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a(n) = floor(A*a(n-1) + B*a(n-2) + C)/p^r, where p^r is the highest power of p dividing floor(A*a(n-1) + B*a(n-2) + C), A=1.0001, B=1.0001, C=1, p=2.
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%I #11 Dec 27 2021 23:52:21

%S 1,1,3,5,9,15,25,41,67,109,177,287,465,753,1219,1973,3193,5167,8361,

%T 6765,1891,8657,5275,6967,3061,5015,8077,6547,7313,6931,7123,1757,

%U 8881,665,9547,5107,229,5337,5567,5453,5511,5483,2749,8233,1373,9607

%N a(n) = floor(A*a(n-1) + B*a(n-2) + C)/p^r, where p^r is the highest power of p dividing floor(A*a(n-1) + B*a(n-2) + C), A=1.0001, B=1.0001, C=1, p=2.

%C Becomes a cyclic sequence whose period is 4793. If A=1, B=1, C=0, p=1, a(1)=a(2)=1 then it is the Fibonacci Sequence.

%Y Cf. A053522, A053523, A028948.

%K nonn

%O 1,3

%A _Yasutoshi Kohmoto_

%E a(33) corrected by _Sean A. Irvine_, Dec 27 2021