 1415, 42258, 1300727, 40820837, 1298493730, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 7 (checked up to p = 101). [Added 23 Oct 2024: More generally, let r be an integer and s a positive integer and define a sequence u(n) by u(n) = [x^(s*n)] A(x)^(r*n). Then we conjecture that the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 7 and positive integers n and k.] - Peter Bala, Sep 12 2021
Inductively define a family of sequences {a(i,n) : n >= 0}, i >= 1, by setting a(1,n) = a(n) and, for i >= 2, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. We conjecture that the sequences {a(i,n) : n >= 0}, i >= 2, also satisfy the supercongruences u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) for primes p >= 7 and positive integers n and k. - Peter Bala, Oct 24 2024

A079489 a(n) = (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1).

Original entry on oeis.org

1, 3, 22, 211, 2306, 27230, 338444, 4362627, 57788170, 781825066, 10757497972, 150073096238, 2117778107732, 30176799215196, 433586825237912, 6274885068167651, 91383942213277530, 1338275570267001458, 19695358741104824036, 291137841642777382330, 4320734864185863437820
Offset: 0

Views

Author

N. J. A. Sloane, Jan 20 2003

Keywords

Comments

a(n) is the number of ordered trees on 2n-1 edges in which every subtree of the root (including its rooting edge) has an even number of edges, except for the leftmost subtree which has an odd number of edges (including its rooting edge). - David Callan, Apr 10 2012
a(n) is the number of 2 X 2n Young tableaux with a wall between the first and second row in each even column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(1)=3:
3 4 2 4 2 3
- - -
1 2, 1 3, 1 4. - Michael Wallner, Mar 09 2022

Crossrefs

Final diagonal of triangle in A078990.

Programs

  • Maple
    a := n -> (2*4^n*binomial(2*n, n) - binomial(4*n + 1, 2*n)) / (n + 1):
    seq(a(n), n = 0..20);  # Peter Luschny, Aug 26 2024
  • Mathematica
    ((Sqrt[2] Sqrt[1 + Sqrt[1 - 16 x]] - Sqrt[1 - 16 x] - 1)/(4 x) + O[x]^20)[[3]] (* Vladimir Reshetnikov, Sep 25 2016 *)
    CoefficientList[Series[-(1 - Sqrt[1 - 4*Sqrt[x]])*(1 - Sqrt[1 + 4*Sqrt[x]])/(4*x), {x,0,50}], x] (* G. C. Greubel, Apr 13 2017 *)
  • PARI
    a(n)=if(n<0,0,polcoeff(serreverse(x*(1-x^2)/(1+x^2)^2+O(x^(2*n+3))),2*n+1))
    
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,binomial(4*m-1,2*m)*x^m/m)+x*O(x^n)),n)} \\ Paul D. Hanna, Dec 30 2010

