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A028270 Central elements in 3-Pascal triangle A028262 (by row).

%I A028270 #31 Jul 02 2025 16:01:55
%S A028270 1,3,8,26,90,322,1176,4356,16302,61490,233376,890188,3409588,13104756,
%T A028270 50517200,195234120,756197910,2934686610,11408741520,44420399100,
%U A028270 173191792620,676104403260,2642356838160,10337529691320,40481034410700
%N A028270 Central elements in 3-Pascal triangle A028262 (by row).
%C A028270 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 A028270 ...1...........2.........6
%C A028270 .1...1.......3...3.....10..10
%C A028270 and the corresponding sums are 3, 8 and 26. - _Amarnath Murthy_, Mar 25 2003
%C A028270 This sequence starting at a(n+2) has Hankel transform A000032(2n+1)*2^n (empirical observation). - _Tony Foster III_, May 20 2016
%F A028270 G.f.: (x+1)/sqrt(1-4*x). - _Vladeta Jovovic_, Jan 08 2004
%F A028270 a(n) = binomial(2n, n)+binomial(2n-2, n-1)=A000984(n)+A000984(n-1). - _Emeric Deutsch_, Apr 20 2004
%F A028270 a(n) = 2binomial(2n-1, n-1)+binomial(2n-2, n-1). - _Emeric Deutsch_, Apr 20 2004
%F A028270 a(n) = (n+1)*C(n) + n*C(n-1), C = Catalan number (A000108). - _Gary W. Adamson_, Dec 28 2007
%F A028270 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
%F A028270 D-finite with recurrence n*a(n) -3*n*a(n-1) +2*(-2*n+5)*a(n-2)=0. - _R. J. Mathar_, May 01 2024
%p A028270 seq(binomial(2*n,n)+binomial(2*n-2,n-1),n=0..24);
%p A028270 seq(2*binomial(2*n-1,n-1)+binomial(2*n-2,n-1),n=1..24);
%Y A028270 Cf. A081494, A081495, A081496, A000984.
%Y A028270 Cf. A000108.
%K A028270 nonn,easy
%O A028270 0,2
%A A028270 _Mohammad K. Azarian_
%E A028270 More terms from _James Sellers_