A245738 Number of compositions of n into parts 1 and 2 with both parts present.

2, 3, 7, 11, 20, 32, 54, 87, 143, 231, 376, 608, 986, 1595, 2583, 4179, 6764, 10944, 17710, 28655, 46367, 75023, 121392, 196416, 317810, 514227, 832039, 1346267, 2178308, 3524576, 5702886, 9227463, 14930351, 24157815, 39088168, 63245984, 102334154, 165580139, 267914295, 433494435, 701408732, 1134903168, 1836311902

Offset

3,1

Links

Colin Barker, Table of n, a(n) for n = 3..1000

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

Formula

G.f.: 1+1/(1-x-x^2)-1/(1-x)-1/(1-x^2).

a(n) = A052952(n-4)+2*A052952(n-3). - R. J. Mathar, Aug 05 2014

From Colin Barker, Jul 13 2017: (Start)

a(n) = (-20 + sqrt(5)*(-(1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n))/2^n) / 10 for n even.

a(n) = (-10 + sqrt(5)*(-(1-sqrt(5))^(1+n) + (1+sqrt(5))^(1+n))/2^n) / 10 for n odd.

a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4) for n>6. (End)

a(n) = Sum_{i=1..floor((n-1)/2)} C(n-i,i). - Wesley Ivan Hurt, Sep 19 2017

a(n) = A000045(n+1) - A000034(n+1). - J. M. Bergot and Robert Israel, Oct 11 2021

Example

a(9) = 54. The tuples are (22221) = 5!/4! = 5, (222111) = 6!/3!/3! = 20, (2211111) = 7!/5!/2! = 21, (21111111) = 8!/7! = 8.

Mathematica

LinearRecurrence[{1, 2, -1, -1}, {2, 3, 7, 11}, 50] (* Harvey P. Dale, Dec 20 2014 *)

Prog

(PARI) Vec(1+1/(1-x-x^2)-1/(1-x)-1/(1-x^2)+O(x^66)) \\ Joerg Arndt, Aug 04 2014

Crossrefs

Cf. A245332, A245492, A245487, A245527.

Sequence in context: A159262 A160434 A139630 * A368032 A265093 A133044

Adjacent sequences: A245735 A245736 A245737 * A245739 A245740 A245741

Keyword

nonn,easy

Author

David Neil McGrath, Jul 31 2014

Status

approved