Formula

Series reversion of x(1-x^2)/(1+x^2)^2 expanded in odd powers of x. [Previous name.]
If x = y*(1-y^2)/(1+y^2)^2 then y = x + 3*x^3 + 22*x^5 + 211*x^7 + 2306*x^9 + ...
G.f. A(x) satisfies x*A(x^2) = (C(x) - C(-x))/(C(x) + C(-x)) where C(x) is g.f. of the Catalan numbers A000108.
a(n) = Sum_{k=0..2n} (-1)^k * A000108(2*n-k) * A000108(k). - David Callan, Aug 16 2006
a(n) = ((2^(4n+2))/Gamma(1/2)) * ((Gamma(n+1/2)/(2*Gamma(n+2))) - Gamma(2n+3/2)/Gamma(2n+3)). [David Dickson (dcmd(AT)unimelb.edu.au), Nov 10 2009]
G.f.: exp( Sum_{n>=1} C(4n-1,2n)*x^n/n ). - Paul D. Hanna, Dec 30 2010
G.f.: C(sqrt(x))*C(-sqrt(x)) where C(x) is the g.f. for the Catalan numbers A000108. - David Callan, Apr 10 2012
D-finite with recurrence n*(n+1)*(2*n+1)*a(n) -2*n*(32*n^2-32*n+11)*a(n-1) +16*(4*n-5)*(4*n-3)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Nov 29 2012
a(n) ~ (2-sqrt(2))*16^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 20 2013
a(n) = 2^(2*n+1)*Catalan(n) - Catalan(2*n+1) (see Regev). It follows that the 2-adic valuations of a(n) and Catalan(n) are equal. In particular, a(n) is odd iff n is of the form 2^m - 1. - Peter Bala, Aug 02 2016
G.f.: (sqrt(2) * sqrt(1 + sqrt(1-16*x)) - sqrt(1-16*x) - 1)/(4*x). - Vladimir Reshetnikov, Sep 25 2016
G.f. A(x) satisfies A(x^2) = C(x)^2*r(-x*C(x)^2), where C(x) is g.f. of the Catalan numbers A000108, and r(x) is g.f. of the large Schröder numbers A006318. - Alexander Burstein, Nov 21 2019
From Peter Bala, Sep 14 2021: (Start)
A(x) = exp( Sum_{n >= 1} (1/2)*binomial(4*n,2*n)*x^n/n ).
1 + x*A(x) is the o.g.f. of A066357.
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 3, 53, 1056, 22181, 480003, 10588508, 236720424, ...] and satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 3. See A333563. Cf. A060941. (End)
From Peter Bala, Oct 23 2024: (Start)
For integer r and positive integer s, define a sequence {u(n) : n >= 0} by setting u(n) = [x^(s*n)] A(x)^(r*n). We conjecture that the supercongruence u(n*p^k) == u(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k.
Let B(x) = 1/x * series_reversion(x*A(x)). Define a sequence {v(n) : n >= 0} by setting v(n) = [x^(s*n)] B(x)^(r*n). We conjecture that the supercongruence v(n*p^k) == v(n*p^(k-1)) (mod p^(3*k)) holds for all primes p >= 5 and for all positive integers n and k. (End)

Extensions

New name by Peter Luschny, Aug 26 2024

A300389 The number of paths of length 13*n from the origin to the line y = 2*x/11 with unit East and North steps that stay below the line or touch it.

Original entry on oeis.org

1, 6, 593, 87143, 15149546, 2891511017, 585739005066, 123655688922720, 26908765569970320, 5993187329634638043, 1359541058523676017369, 313029501692713279534165, 72965556751635426636633639, 17184586991024424745328563477, 4083065013894860643162116395527
Offset: 0

Views

Author

Bryan T. Ek, Mar 04 2018

Keywords

Comments

Equivalent to nonnegative walks from (0,0) to (13*n,0) with step set [1,2], [1,-11].

Examples

			For n=1, the possible walks are EEEEEEEEEEENN, EEEEEEEEEENEN, EEEEEEEEENEEN, EEEEEEEENEEEN, EEEEEEEENEEEEN, EEEEEEENEEEEN.
		

Crossrefs

Programs

  • Mathematica
    m = 15;
    Exp[Sum[(1/13) Binomial[13n, 2n] x^n/n, {n, 1, m}]] + O[x]^m // CoefficientList[#, x]& (* Jean-François Alcover, Feb 26 2020, after Peter Bala *)

