A338946 Lengths of Cunningham chains of the second kind that are sorted by first prime in the chain.
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, 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, 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
Offset: 1
Keywords
Examples
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. 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. 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.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Chris K. Caldwell, Cunningham Chain (PrimePages, Prime Glossary).
- Wikipedia, Cunningham chain.
Programs
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Mathematica
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]
Comments