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 A136422 Floor((x^n - (1-x)^n)/sqrt(3)+.5) where x = (sqrt(3)+1)/2. 0

%I #6 Sep 10 2016 15:30:45

%S 1,1,2,2,3,4,5,7,10,13,18,24,33,45,62,85,116,158,216,296,404,551,753,

%T 1029,1406,1920,2623,3583,4895,6687,9134,12477,17044,23283,31805,

%U 43447,59349,81072,110747,151283,206657,282298,385626,526775,719589,982976

%N Floor((x^n - (1-x)^n)/sqrt(3)+.5) where x = (sqrt(3)+1)/2.

%C This is analogous to the closed form of the formula for the n-th Fibonacci number. Even before truncation, these numbers are rational and the decimal part always ends in 5. For x=(sqrt(3)+1)/2, a(n)/a(n-1) -> x.

%F The general form of x is (sqrt(r)+1)/2, r=1,2,3..

%F a(n) = floor(b(n)/2^n) where b(n) = 2*A002605(n)+2^(n-1). - _R. J. Mathar_, Sep 10 2016

%o (PARI) g(n,r) = for(m=1,n,print1(fib(m,r)",")) fib(n,r) = x=(sqrt(r)+1)/2;floor((x^n-(1-x)^n)/sqrt(r)+.5)

%K nonn

%O 1,3

%A _Cino Hilliard_, Apr 01 2008

%E a(3) corrected by _R. J. Mathar_, Sep 10 2016

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Last modified June 10 17:16 EDT 2023. Contains 363206 sequences. (Running on oeis4.)