Formula

G.f. satisfies: f = f^78*t^6 + 5*f^67*t^5 - f^66*t^5 + 6*f^65*t^5 + 10*f^56*t^4 - 4*f^55*t^4 + 20*f^54*t^4 - 5*f^53*t^4 + 15*f^52*t^4 + 10*f^45*t^3 - 6*f^44*t^3 + 24*f^43*t^3 - 12*f^42*t^3 + 30*f^41*t^3 - 10*f^40*t^3 + 20*f^39*t^3 + 5*f^34*t^2 - 4*f^33*t^2 + 12*f^32*t^2 - 9*f^31*t^2 + 18*f^30*t^2 - 12*f^29*t^2 + 20*f^28*t^2 - 10*f^27*t^2 + 15*f^26*t^2 + f^23*t - f^22*t + 2*f^21*t - 2*f^20*t + 3*f^19*t - 3*f^18*t + 4*f^17*t - 4*f^16*t + 5*f^15*t - 5*f^14*t + 6*f^13*t + 1.
From Peter Bala, Jan 03 2019: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/13)*binomial(13*n, 2*n)*x^n/n ) - Bizley.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/13)*binomial(13*n-13*k, 2*n-2*k)*a(k) for n >= 1. (End)
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 6, 1222, 282993, 69239846, 17468997381, 4494716943847, 1172353182893367, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 except for p = 11 and p = 13 (checked up to p = 101). - Peter Bala, Sep 14 2021
a(n) ~ c * 13^(13*n) / (n^(3/2) * 2^(2*n) * 11^(11*n)), where c = 0.0250562444901910770802983936320823301923793538303930752981380507191770309... - Vaclav Kotesovec, Sep 16 2021

A182960 G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ).

Original entry on oeis.org

1, 5, 95, 2496, 76063, 2524161, 88534548, 3228482908, 121171012431, 4649906785719, 181614908182551, 7196014051078368, 288537887780406468, 11686156771344086156, 477379538839242423528, 19645977861506470428324
Offset: 0

Views

Author

Paul D. Hanna, Dec 31 2010

Keywords

Comments

The logarithmic derivative of this sequence is a bisection of the logarithmic derivative of A001764 (ternary trees).
To see this, compare the g.f. of this sequence with g.f. of A001764:
exp(Sum_{n>=1} C(3n-1,n-1)*x^n/n) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 +...

Examples

			G.f.: A(x) = 1 + 5*x + 95*x^2 + 2496*x^3 + 76063*x^4 + 2524161*x^5 +...
log(A(x)) = 5*x + 165*x^2 + 6188*x^3 + 245157*x^4 + 10015005*x^5 +...+ A025174(2n)*x^n/n +...
G.f. satisfies: A(x) = G(x*A(x)^3) where G(x) begins:
G(x) = 1 + 5*x + 20*x^2 + 96*x^3 + 528*x^4 + 3136*x^5 + 19584*x^6 +...
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Exp[ Sum[ Binomial[6 i - 1, 2 i - 1]*x^i/i, {i, n}]]; CoefficientList[ Series[f@ 15, {x, 0, 15}], x]  (* Robert G. Wilson v, Dec 31 2010 *)
  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,binomial(6*m-1,2*m-1)*x^m/m)+x*O(x^n)),n)}
    
  • PARI
    {a(n)=polcoeff((serreverse(x*(1-2*x+sqrt(1-8*x+x*O(x^n)))^3/(8*(1+x)^6))/x)^(1/3),n)}

Formula

G.f. A(x) satisfies: A(x^2) = F(x)*F(-x) where F(x) = 1 + x*F(x)^3 = Sum_{n>=0} C(3n,n)*x^n/(2n+1) is the g.f. of A001764.
Let G(x) = 2*(1+x)^2/(1-2*x+sqrt(1-8*x)), then g.f. A(x) satisfies:
* A(x) = G(x*A(x)^3) and A(x/G(x)^3) = G(x);
* A(x) = [Series_Reversion( x/G(x)^3 ) / x]^(1/3).
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 5, 215, 10463, 537287, 28435880, 1534398353, 83920389642, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 (checked up to p = 101). - Peter Bala, Sep 14 2021
From Vaclav Kotesovec, Sep 16 2021: (Start)
Recurrence: 16*n*(2*n + 1)*(4*n - 1)*(4*n + 1)*(27*n - 31)*a(n) = 18*(69984*n^5 - 220320*n^4 + 267705*n^3 - 156510*n^2 + 42021*n - 3680)*a(n-1) - 59049*(n-1)*(2*n - 3)*(3*n - 5)*(3*n - 4)*(27*n - 4)*a(n-2).
a(n) ~ sqrt((sqrt(2) - 1)^(2/3) + (sqrt(2) + 1)^(2/3) - 2) * 3^(6*n + 3/2) / (sqrt(Pi) * n^(3/2) * 2^(4*n + 7/2)). (End)

A300387 The number of paths of length 9*n from the origin to the line y = 2*x/7 with unit East and North steps that stay below the line or touch it.

