%I #7 Nov 17 2020 19:29:12
%S 3,2,1,1,3,1,1,2,1,1,1,1,1,1,1,3,1,1,2,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,
%T 1,1,1,2,2,1,1,2,1,1,1,1,1,1,1,2,1,1,1,2,1,1,3,2,1,1,1,1,2,1,2,1,1,1,
%U 1,1,1,1,3,1,1,1,1,1,1,1,1,3,1,1,1,1,2,1,1,1,1,2,1,1,1,2,2,1,3,1,1,1,1,1,1
%N Lengths of Cunningham chains of the second kind that are sorted by first prime in the chain.
%C Row lengths of A338944.
%H Michael De Vlieger, <a href="/A338946/b338946.txt">Table of n, a(n) for n = 1..10000</a>
%H Chris K. Caldwell, <a href="https://primes.utm.edu/glossary/page.php?sort=CunninghamChain">Cunningham Chain</a> (PrimePages, Prime Glossary).
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Cunningham_chain">Cunningham chain</a>.
%e We start with p = 2. Since 2(2) - 1 = 3 is prime, and further 2(3) - 1 = 5 is prime, but 2(5) - 1 is composite, we have chain length 3, so a(1) = 3.
%e p = 7 is the smallest prime that hasn't appeared in a chain thus far; since 2(7) - 1 = 13 is prime but 2(13) - 1 = 25 is composite, we have a chain of length 2, so a(2) = 2.
%e p = 11 is the smallest prime that hasn't appeared in a chain; 2(11) - 1 = 21 is composite, so we have a singleton chain, thus a(3) = 1, etc.
%t Block[{a = {2}, b = {}, j = 0, k, p}, Do[k = Length@ b + 1; If[PrimeQ@ a[[-1]], AppendTo[a, 2 a[[-1]] - 1]; j++, While[! FreeQ[a, Set[p, Prime[k]]], k++]; AppendTo[b, j]; Set[j, 0]; Set[a, Append[a[[1 ;; -2]], p]]], 10^3}; b]
%Y Cf. A075712, A338944, A338945.
%K nonn
%O 1,1
%A _Michael De Vlieger_, Nov 17 2020