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A162540
a(n) = (2*n+1)*(2*n+3)*(2*n+5)/3.
2
5, 35, 105, 231, 429, 715, 1105, 1615, 2261, 3059, 4025, 5175, 6525, 8091, 9889, 11935, 14245, 16835, 19721, 22919, 26445, 30315, 34545, 39151, 44149, 49555, 55385, 61655, 68381, 75579, 83265, 91455, 100165, 109411, 119209, 129575, 140525, 152075, 164241
OFFSET
0,1
LINKS
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.
M. R. Sepanski, On Divisibility of Convolutions of Central Binomial Coefficients, Electronic Journal of Combinatorics, 21 (1) 2014, #P1.32.
FORMULA
a(n) = A061550(n)/3 = A077415(2*n+3).
From R. J. Mathar, Jul 16 2009: (Start)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3.
G.f.: (5 + 15*x - 5*x^2 + x^3)/(x-1)^4. (End)
a(n) = 5*Pochhammer(7/2,n)/Pochhammer(1/2,n). Hence e.g.f. is 5* 1F1(7/2;1/2;x), with 1F1 being the confluent hypergemetric function (also known as Kummer's). - Stanislav Sykora, May 26 2016
E.g.f.: (8*x^3 + 60*x^2 + 90*x + 15)*exp(x)/3. - Robert Israel, May 27 2016
From Amiram Eldar, Jan 09 2021: (Start)
Sum_{n>=0} 1/a(n) = 1/4.
Sum_{n>=0} (-1)^n/a(n) = 3*Pi/8 - 1 = A093828 - 1. (End)
MAPLE
A162540:=n->(2*n+1)*(2*n+3)*(2*n+5)/3: seq(A162540(n), n=0..80); # Wesley Ivan Hurt, May 28 2016
MATHEMATICA
Table[((2n+1)(2n+3)(2n+5))/3, {n, 0, 40}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {5, 35, 105, 231}, 40] (* Harvey P. Dale, Nov 06 2011 *)
PROG
(Magma) [(2*n+1)*(2*n+3)*(2*n+5)/3: n in [0..40]]; // Vincenzo Librandi, Nov 16 2011
(PARI) Vec((5+15*x-5*x^2+x^3)/(x-1)^4 + O(x^100)) \\ Altug Alkan, Oct 26 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jacob Landon (jacoblandon(AT)aol.com), Jul 05 2009
EXTENSIONS
Offset corrected, definition clarified by R. J. Mathar, Jul 16 2009
STATUS
approved