Original entry on oeis.org

1, 4, 178, 11654, 900239, 76266406, 6853777795, 641688752961, 61916364799849, 6113859987916630, 614832988424140624, 62752222758863566993, 6483650829899569496380, 676834416167597357806799, 71278487569046416052210050, 7563527671079260544924794587, 807900192360879042402313084390
Offset: 0

Views

Author

Bryan T. Ek, Mar 04 2018

Keywords

Comments

Equivalent to nonnegative walks from (0,0) to (9*n,0) with step set [1,2], [1,-7].

Examples

			For n=1, the possible walks are EEEEEEENN, EEEEEENEN, EEEEENEEN, EEEENEEEN.
		

Crossrefs

Programs

  • Mathematica
    terms = 17; f[_] = 0;
    Do[f[t_] = f[t]^36 t^4 + 3 f[t]^29 t^3 - f[t]^28 t^3 + 4 f[t]^27 t^3 + 3 f[t]^22 t^2 - 2 f[t]^21 t^2 + 6 f[t]^20 t^2 - 3 f[t]^19 t^2 + 6 f[t]^18 t^2 + f[t]^15 t - f[t]^14 t + 2 f[t]^13 t - 2 f[t]^12 t + 3 f[t]^11 t - 3 f[t]^10 t + 4 f[t]^9 t + 1 + O[t]^terms, {terms}];
    CoefficientList[f[t], t] (* Jean-François Alcover, Dec 04 2018 *)
    nmax = 20; CoefficientList[Series[Exp[Sum[Binomial[9*k, 2*k]*x^k/(9*k), {k, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 16 2021 *)

Formula

G.f. satisfies: f=f^36*t^4+3*f^29*t^3-f^28*t^3+4*f^27*t^3+3*f^22*t^2-2*f^21*t^2+6*f^20*t^2-3*f^19*t^2+6*f^18*t^2+f^15*t-f^14*t+2*f^13*t-2*f^12*t+3*f^11*t-3*f^10*t+4*f^9*t+1.
From Peter Bala, Jan 03 2019: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/9)*binomial(9*n, 2*n)*x^n/n ) - Bizley. Cf. A274244.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/9)*binomial(9*n-9*k, 2*n-2*k)*a(k) for n >= 1. (End)
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 4, 372, 39298, 4384884, 504464254, 59183637186, 7038517648243, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 11 (checked up to p = 101). - Peter Bala, Sep 14 2021
a(n) ~ c * 3^(18*n) / (n^(3/2) * 2^(2*n) * 7^(7*n)), where c = 0.0389180896257538883301359279112039841187646397413254619045749515282872957... - Vaclav Kotesovec, Sep 16 2021

A300388 The number of paths of length 11*n from the origin to the line y = 2*x/9 with unit East and North steps that stay below the line or touch it.

Original entry on oeis.org

1, 5, 345, 35246, 4255288, 563796161, 79264265868, 11612106079203, 1753402118587333, 270965910076404428, 42648418241303137766, 6813002989827352100145, 1101807202785456951146158, 180034116076502209139781574, 29677341363243548521326632028, 4929368173228370040701922315332
Offset: 0

Views

Author

Bryan T. Ek, Mar 04 2018

Keywords

Comments

Equivalent to nonnegative walks from (0,0) to (11*n,0) with step set [1,2], [1,-9].

Examples

			For n=1, the walks are EEEEEEEEENN, EEEEEEEENEN, EEEEEEENEEN, EEEEEENEEEN, EEEEENEEEEN.
		

