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Least prime p_k such that (p_k)^n has p_{k+1} as substring.
1

%I #14 Feb 09 2018 11:37:49

%S 23,11,37,2,7,5,3,41,3,13,3,3,2,2,2,2,5,5,5,3,2,2,3,2,3,2,2,2,2,2,3,2,

%T 3,2,2,2,3,2,3,2,2,2,17,2,2,2,3,2,3,2,2,3,2,2,2,3,2,2,3,2,2,2,2,2,2,2,

%U 2,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,2

%N Least prime p_k such that (p_k)^n has p_{k+1} as substring.

%C It appears that a(n) = 2 for n>153. In other words, for n>153, 3 is always a substring of 2^n. Is there any proof? See A035058.

%e 23^2 = 529 and 29 is the prime after 23.

%e 11^3 = 1331 and 13 is the prime after 11.

%e 37^4 = 1874161 and 41 is the prime after 37.

%p P:=proc(q) local a,b,h,k,n,ok; for h from 2 to q do ok:=1; for n from 1 to q do

%p if ok=1 then a:=ithprime(n); b:=nextprime(a); for k from 1 to ilog10(a^h)-ilog10(b)+1 do

%p if b=trunc(a^h/10^(k-1)) mod 10^(ilog10(b)+1) then print(a); ok:=0; break;

%p fi; od; fi; od; od; end: P(10^6);

%Y Cf. A035058, A052073, A052075, A294088.

%K nonn,easy,base

%O 2,1

%A _Paolo P. Lava_, Feb 09 2018