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A245024
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Even numbers n for which lpf(n-1) < lpf(n-3), where lpf = least prime factor.
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13
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10, 16, 22, 26, 28, 34, 40, 46, 50, 52, 56, 58, 64, 70, 76, 82, 86, 88, 92, 94, 100, 106, 112, 116, 118, 124, 130, 134, 136, 142, 146, 148, 154, 160, 166, 170, 172, 176, 178, 184, 190, 196, 202, 206, 208, 214, 220, 226, 232, 236, 238, 244, 250, 254, 256, 260
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OFFSET
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1,1
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COMMENTS
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By the definition, either a(n)==1 (mod 3) or, for every pair of primes (p,q), p>q>=3, a(n)==1 (mod p) and a(n) not==3 (mod q).
Conjecture: All differences are 2,4 or 6 such that no two consecutive terms 2 (...,2,2,...), no two consecutive terms 4, while consecutive terms 6 occur 1,2,3 or 4 times; also consecutive pairs of terms 4,2 appear 1,2,3 or 4 times.
The first comment is wrong as stated. This would fix it: for every pair of primes (p,q), p>q>=3, if a(n)==1 (mod p) then a(n) not==3 (mod q). Divisibility by 3 means 6m+4 is in the sequence for all m>0, and 6m never is, while 6m+2 is undetermined. Divisibility by 5 means 30m+26 is always in the sequence, and 30m+8 never is. This proves the above conjecture. - Jens Kruse Andersen, Jul 13 2014
Note that the sequence {a(n)-3} contains all odd primes, except for lesser primes in twin primes pairs (A001359). Other terms of {a(n)-3} are 25,49,55,85,91,... - Vladimir Shevelev, Jul 15 2014
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LINKS
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MAPLE
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lpf:= n -> min(numtheory:-factorset(n)):
select(n -> lpf(n-1) < lpf(n-3), [seq(2*k, k=3..1000)]); # Robert Israel, Jul 15 2014
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MATHEMATICA
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lpf[n_] := FactorInteger[n][[1, 1]];
Reap[For[n = 6, n <= 300, n += 2, If[lpf[n-1] < lpf[n-3], Sow[n]]]][[2, 1]] (* Jean-François Alcover, Feb 25 2019 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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