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A261309
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a(n+1) = abs(a(n) - gcd(a(n), 9n+8)), u(1) = 1.
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2
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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
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1,3
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COMMENTS
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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.)
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LINKS
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EXAMPLE
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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.
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PROG
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(PARI) print1(a=1); for(n=1, 99, print1(", ", a=abs(a-gcd(a, 9*n+8))))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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