Crossrefs

Programs

  • Mathematica
    terms = 16; f[_] = 0;
    Do[f[t_] = f[t]^55 t^5 + 4 f[t]^46 t^4 - f[t]^45 t^4 + 5 f[t]^44 t^4 + 6 f[t]^37 t^3 - 3 f[t]^36 t^3 + 12 f[t]^35 t^3 - 4 f[t]^34 t^3 + 10 f[t]^33 t^3 + 4 f[t]^28 t^2 - 3 f[t]^27 t^2 + 9 f[t]^26 t^2 - 6 f[t]^25 t^2 + 12 f[t]^24 t^2 - 6 f[t]^23 t^2 + 10 f[t]^22 t^2 + f[t]^19 t - f[t]^18 t + 2 f[t]^17 t - 2 f[t]^16 t + 3 f[t]^15 t - 3 f[t]^14 t + 4 f[t]^13 t - 4 f[t]^12 t + 5 f[t]^11 t + 1 + O[t]^terms, {terms}];
    CoefficientList[f[t], t] (* Jean-François Alcover, Dec 04 2018 *)
    nmax = 20; CoefficientList[Series[Exp[Sum[Binomial[11*k, 2*k]*x^k/(11*k), {k, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 16 2021 *)

Formula

G.f. satisfies: f = f^55*t^5 + 4*f^46*t^4 - f^45*t^4 + 5*f^44*t^4 + 6*f^37*t^3 - 3*f^36*t^3 + 12*f^35*t^3 - 4*f^34*t^3 + 10*f^33*t^3 + 4*f^28*t^2 - 3*f^27*t^2 + 9*f^26*t^2 - 6*f^25*t^2 + 12*f^24*t^2 - 6*f^23*t^2 + 10*f^22*t^2 + f^19*t - f^18*t + 2*f^17*t - 2*f^16*t + 3*f^15*t - 3*f^14*t + 4*f^13*t - 4*f^12*t + 5*f^11*t + 1.
From Peter Bala, Jan 03 2019: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/11)*binomial(11*n, 2*n)*x^n/n ) - Bizley. Cf. A274256.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/11)*binomial(11*n-11*k, 2*n-2*k)*a(k) for n >= 1. (End)
The sequence defined by b(n) := [x^n] A(x)^n begins [1, 5, 715, 116213, 19954187, 3532860880, 637870220023, 116749388814357, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 except for p = 11 (checked up to p = 101). - Peter Bala, Sep 14 2021
a(n) ~ c * 11^(11*n) / (n^(3/2) * 2^(2*n) * 3^(18*n)), where c = 0.0304820662333129164912550234496338371466905844787974500412037592866845093... - Vaclav Kotesovec, Sep 16 2021

A381757 Expansion of exp( Sum_{k>=1} binomial(7*k-1,2*k-1) * x^k/k ).

Original entry on oeis.org

1, 6, 161, 6062, 265868, 12720904, 643915209, 33905228350, 1838102210977, 101910583801012, 5751779249830131, 329359930638541776, 19087504000780665541, 1117418973753045781944, 65982722733895652916539, 3925378032146863676341770, 235048328495265879957413946
Offset: 0

Views

Author

Seiichi Manyama, Mar 06 2025

Keywords

Crossrefs

Programs

  • PARI
    my(N=20, x='x+O('x^N)); Vec(exp(sum(k=1, N, binomial(7*k-1, 2*k-1)*x^k/k)))

Formula

a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} binomial(7*k-1,2*k-1) * a(n-k).
G.f.: B(x)^2, where B(x) is the g.f. of A300386.

A300390 The number of paths of length 7*n from the origin to the line y = 3*x/4 with unit east and north steps that stay below the line or touch it.

Original entry on oeis.org

1, 5, 227, 15090, 1182187, 101527596, 9247179818, 877362665128, 85783306955099, 8582893111512001, 874542924575207352, 90437361732467946334, 9467275300762187682554, 1001309098267187214993056, 106836493655355495755649544, 11485688815900189437990930096, 1242964338344397490958154292155
Offset: 0

Views

Author

Bryan T. Ek, Mar 05 2018

Keywords

Comments

Equivalent to nonnegative walks from (0,0) to (7*n,0) with step set [1,3], [1,-4].

Examples

			For n=1, the possible walks are EEEENNN, EEENENN, EENEENN, EEENNEN, EENENEN.
		

