OFFSET
0,3
COMMENTS
The offset 0 is a choice. Another sequence could begin with A001477 instead of A000027. The Akiyama-Tanigawa transform applied to 1/(n+1) and 1/2^n are the second Bernoulli numbers A164555(n)/A027642(n) and the second (fractional) Euler numbers A198631(n)/A006519(n+1). (The first Euler numbers are not in the OEIS). Hence a(n).
a(2n+1) - a(2n) = 2^n -n -1 = 0, 0, 1, 4, 11,... = A000295(n) (Eulerian numbers).
a(2n+1) + a(2n) = 2^n +n +1 = A005126(n).
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-5,0,2).
FORMULA
a(n) = ((n+1) mod 2) * (n/2 + 1) + (n mod 2) * 2^((n-1)/2). - Wesley Ivan Hurt, Jun 29 2014
G.f.: (1 + x - x^2) * (1 - x^2 - x^3) / ((1 - x^2)^2 * (1 - 2*x^2)). - Michael Somos, Jun 30 2014
EXAMPLE
G.f. = 1 + x + 2*x^2 + 2*x^3 + 3*x^4 + 4*x^5 + 4*x^6 + 8*x^7 + 5*x^8 + ...
MAPLE
A242971:=n->((n+1) mod 2)*(n/2 + 1) + (n mod 2) * 2^((n-1)/2); seq(A242971(n), n=0..50); # Wesley Ivan Hurt, Jun 29 2014
MATHEMATICA
Table[Mod[n + 1, 2] (n/2 + 1) + Mod[n, 2] 2^((n - 1)/2), {n, 0, 50}] (* Wesley Ivan Hurt, Jun 29 2014 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul Curtz, Jun 22 2014
STATUS
approved