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%I #15 Jul 19 2021 01:23:12
%S 1,3,8,34,55,144,610,233,12166,2584,4181,68260,46368,75025,3917414,
%T 464656,1346269,16349962
%N a(n) is the smallest number k such that k^2+1 divided by its largest prime factor is equal to F(2*n-1) for n > 0, or 0 if no such k exists, where F(n) is the Fibonacci sequence.
%C a(n) is the smallest number k such that A248516(k) = A001519(n) for n > 0, or 0 if no such k exists, where A001519(n) = F(2*n-1) (bisection of the Fibonacci sequence), with F(n) = A000045(n).
%C We observe that a(2 + 3m) = A001519(1 + 3m) = A000045(1 + 6m) for m = 2, 3, 4, 5. For n = 6, this property no longer works.
%C For k > 0, a(3k - 1) is odd, a(3k) and a(3k+1) are even.
%C We observe that a(n)^2 + 1 is the product of two prime Fibonacci numbers for n = 2, 3, 4, 6, 7.
%C The first 18 terms of the sequence are Fibonacci numbers, except a(9), a(12), a(15), a(16) and a(18).
%C The corresponding sequence b(n) = (a(n)^2+1)/ A001519(n) is 2, 5, 13, 89, 89, 233, 1597, 89, 92681, 1597, 1597, 162593, 28657, 28657, 29842993, 160373, 514229. We observe that a majority of terms of b(n) are prime Fibonacci numbers, except b(9), b(12), b(15) and b(16).
%e a(4) = 34 because 34^2 + 1 = 13*89 = 1157, and 1157/89 = 13 = A248516(34) = A001519(4).
%e A curiosity: a(22) = 1134903170 = F(45) with F(45)^2 + 1 = F(43)*F(47) where F(43) and F(47) are prime Fibonacci numbers.
%p with(numtheory):with(combinat,fibonacci):
%p nn:=100:n0:=20:
%p for n from 1 to n0 do:
%p ii:=0:
%p for m from 1 to 10^10 while(ii=0) do:
%p x:=m^2+1:y:=factorset(x):n1:=nops(y):
%p z:=x/y[n1]:
%p if z = fibonacci(2*n-1)
%p then
%p ii:=1:printf(`%d %d \n`,n,m):
%p else
%p fi:
%p od:
%p od:
%o (PARI) a(n) = {my(k=1, f=fibonacci(2*n-1)); while ((k^2+1)/vecmax(factor(k^2+1)[,1]) != f, k++); k;} \\ _Michel Marcus_, Nov 30 2020
%Y Cf. A000045, A001519, A001906, A002522, A005478, A014442, A245236, A245306, A248516, A281618.
%K nonn,hard
%O 1,2
%A _Michel Lagneau_, Nov 30 2020
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