A065423 Number of ordered length 2 compositions of n with at least one even summand.

0, 0, 2, 1, 4, 2, 6, 3, 8, 4, 10, 5, 12, 6, 14, 7, 16, 8, 18, 9, 20, 10, 22, 11, 24, 12, 26, 13, 28, 14, 30, 15, 32, 16, 34, 17, 36, 18, 38, 19, 40, 20, 42, 21, 44, 22, 46, 23, 48, 24, 50, 25, 52, 26, 54, 27, 56, 28, 58, 29, 60, 30, 62, 31, 64, 32, 66, 33, 68, 34, 70, 35, 72, 36, 74

Offset

1,3

Links

Stefano Spezia, Table of n, a(n) for n = 1..10000

Mircea Merca and Emil Simion, n-Color Partitions into Distinct Parts as Sums over Partitions, Symmetry (2023) Vol. 15, Iss. 11.

Index entries for two-way infinite sequences

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

Formula

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

a(n) = floor((n-1)/2) + (n is odd)*floor((n-1)/2).

a(n+2) = Sum_{k=0..n} (gcd(n, k) mod 2). - Paul Barry, May 02 2005

a(n) = Sum_{i=1..n-1} (-1)^i (floor(i/2) + ((i+1) mod 2)). - Olivier Gérard, Jun 21 2007

a(n) = A210530(n,4)/2 for n>2. - Boris Putievskiy, Jan 29 2013

a(n) = (3*n-4-n*(-1)^n)/4. - Boris Putievskiy, Jan 29 2013, corrected Jan 24 2022

a(n) = A026741(n)-1. - Wesley Ivan Hurt, Jun 23 2013

E.g.f.: 1 + (x - 1)*cosh(x) + (x - 2)*sinh(x)/2. - Stefano Spezia, Dec 17 2023

Example

a(7) = 6 because we can write 7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1; a(8) = 3 because we can write 8 = 2+6 = 4+4 = 6+2.

Maple

A065423 := proc(n)
    (3*n-4-(-1)^n*n)/4 ;
end proc:
seq(A065423(n), n=1..40) ; # R. J. Mathar, Jan 24 2022

Mathematica

LinearRecurrence[{0, 2, 0, -1}, {0, 0, 2, 1}, 100] (* Harvey P. Dale, May 14 2014 *)

Prog

(PARI) a(n)=n-=2; if(n%2, n+1, n/2)

Crossrefs

Cf. A026741, A097140 (first differences), A030451 (absolute first differences), A210530.

Sequence in context: A130107 A107130 A194747 * A239242 A340621 A008733

Adjacent sequences: A065420 A065421 A065422 * A065424 A065425 A065426

Keyword

nonn,easy

Author

Len Smiley, Nov 23 2001

Status

approved