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%I #12 Oct 13 2023 06:48:37
%S 3,2,2,2,2,2,4,2,3,3,2,2,3,3,4,2,5,4,3,2,3,4,2,2,2,2,3,3,2,5,2,2,3,2,
%T 3,2,2,3,8,2,4,2,2,2,2,2,3,3,3,6,3,4,2,2,3,3,2,2,4,2,2,3,6,2,3,2,2,2,
%U 5,2,2,2,2,2,2,2,5,3,2,4,3,5,3,3,2,8,2,2,2,2,3,8,4,3,3,3,4,2,3,8,2,3,3,5,3
%N Run lengths of 2 or larger for consecutive prime numbers in A006577.
%C This sequence is given only for n <=5000 with max(s(n)) = 10. But we can find long sequences of primes, for example,length(s(12956))= 55, and corresponding to A006577(282984 + k), k = 0,1,...,54. We obtain a sequence of 55 consecutive primes numbers given in the example below.
%e a(1) = 3 represents the run (7, 2, 5).
%e a(2) = 2 represents the run (3, 19).
%e a(3)=2 represents the run (17, 17).
%e a(7) = 4 represents the run (19, 19, 107, 107).
%e a(12956) = 55 represents the run (83, 251, 83, 251, 127, 127, 127, 251, 83, 83, 83, 83, 83, 83, 83, 83, 83, 251, 83, 83, 83, 83, 83, 83, 101, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 251, 251, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83, 83)
%p nn:=2000:T:=array(1..nn):for n from 1 to nn do: m:=n:for p from 0 to 1000 while (m<>1) do: if irem(m,2)=1 then m:=3*m+1:else m:=m/2:fi:od:T[n]:=p:od:ii:=1:for i from 1 to nn do:if type(T[i],prime)=true and type(T[i+1],prime)=true then ii:=ii+1:else if ii<>1 then printf(`%d, `, ii):ii:=1:else fi:fi:od:
%Y Cf. A174538, A006577, A066906.
%K nonn,less
%O 1,1
%A _Michel Lagneau_, Mar 22 2010
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