The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A065096 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

A065096

Sums of lists produced by a variant of the iteration that produces the Catalan numbers: start with 0 and at each iteration replace each integer k with the list 0,1,...,k-1,k,k+1,k,k-1,...,1,0 and let a(n) be the sum of the resulting (flattened) list after n iterations.
(history; published version)

#47 by Michael De Vlieger at Fri Sep 01 14:13:19 EDT 2023

STATUS

reviewed

approved

#46 by Joerg Arndt at Fri Sep 01 11:25:05 EDT 2023

STATUS

proposed

reviewed

#45 by Michel Marcus at Fri Sep 01 11:22:32 EDT 2023

STATUS

editing

proposed

#44 by Michel Marcus at Fri Sep 01 11:22:19 EDT 2023

LINKS

S. B. Ekhad, and M. Yang, <a href="http://sites.math.rutgers.edu/~zeilberg/tokhniot/oMathar1maple12.txt">Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences</a>, (2017).

MATHEMATICA

Table[Range[n, 0, -1].Table[a[n, k], {k, 0, n}], {n, 0, 36}] (* with a[n, k] as defined in A033877. *)

STATUS

proposed

editing

#43 by Peter Bala at Fri Sep 01 08:48:15 EDT 2023

STATUS

editing

proposed

#42 by Peter Bala at Wed Aug 30 18:13:14 EDT 2023

FORMULA

a(n) = Sum_{k = 0..n+-1} 2^(k+1)/(n+1) * binomial(n+1, k)*binomial(n+1, k+2).

MAPLE

a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 else (3*n*(2*n+1)*a(n-1) - n*(n-1)*a(n-2))/((n+3)*(n-1)) end if; end:

#41 by Peter Bala at Wed Aug 30 15:27:01 EDT 2023

FORMULA

(n+3)*(n-1)*a(n) = 3*n*(2*n+1)*a(n-1) - n*(n-1)*a(n-2) with a(0) = 0 and a(1) = 1. (End)

G.f. A(x) satisfies the algebraic equation 4*x^3*A(x)^2 - (5*x^2 - 6*x + 1)*A(x) + x = 0 and the differential equation

(3*x^4 - 19*x^3 + 9*x^2 - x)*dA/dx + (3*x^3 - 29*x^2 + 21*x - 3)*A(x) + 4*x = 0 with A(0) = 0. (End)

KEYWORD

nonn,easy,changed

#40 by Peter Bala at Wed Aug 30 15:16:41 EDT 2023

FORMULA

From Peter Bala, Aug 30 2023: (Start)

a(n) = Sum_{k = 0..n+1} 2^(k+1)/(n+1) * binomial(n+1, k)*binomial(n+1, k+2).

(n+3)*(n-1)*a(n) = 3*n*(2*n+1)*a(n-1) - n*(n-1)*a(n-2) with a(0) = 0 and a(1) = 1. (End)

MAPLE

a := proc(n) option remember; if n=0 then 0 elif n = 1 then 1 else (3*n*(2*n+1)*a(n-1) - n*(n-1)*a(n-2))/((n+3)*(n-1)) end if; end:

seq(a(n), n = 0..20); # Peter Bala, Aug 30 2023

STATUS

approved

editing

#39 by Michel Marcus at Mon Sep 09 01:45:26 EDT 2019

STATUS

reviewed

approved

#38 by Joerg Arndt at Mon Sep 09 01:29:55 EDT 2019

STATUS

proposed

reviewed