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%I #13 Jul 16 2021 08:36:36
%S 5,35,235,335,2335,3335,23335,32335,33335,72335,233335,323335,333335,
%T 372335,572335,723335,2333335,2372335,2723335,3233335,3323335,3333335,
%U 3572335,3723335,7233335,7323335,7372335,7572335,22372335,23333335
%N Numbers k with d digits such that all digits of k and the last d+1 digits of k^2 are prime.
%C Any term with d digits is the concatenation of a prime digit and an earlier term (with d-1 digits).
%C The sequence is infinite since it contains subsequences b(n) = (10^n-1)/3+2 = (5,35,335,3335,...), c(n) = 23*10^n+b(n) = (235,2335,23335,...), d(n) = 3233*10^n+b(n), e(n) = 7233*10^n+b(n) etc.
%F For all n, a(n) == (5 mod 10).
%F For a(n) > 5, a(n) == 35 (mod 100).
%F For a(n) > 35, a(n) == 235 or 335 (mod 1000).
%F For a(n) > 335, a(n) == 2335 or 3335 (mod 10^4).
%t Select[Range[5,24000000,5],And@@PrimeQ[IntegerDigits[#]]&& And@@ PrimeQ[ Take[ IntegerDigits[#^2],-(IntegerLength[#]+1)]]&] (* _Harvey P. Dale_, Dec 31 2012 *)
%o (PARI) last=[0]; {for( d=1,8, new=[]; forprime( p=0,9, for( k=1,#last, is_A046034((p*10^(d-1)+last[k])^2%10^(d+1)+20*10^d) & new=concat( new, p*10^(d-1)+last[k]))); print1(last=new,","))} /* for slightly more efficient code see A154779 */
%Y Subsequence of A046034; contains A153025 as a subsequence.
%K base,easy,nice,nonn
%O 1,1
%A _M. F. Hasler_, Jan 23 2009
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