

A219328


Number of different prime divisors >= prime(n) of sums of two consecutive terms of sequence {f_n(k)} defined in A224523.


2



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, 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, 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
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OFFSET

1,4


COMMENTS

a(n) shows that it is sufficient to choose a(n) primes >= prime(n) such that Fibonaccilike sequence without multiples of these primes is periodic (see comment in A078414).


LINKS

Peter J. C. Moses, Table of n, a(n) for n = 1..4000


EXAMPLE

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 Fibonaccilike sequence without multiples of 11,13,19. So a(4)=3.
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 Fibonaccilike sequence without multiples of 659, 997, 1597, or 1987. Since 659 = prime(120), then in the considered interval every a(n)=4.


CROSSREFS

Cf. A078414, A078412, A214684, A219255, A224382, A224523.
Sequence in context: A259897 A091849 A035567 * A108688 A266688 A285286
Adjacent sequences: A219325 A219326 A219327 * A219329 A219330 A219331


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Apr 11 2013


EXTENSIONS

Corrections and terms beginning a(37) were calculated by Peter J. C. Moses, Apr 19 2013


STATUS

approved



