[seqfan] Add n to the composite a(n): the result is a composite too
Eric Angelini
Eric.Angelini at kntv.be
Sun Jan 25 23:12:25 CET 2009
Hello SeqFans,
1-> a(1) = 8
2-> a(n) is composite
3-> a(n)+n is composite
4-> a(n+1) is the smallest possible integer
... this set of rules gives S1 (I think):
S1 = 8 4 6 4 4 4 8 4 6 4 4 4 8 4 6 4 4 4 6 4 ...
n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ...
a(n)+n = 9 6 9 8 9 10 15 12 15 14 15 16 21 18 21 20 21 22 25 24 ...
We could cut the three seq above in vertical slices, each one
showing the first pair of composites having n as difference.
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If we insert the rule 3bis: a(n)<a(n+1), we get (I think):
S2 = 8 10 12 14 15 16 18 20 21 22 24 26 27 28 30 32 33 34 35 ...
n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...
a(n)+n = 9 12 15 18 20 22 25 28 30 32 35 38 40 42 45 48 50 52 54 ...
S2 is not A080257; after 34, 35... S2 has 36, 39, 40, 42, ...
A080257 has 36, 38, 39, 40, 42, ...
... the term 38 is forbidden in S2 because 38 + n = 59, which is prime.
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Why not play accordingly with primes? We have to write new rules:
1-> a(1) = 3
2-> a(n) is prime
3-> a(n)+2n is prime
4-> a(n+1) is the smallest possible integer
... this new set of rules gives S3 (I think):
S3 = 3 3 5 3 3 5 3 3 5 3 7 5 3 3 7 5 3 5 3 3 ...
2n = 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 ...
a(n)+2n = 5 7 11 11 13 17 17 19 23 23 29 29 29 31 37 37 37 41 41 43 ...
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Again, if we add the 3bis rule a(n)<a(n+1), we get (I guess):
S4 = 3 7 11 23 31 41 47 67 71 89 ...
2n = 2 4 6 8 10 12 14 16 18 20 ...
a(n)+2n = 5 11 17 31 41 53 61 83 89 103 ...
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Are those 4*2=8 possible seq worth the OEIS?
Best,
E.
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