|
%I #28 May 21 2016 03:22:49
%S 1,3,8,26,90,322,1176,4356,16302,61490,233376,890188,3409588,13104756,
%T 50517200,195234120,756197910,2934686610,11408741520,44420399100,
%U 173191792620,676104403260,2642356838160,10337529691320,40481034410700
%N Central elements in 3-Pascal triangle A028262 (by row).
%C Or, start with Pascal's triangle; a(n) is the sum of the numbers on the periphery of the n-th central triangle containing exactly 3 numbers. The first three triangles are
%C ...1...........2.........6
%C .1...1.......3...3.....10..10
%C and the corresponding sums are 3, 8 and 26. - _Amarnath Murthy_, Mar 25 2003
%C This sequence starting at a(n+2) has Hankel transform A000032(2n+1)*2^n (empirical observation). - _Tony Foster III_, May 20 2016
%F G.f.: (x+1)/sqrt(1-4*x). - _Vladeta Jovovic_, Jan 08 2004
%F a(n) = binomial(2n, n)+binomial(2n-2, n-1)=A000984(n)+A000984(n-1). - _Emeric Deutsch_, Apr 20 2004
%F a(n) = 2binomial(2n-1, n-1)+binomial(2n-2, n-1) = binomial(2n, n)+binomial(2n-2, n-1) = A000984(n)+A000984(n-1). - _Emeric Deutsch_, Apr 20 2004
%F a(n) = (n+1)*C(n) + n*C(n-1), C = Catalan number (A000108). - _Gary W. Adamson_, Dec 28 2007
%F G.f.: G(0) where G(k)= 1 + x/(1 - (4*k+2)/((4*k+2) + (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - _Sergei N. Gladkovskii_, Jul 24 2012
%p seq(binomial(2*n,n)+binomial(2*n-2,n-1),n=0..24);
%p seq(2*binomial(2*n-1,n-1)+binomial(2*n-2,n-1),n=1..24);
%Y Cf. A081494, A081495, A081496, A000984.
%Y Cf. A000108.
%K nonn,easy
%O 0,2
%A _Mohammad K. Azarian_
%E More terms from _James A. Sellers_
|
