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A368231 Lexicographically earliest infinite sequence of distinct positive numbers such that, for n>3, a(n) has a common factor with a(n-1) but not with a(n-2) or n.

%I A368231 #11 Dec 20 2023 08:03:46
%S A368231 1,15,35,77,143,65,30,21,91,221,85,55,33,39,182,133,95,115,69,51,170,
%T A368231 145,203,119,102,45,155,341,154,161,207,57,190,185,407,187,153,63,217,
%U A368231 403,130,205,123,87,319,209,247,299,138,93,589,323,238,259,111,75,70,287,451,253,230,195,377
%N A368231 Lexicographically earliest infinite sequence of distinct positive numbers such that, for n>3, a(n) has a common factor with a(n-1) but not with a(n-2) or n.
%C A368231 This is a variation of the Enots Wolley sequence A336957 and A360519, with an additional restriction that no term a(n) can have a common factor with n. For the sequence to be infinite a(n) must always have a prime factor that is not a factor of a(n-1)*(n+1). See the examples below.
%C A368231 Other than no term being a prime or prime power, see A336957, no term can be an even number with only two distinct prime factors. Clearly no term a(2*k) can be even, so if we assume that a(2*k+1) = 2^n*p^m, with n and m>=1, then a(2*k) must have p as a factor. But as a(2*k+2) must share a factor with a(2*k+1) and cannot have 2 as a factor, it must also have p as a factor. However that is not allowed as a(n) cannot share a factor with a(n-2), so no term can be even with only two distinct prime factors. Therefore the smallest even number is a(7) = 30.
%H A368231 Scott R. Shannon, <a href="/A368231/b368231.txt">Table of n, a(n) for n = 1..10000</a>
%H A368231 Scott R. Shannon, <a href="/A368231/a368231.png">Image of the first 100000 terms</a>. The green line is a(n) = n.
%e A368231 a(2) = 15 as 15 is the smallest number that is not a prime power and does not have 2 as a factor.
%e A368231 a(3) = 35 as a(3) is chosen so it shares a factor with a(2) = 3*5 while not having 3 as a factor; it therefore must be a multiple of 5 while not being a power of 5. The smallest number meeting those criteria is 10, but a(2)*(3+1) = 15*4 = 60, and 10 has no prime factor not in 60, so choosing 10 would mean a(4) would not exist. The next smallest available number is 35.
%e A368231 a(4) = 77 as a(4) must be a multiple of 7 but not a power of 7, not a multiple of 2, 3 or 5, while having a prime factor not in 35*(4+1) = 165. The smallest number satisfying these criteria is 77.
%Y A368231 Cf. A367741, A336957, A360519, A098550, A064413, A027748.
%K A368231 nonn
%O A368231 1,2
%A A368231 _Scott R. Shannon_, Dec 18 2023