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Search: id:A108763
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| A108763 |
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If n-th prime is 8m+1, then a(n) = 8m+3. If n-th prime is 8m+3, then a(n) = 8m+5. If n-th prime is 8m+5, then a(n) = 8m+7. If n-th prime is 8m+7, then a(n) = 8m+1. |
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+0
1
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| 5, 7, 1, 13, 15, 19, 21, 17, 31, 25, 39, 43, 45, 41, 55, 61, 63, 69, 65, 75, 73, 85, 91, 99, 103, 97, 109, 111, 115, 121, 133, 139, 141, 151, 145, 159, 165, 161, 175, 181, 183, 185, 195, 199, 193, 213, 217, 229, 231, 235, 233, 243, 253, 259, 257, 271, 265, 279, 283
(list; graph; listen)
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OFFSET
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2,1
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COMMENT
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a(n) is asymptotically equal to prime(n). There are infinitely many a(n) > a(n) and infinitely many a(n+1) < a(n). There are arbitrarily large gaps between a(n) and a(n+1) for sufficiently large n. If Goldbach's conjecture is true, this sequence contains infinitely many primes. Conjecture: for any whole number k, there exists at least one a(n) with exactly k prime factors (for k = 3, for instance, 8*20 + 3 = 163 which is prime and the sequence includes 8*20 + 5 = 3 x 5 x 11).
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FORMULA
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a(n)= 8*quotient(nth prime, 8) + mod(2 + n-th prime, 8) = 2 + n-th prime unless mod( n-th prime, 8) = 7 then -6 + n-th prime.
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EXAMPLE
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Prime(1) = 2, which is not of the form 8m+1, 8m+3, 8m+5, nor 8m+7, hence this sequence starts with a(2) = 5 because prime(2) = 8*0 + 3, so 8*0+5 = 5.
Prime(11) = 31 = 8*3 + 7, so a(11) = 8*3 + 1 = 25 = 5^2, which happens not to be a prime.
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MATHEMATICA
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Table[p = Prime[n]; If[ Mod[p, 8] == 7, p - 6, p + 2], {n, 2, 61}] (from Robert G. Wilson v (rgwv(AT)rgwv.com), Jun 24 2005)
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CROSSREFS
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Cf. A107323, A108245.
Sequence in context: A023571 A155066 A145737 this_sequence A061415 A087455 A117759
Adjacent sequences: A108760 A108761 A108762 this_sequence A108764 A108765 A108766
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KEYWORD
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easy,nonn
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AUTHOR
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Jonathan Vos Post (jvospost3(AT)gmail.com), Jun 19 2005
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EXTENSIONS
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Edited, corrected and extended by Robert G. Wilson v (rgwv(AT)rgwv.com), Jun 24 2005
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