

A261309


a(n+1) = abs(a(n)  gcd(a(n), 9n+8)), u(1) = 1.


2



1, 0, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 269, 268, 267, 266, 265, 264, 263, 262, 261, 260, 259, 258, 257, 256, 255, 254, 253, 252, 251, 250, 249, 248, 247, 246, 245, 244, 243, 242, 241, 240, 239, 238, 237, 236, 235, 234, 233, 232, 231
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OFFSET

1,3


COMMENTS

It is conjectured that for all n > 2, u(n) = 0 implies that u(n+1) = 9n+8 is prime, cf. A186261. (This is the sequence {u(n)} mentioned there.)


LINKS

Table of n, a(n) for n=1..68.


EXAMPLE

a(2) = a(1)  gcd(a(1),9+8) = 1  1 = 0.
a(3) = a(2)  gcd(a(2),9*2+8) = gcd(0,26) = 26.
a(3+26) = a(29) = 0 and a(29+1) = gcd(0,9*29+8) = 269 is prime.


PROG

(PARI) print1(a=1); for(n=1, 99, print1(", ", a=abs(agcd(a, 9*n+8))))


CROSSREFS

Cf. A261301  A261310, A186253  A186263, A106108.
Sequence in context: A220087 A022982 A023468 * A122163 A109341 A010865
Adjacent sequences: A261306 A261307 A261308 * A261310 A261311 A261312


KEYWORD

nonn


AUTHOR

M. F. Hasler, Aug 14 2015


STATUS

approved