Crossrefs

Programs

  • Mathematica
    m = 17; f = 0; Do[f = f^35*t^5 - f^31*t^4 + f^30*t^4 - f^29*t^4 + 5*f^28*t^4 - f^25*t^3 + f^24*t^3 + 3*f^23*t^3 - 4*f^22*t^3 + 10*f^21*t^3 + f^19*t^2 - f^18*t^2 + 5*f^17*t^2 + 3*f^16*t^2 - 6*f^15*t^2 + 10*f^14*t^2 + f^13*t - f^12*t + 3*f^10*t + f^9*t - 4*f^8*t + 5*f^7*t + 1 + O[t]^m, {m}]; CoefficientList[f, t] (* Jean-François Alcover, Feb 18 2019 *)

Formula

G.f. satisfies: f = f^35*t^5 - f^31*t^4 + f^30*t^4 - f^29*t^4 + 5*f^28*t^4 - f^25*t^3 + f^24*t^3 + 3*f^23*t^3 - 4*f^22*t^3 + 10*f^21*t^3 + f^19*t^2 - f^18*t^2 + 5*f^17*t^2 + 3*f^16*t^2 - 6*f^15*t^2 + 10*f^14*t^2 + f^13*t - f^12*t + 3*f^10*t + f^9*t - 4*f^8*t + 5*f^7*t + 1.
From Peter Bala, Jan 03 2019: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/7)*binomial(7*n, 3*n)*x^n/n ) - Bizley.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/7)*binomial(7*n-7*k, 3*n-3*k)*a(k) for n >= 1. (End)

A300391 The number of paths of length 8*n from the origin to the line y = 3*x/5 with unit east and north steps that stay below the line or touch it.

Original entry on oeis.org

1, 7, 525, 58040, 7574994, 1084532963, 164734116407, 26070940600055, 4252443527211637, 709846349042619913, 120679177855928146859, 20822762876863605793639, 3637213213067542990001936, 641912742432770594132245835, 114287840570892852593437353124, 20502971288127330644273350110698
Offset: 0

Views

Author

Bryan T. Ek, Mar 05 2018

Keywords

Comments

Equivalent to nonnegative walks from (0,0) to (8*n,0) with step set [1,3], [1,-5].

Examples

			For n=1, the possible walks are EEEEENNN, EEEENENN, EEEENNEN, EEENEENN, EEENENEN, EENEEENN, EENEENEN.
		

Crossrefs

Formula

G.f. f satisfies f = t^7*f^56 - 2*t^6*f^51 + t^6*f^50 - t^6*f^49 + 7*t^6*f^48 + t^5*f^46 - t^5*f^45 - 3*t^5*f^43 + 5*t^5*f^42 - 6*t^5*f^41 + 21*t^5*f^40 - 3*t^4*f^37 - 3*t^4*f^36 + 8*t^4*f^35 + 10*t^4*f^34 - 15*t^4*f^33 + 35*t^4*f^32 - 2*t^3*f^31 + 2*t^3*f^30 - 9*t^3*f^28 + 22*t^3*f^27 + 10*t^3*f^26 - 20*t^3*f^25 + 35*t^3*f^24 + 3*t^2*f^22 + 5*t^2*f^21 - 9*t^2*f^20 + 18*t^2*f^19 + 5*t^2*f^18 - 15*t^2*f^17 + t*(21*t + 1)*f^16 - t*f^15 + 3*t*f^13 - 3*t*f^12 + 5*t*f^11 + t*f^10 - 6*t*f^9 + 7*t*f^8 + 1.
From Peter Bala, Jan 03 2019: (Start)
O.g.f.: A(x) = exp( Sum_{n >= 1} (1/8)*binomial(8*n, 3*n)*x^n/n ) - Bizley.
Recurrence: a(0) = 1 and a(n) = (1/n) * Sum_{k = 0..n-1} (1/8)*binomial(8*n-8*k, 3*n-3*k)*a(k) for n >= 1. (End)
Showing 1-9 of 9 results.