A224523 -id:A224523 - cp's OEIS frontend

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

Vladimir Shevelev, Apr 11 2013

nonn

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).

			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.

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

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

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

A219255 Numbers n for which prime(n) divides at least one sum of two consecutive terms in sequence {f_m(k)}, defined in A224523.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 8, 10, 17, 23, 24, 25, 34, 36, 37, 45, 51, 120, 124, 219, 231, 244, 251, 2034, 2057, 3121, 4176, 5185, 9492

Offset: 1

Vladimir Shevelev, Apr 11 2013

Places of the last elements of runs of the same terms in A224523 are in the sequence. Whether these sequences coincide one to another?

			7 is in the sequence, since prime(7)=17 and sequence {f_7(k)} is periodic with period {1,1,2,3,5,8,13,21,2,1,3,4,7,11,18}, such that sum of terms 13+21=34 is divisible by 17.

Cf. A078414, A078412, A214684, A219328, A224523.

Rest[Flatten[Position[Differences[A224523], Except[0]]]] (* Peter J. C. Moses, Apr 13 2013 *)

cp's OEIS Frontend

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

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