A219328 Number of different prime divisors >= prime(n) of sums of two consecutive terms of sequence {f_n(k)} defined in A224523.

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

Offset: 1

Author

Vladimir Shevelev, Apr 11 2013

Keywords

• nonn

Comments

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).

Examples

			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.
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.
		

Links

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

Crossrefs

Cf. A078414, A078412, A214684, A219255, A224382, A224523.

Extensions

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