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%I #32 Jul 15 2014 20:59:17
%S 1,1,1,3,3,3,4,7,3,3,3,3,3,3,3,3,3,9,9,9,9,9,9,18,14,6,6,6,6,6,6,6,6,
%T 6,9,9,12,19,19,19,19,19,19,19,19,5,5,5,5,5,5,4,4,4,4,4,4,4,4,4,4,4,4,
%U 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,6
%N Number of different prime divisors >= prime(n) of sums of two consecutive terms of sequence {f_n(k)} defined in A224523.
%C a(n) shows that it is sufficient to choose a(n) primes >= prime(n) such that Fibonacci-like sequence without multiples of these primes is periodic (see comment in A078414).
%H Peter J. C. Moses, <a href="/A219328/b219328.txt">Table of n, a(n) for n = 1..4000</a>
%e 1) In case n=4, the sequence {f_4(k)} has period {1,1,2,3,5,8,1,9,10}. We see that only sums of consecutive terms 5+8=13, 9+10=19, 10+1=11 have divisors >= prime(4)=7. Thus {f_4(k)} is the Fibonacci-like sequence without multiples of 11,13,19. So a(4)=3.
%e 2) In cases 52 <= n <= 120, prime(n) >= prime(52) = 239, every sequence {f_n(k)} has period {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1, 988, 989, 3, 992, 995, 1, 996}. It is Fibonacci-like sequence without multiples of 659, 997, 1597, or 1987. Since 659 = prime(120), then in the considered interval every a(n)=4.
%Y Cf. A078414, A078412, A214684, A219255, A224382, A224523.
%K nonn
%O 1,4
%A _Vladimir Shevelev_, Apr 11 2013
%E Corrections and terms beginning a(37) were calculated by _Peter J. C. Moses_, Apr 19 2013
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