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.

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