A245024 Even numbers n for which lpf(n-1) < lpf(n-3), where lpf = least prime factor.

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

Offset: 1

Author

Vladimir Shevelev, Jul 10 2014

Keywords

• nonn

Comments

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.

Conjecture is verified up to n = 2.5*10^7. - Vladimir Shevelev and Peter J. C. Moses, Jul 11 2014

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

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A020639, A242719, A242720, A243937.

Programs

Maple

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

Mathematica

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 *)

Extensions

More terms from Peter J. C. Moses, Jul 10 2014