A179903 (1, 3, 5, 7, 9, ...) convolved with (1, 0, 3, 5, 7, 9, ...).

1, 3, 8, 21, 46, 87, 148, 233, 346, 491, 672, 893, 1158, 1471, 1836, 2257, 2738, 3283, 3896, 4581, 5342, 6183, 7108, 8121, 9226, 10427, 11728, 13133, 14646, 16271, 18012, 19873, 21858, 23971, 26216, 28597, 31118, 33783, 36596, 39561, 42682, 45963

Offset

0,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]

Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Formula

G.f.: (1 + 3*x + 8*x^2 + 21*x^3 + ...) = (1 + 3*x + 5*x^2 + 7*x^3 + 9*x^4 + ...) * (1 + 3*x^2 + 5*x^3 + 7*x^4 + 9*x^5 + ...).

From R. J. Mathar, Aug 13 2010: (Start)

a(n) = 2 + A005900(n), n > 0.

G.f.: -(1 + x)*(x^3 - 4*x^2 + 2*x - 1)/(x - 1)^4. (End)

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jul 04 2012

Example

a(5) = 46 = (9, 7, 5, 3, 1) dot (1, 0, 3, 5, 7) = 9 + 0 + 15 + 15 + 7.

Mathematica

CoefficientList[Series[-(1+x)*(x^3-4*x^2+2*x-1)/(x-1)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 04 2012 *)
Join[{1}, Table[ListConvolve[Range[1, 2n+1, 2], Join[{1, 0}, Range[3, 2n-1, 2]]], {n, 50}]// Flatten] (* or *) LinearRecurrence[{4, -6, 4, -1}, {1, 3, 8, 21, 46}, 50] (* Harvey P. Dale, Jan 30 2023 *)

Prog

(Magma) I:=[1, 3, 8, 21, 46]; [n le 5 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jul 04 2012

Crossrefs

Cf. A005900.

Sequence in context: A101332 A007773 A071078 * A363601 A193045 A238831

Adjacent sequences: A179900 A179901 A179902 * A179904 A179905 A179906

Keyword

nonn,easy

Author

Gary W. Adamson, Jul 31 2010

Status

approved