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
Examples
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.
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).
Programs
-
C
int a(int n){n--;return n>>(n&1);} // Mia Boudreau, Aug 27 2025 -
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 *) -
PARI
a(n)=n-=2;if(n%2,n+1,n/2)
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
a(n) = floor((n-1) / 2^mod(n-1,2)). - Mia Boudreau, Aug 27 2025
E.g.f.: 1 + (x - 1)*cosh(x) + (x - 2)*sinh(x)/2. - Stefano Spezia, Dec 17 2023
Comments