0,4

a(n) = 2^n + 1 - 2*Fibonacci(n+1) = 2^n + 1 + Fibonacci(n) - Fibonacci(n+3) = 2^n + 1 - Fibonacci(n) - Lucas(n). a(n) = 2(2^(n-1) - Fibonacci(n+1)) + 1, for n > 0. a(n) = A000051(n) - A006355(n+2) = A000051(n) - A000045(n) - A000032(n). a(n) = A101220(2,2,n-1) - A101220(1,1,n-3), for n > 2. a(n) = A008466(n) - A000071(n-1), for n > 0. a(n) = 2*A008466(n-1) + 1, for n > 0.

a(n) = 2*A101220(2,2,n-2) + 1, for n > 1. a(n) = Sum[2^(n-k)Fibonacci(k) - Fibonacci(k-2),{k,0,n}] = antidiagonal sums of A118654. a(n+1) - a(n) = 2(2^(n-1) - Fibonacci(n)), for n > 0. a(n+1) - a(n) = 2*A027934(n-2), for n > 1. a(n+1) - a(n) = 2*A101220(1,2,n-1), for n > 0. a(0) = 0; a(1) = 1; a(n) = a(n-1) + a(n-2) + 2^(n-2) - 1, for n > 1. a(0) = 0; a(1) = 1; a(2) = 1; a(3) = 3; a(n) = 4*a(n-1) - 4*a(n-2) - a(n-3) + 2*a(n-4), for n > 3. O.g.f. = (x(1-3x+3x^2))/((1-x)(1-2x)(1-x-x^2)).

a(n)=1+2^n-[1/2+(1/2)*sqrt(5)]^n-(1/5)*[1/2+(1/2)*sqrt(5)]^n*sqrt(5)+(1/5)*sqrt(5)*[1/2-(1/2) *sqrt(5)]^n-[1/2-(1/2)*sqrt(5)]^n, with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Oct 07 2008]

Table[2^n + 1 - 2 Fibonacci[n + 1], {n, 0, 30}]

Cf. A000032, A000045, A000051, A000071, A006355, A008466, A027934, A101220, A118654.

Sequence in context: A026396 A141199 A003478 this_sequence A127984 A157029 A077927

Adjacent sequences: A119584 A119585 A119586 this_sequence A119588 A119589 A119590

nonn

Ross La Haye (rlahaye(AT)new.rr.com), May 31 2006, Jun 27 2007