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Search: id:A164901
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| A164901 |
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a(1)=1, a(2) = 2. For n >=3, a(n) = the smallest integer > a(n-1) that is coprime to every sum of any two distinct earlier terms of this sequence. |
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6
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| 1, 2, 4, 7, 13, 19, 29, 37, 43, 53, 59, 67, 73, 79, 89, 97, 103, 109, 127, 137, 149, 157, 163, 173, 179, 191, 197, 211, 223, 229, 239, 251, 257, 263, 269, 277, 283, 293, 307, 313, 331, 337, 347, 353, 359, 367, 373, 379, 389, 397, 409, 419, 431, 439, 449, 457
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Except for 1 & 4, all the rest of the terms are prime. They have a density of about 4/5 of the primes for the first 2500 terms. Primes not a member of this sequence: 3, 5, 11, 17, 23, 31, 41, 47, 61, 71, 83, 101, 107, 113, 131, 139, 151, 167, 181, 193, 199, ..., . [From Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 31 2009]
Conjecture: No two consecutive primes are absent from this sequence (other than 3 and 5). (Equivalently, if p < q < r are consecutive primes and q > 5 is not a term, then p and r are terms.). See A164980. [From Rick L. Shepherd (rshepherd2(AT)hotmail.com), Sep 03 2009]
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LINKS
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Robert G. Wilson v, Table of n, a(n) for n = 1..2500 . [From Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 31 2009]
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EXAMPLE
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The first 4 terms are 1,2,4,7. The sums of every pair of distinct terms are: 1+2=3, 1+4=5, 2+4=6, 7+1=8, 7+2=9, and 7+4=11. So, we are looking for the smallest integer > 7 that is coprime to 3, 5, 6, 8, 9, and 11. This number, which is a(5), is 13.
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MATHEMATICA
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a[1] = 1; a[2] = 2; a[n_] := a[n] = Block[{k = a[n - 1] + 1, t = Plus @@@ Subsets[Array[a, n - 1], {2}]}, While[ Union@ GCD[t, k] != {1}, k++ ]; k]; Array[a, 59] [From Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 31 2009]
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CROSSREFS
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A164902, A164903
Cf. A164980. [From Rick L. Shepherd (rshepherd2(AT)hotmail.com), Sep 03 2009]
Sequence in context: A163522 A002466 A162842 this_sequence A109853 A127977 A112997
Adjacent sequences: A164898 A164899 A164900 this_sequence A164902 A164903 A164904
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), Aug 30 2009
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EXTENSIONS
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a(12) & onward from Robert G. Wilson v (rgwv(AT)rgwv.com), Aug 31 2009
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