A096982 Define a(1)=0. Then define b(1)=1, b(2)=1 and for k > 2, b(k) = (b(k-2) + b(k-1)) (mod n). This gives for each n a cyclic repetitive sequence; a(n) is the least k > 1 for the first 1 of the sequence 1,1,....

0, 4, 9, 7, 21, 25, 17, 13, 25, 61, 11, 25, 29, 49, 41, 25, 37, 25, 19, 61, 17, 31, 49, 25, 101, 85, 73, 49, 15, 121, 31, 49, 41, 37, 81, 25, 77, 19, 57, 61, 41, 49, 89, 31, 121, 49, 33, 25, 113, 301, 73, 85, 109, 73, 21, 49, 73, 43, 59, 121, 61, 31, 49, 97, 141, 121, 137

Offset

1,2

Comments

This sequence gives a great number of primes or primes squared: For n = P(i)# = 2, 6, 30, 210, 2310, 30030, 510510, 9699690, 223092870 we get a(P(i)#) = 2^2, 5^2, 11^2, 241, 241, 41^2, 71^2, 71^2, 71^2.

Links

Table of n, a(n) for n=1..67.

Formula

a(n) = A001175(n) + 1 for n > 1. - T. D. Noe, Jan 09 2009

Example

b(1)=1, b(2)=1; for n=2 the sequence is 1,1,0,1,1,0,1,1,0,... so a(2)=4 as for k=4 you recover 1,1,0.
For n=3 the sequence is 1,1,2,0,2,2,1,0,1,1,2,0,... so a(3)=9 as for k=9 you recover 1,1,2.

Crossrefs

Sequence in context: A353071 A365906 A256174 * A212877 A205297 A159643

Adjacent sequences: A096979 A096980 A096981 * A096983 A096984 A096985

Keyword

nonn

Author

Pierre CAMI, Aug 19 2004

Status

approved