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A353280 n is a term if n = 0 or n does not divide F(n, k) for all k >= 0, where F(n, k) are the Fibonacci numbers A352744.

%I A353280 #10 Apr 11 2022 12:55:41
%S A353280 0,5,6,10,12,15,18,20,24,25,30,35,36,40,42,45,48,50,54,55,56,60,65,66,
%T A353280 70,72,75,78,80,84,85,90,91,95,96,100,102,105,108,110,112,114,115,120,
%U A353280 125,126,130,132,135,138,140,144,145,150,153,155,156,160,162,165
%N A353280 n is a term if n = 0 or n does not divide F(n, k) for all k >= 0, where F(n, k) are the Fibonacci numbers A352744.
%C A353280 n is a term if 0 is not a term of the sequence A352747(n, .). Since A352747(n, .) is for all n a pure periodic sequence, it is sufficient to require that 0 is not a term of period(A352747(n, .)). Since the length of the period is <= n, the condition can be checked in a finite number of steps.
%C A353280 The multiples of 5 and 6 (A093509) are a subsequence. The terms not of this form start 56, 91, 112, ..., and are in A353281.
%e A353280 period(A352747(6, .)) = (5, 1, 3) is zero-free, therefore 6 is a term of a.
%e A353280 period(A352747(7, .)) = (1, 0, 6, 5, 4, 3, 2), thus 7 is not a term of a.
%p A353280 f := n -> combinat:-fibonacci(n): F := (n, k) -> (n-1)*f(k) + f(k+1):
%p A353280 df := n -> denom(f(n)/n) - 1: period := n -> [seq(modp(F(k,n), n), k = 0..df(n))]:
%p A353280 isA353280 := n -> n = 0 or not member(0, period(n)):
%p A353280 select(isA353280, [$(0..166)]);
%o A353280 (SageMath)
%o A353280 def F(n, k): return (n - 1)*fibonacci(k) + fibonacci(k + 1)
%o A353280 def df(n): return denominator(fibonacci(n) / n)
%o A353280 def period(n): return (Integer(n).divides(F(k, n)) for k in range(df(n)))
%o A353280 def isA353280(n): return n == 0 or not any([k == True for k in period(n)])
%o A353280 def A353280List(upto): return [n for n in range(upto + 1) if isA353280(n)]
%o A353280 print(A353280List(165))
%Y A353280 a = A093509 union A353281.
%Y A353280 Cf. A352744, A352747.
%K A353280 nonn
%O A353280 1,2
%A A353280 _Peter Luschny_, Apr 09 2022