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A028270 Central elements in 3-Pascal triangle A028262 (by row). 1
1, 3, 8, 26, 90, 322, 1176, 4356, 16302, 61490, 233376, 890188, 3409588, 13104756, 50517200, 195234120, 756197910, 2934686610, 11408741520, 44420399100, 173191792620, 676104403260, 2642356838160, 10337529691320, 40481034410700 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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

...1...........2.........6

.1...1.......3...3.....10..10

and the corresponding sums are 3, 8 and 26. - Amarnath Murthy, Mar 25 2003

This sequence starting at a(n+2) has Hankel transform A000032(2n+1)*2^n (empirical observation). - Tony Foster III, May 20 2016

LINKS

Table of n, a(n) for n=0..24.

FORMULA

G.f.: (x+1)/sqrt(1-4*x). - Vladeta Jovovic, Jan 08 2004

a(n) = binomial(2n, n)+binomial(2n-2, n-1)=A000984(n)+A000984(n-1). - Emeric Deutsch, Apr 20 2004

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

a(n) = (n+1)*C(n) + n*C(n-1), C = Catalan number (A000108). - Gary W. Adamson, Dec 28 2007

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

MAPLE

seq(binomial(2*n, n)+binomial(2*n-2, n-1), n=0..24);

seq(2*binomial(2*n-1, n-1)+binomial(2*n-2, n-1), n=1..24);

CROSSREFS

Cf. A081494, A081495, A081496, A000984.

Cf. A000108.

Sequence in context: A148818 A242903 A081497 * A124383 A148819 A148820

Adjacent sequences:  A028267 A028268 A028269 * A028271 A028272 A028273

KEYWORD

nonn,easy

AUTHOR

Mohammad K. Azarian

EXTENSIONS

More terms from James A. Sellers

STATUS

approved

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Last modified October 23 01:24 EDT 2018. Contains 316518 sequences. (Running on oeis4.)