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%I #14 Nov 12 2019 09:24:49
%S 2,3,5,7,9,9,9,9,9,9,9,8,9,8,9,9,9,7,9,9,9,9,9,9,9,9,9,9,8,9,6,9,9,9,
%T 9,9,9,9,9,9,9,9,9,9,9,8,9,9,9,9,9,9,9,9,9,9,9,8,7,9,9,9,9,9,9,9,9,9,
%U 9,9,9,9,9,9,8,9,9,9,9,9,9,9,9,9,9,9,9,8,9,9
%N Base at which the first composite occurs in the sequence p=prime(n), p1=base(p,9), p2=base(p1,8), p3=base(p2,7),..., where base(N,b) means N written in base b and read in base 10; a(n)=0 if p1,...,p8 are all prime.
%C Sequence suggested by a question asked in the "primenumbers" group, cf. link.
%C The first 4 occurs for p=26571169, at index n=1657999.
%C A variant of this sequence would have a(n)=10 for nonprime n, and a(prime(n))=A193889(n).
%H J. Brennen, in reply to J. Merickel, <a href="http://groups.yahoo.com/group/primenumbers/message/22932">Problem that should be solvable requiring scientific approach</a> on yahoo group "primenumbers", Aug 07 2011.
%H James Merickel, Jack Brennen and others, <a href="/A193888/a193888.txt">Problem that should be solvable requiring scientific approach</a>, digest of 11 messages in primenumbers Yahoo group, Aug 6 - Aug 7, 2011.
%e a(1)=2 because for p=prime(1)=2, we have p=p1=...=p7=2 all prime, but p8=base(2,2)=10 is composite.
%e a(5)=9 because for p=prime(5)=11, we have already p1=base(11,9)=12 composite.
%e a(18)=7 because for p=prime(18)=61 we have p1=base(61,9)=67 and p2=base(67,8)=103 both prime, but p3=base(103,7)=205 composite.
%e a(1657999)=4 because for p=prime(1657999)=26571169 we have p1=base(p,9)=54887711, p2=base(p1,8)=321302437, p3=base(p2,7)=10651011541, p4=base(p3,6)=4520520050341 and p5=base(p4,5)=1043031011113102331 all prime, but p6=base(p5,4)=321321211302312223013032233323 composite.
%o (PARI) base(n,b)={my(a=n%b,t=1);while(0<n\=b,a+=n%b*t*=10);a}
%o A193889(N)={ N=prime(N);forstep(b=9,2,-1,isprime(N=base(N,b)))||return(b))}
%Y Cf. A193888.
%K nonn,base
%O 1,1
%A _M. F. Hasler_, Aug 07 2011
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