A175899 a(n) = a(n-2) + a(n-3) + 2*a(n-4), with a(1) = 0, a(2) = 2, a(3) = 3, a(4) = 10.

0, 2, 3, 10, 5, 17, 21, 42, 48, 97, 132, 229, 325, 555, 818, 1338, 2023, 3266, 4997, 7965, 12309, 19494, 30268, 47733, 74380, 116989, 182649, 286835, 448398, 703462, 1100531, 1725530, 2700789, 4232985, 6627381, 10384834, 16261944, 25478185, 39901540, 62509797

Offset

1,2

Comments

According to the reference, p divides a(p) for every prime p.

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Eric Pite, Problem 1851, Mathematics Magazine 83 (2010) 303.

Index entries for linear recurrences with constant coefficients, signature (0, 1, 1, 2).

Formula

G.f.: x*(-2*x-3*x^2-8*x^3)/(-1+x^2+x^3+2*x^4). - Harvey P. Dale, Jul 24 2011

a(n) = n*sum(k=1..n/2, sum(j=0..k, binomial(j,n-4*k+2*j)*2^(k-j) * binomial(k,j))/k), n>0. - Vladimir Kruchinin, Oct 21 2011

Maple

a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <2|1|1|0>>^n.
        <<4, 0, 2, 3>>)[1, 1]:
seq(a(n), n=1..50);  # Alois P. Heinz, Oct 21 2011

Mathematica

LinearRecurrence[{0, 1, 1, 2}, {0, 2, 3, 10}, 40] (* Harvey P. Dale, Jul 24 2011 *)

Prog

(Maxima) a(n):=n*sum(sum(binomial(j, n-4*k+2*j)*2^(k-j)*binomial(k, j), j, 0, k)/k, k, 1, n/2); /* Vladimir Kruchinin, Oct 21 2011 */
(Haskell)
a175899 n = a175899_list !! (n-1)
a175899_list = 0 : 2 : 3 : 10 : zipWith (+) (map (* 2) a175899_list)
   (zipWith (+) (tail a175899_list) (drop 2 a175899_list))
-- Reinhard Zumkeller, Mar 23 2012

Crossrefs

Cf. A001608, A001634.

Sequence in context: A266552 A263716 A344457 * A328613 A064946 A078730

Adjacent sequences: A175896 A175897 A175898 * A175900 A175901 A175902

Keyword

nonn,easy

Author

John W. Layman, Oct 11 2010

Status

approved