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%I #14 May 27 2023 10:33:27
%S 23,223,227,337,353,557,577,773,22573,23327,25253,27527,32233,33353,
%T 33377,35353,35537,53777,57557,75577,77377,2222333,2222533,2222537,
%U 2227727,2233337,2233757,2235353,2235557,2237773,2277553,2332333,2525557,2572777,3333773,3335537,3335737
%N Let p be an element of A110028. Let L(p) be the sorted list of digits of p and let LL be the set of all L(p) with duplicates removed and ordered lexicographically. Then a(n) is the first element of A110028 such that L(a(n))=LL(n).
%C Terms have at least one 3 or at least one 7. - _David A. Corneth_, May 26 2023
%H David A. Corneth, <a href="/A117608/b117608.txt">Table of n, a(n) for n = 1..10871</a> (terms <= 10^43)
%H David A. Corneth, <a href="/A117608/a117608.gp.txt">PARI program</a>
%e a(4)=337 since 337, 373 and 733 all have the same sorted list of digits [3,3,7].
%p a:=proc(b,n) local nn: nn:=convert(n, base, b): if isprime(n) and isprime(nops(nn)) and andmap(isprime,nn) and isprime(convert(nn,`+`)) then n else fi end: L:=[seq(a(10,k), k=1..10^5)]; U:=[]: for z to 1 do A:=L; for x in L do l:=sort(convert(x,base,10)); m:=[selectremove(proc(z) sort(convert(z,base,10))=l end, A)]; if not m[1]=[] then U:=[op(U),min(op(m[1]))]; fi; if m[2]=[] then break else A:=m[2]; fi od od; U;
%Y Cf. A000040, A062088, A110028.
%K nonn,easy,base
%O 1,1
%A _Walter Kehowski_, Apr 06 2006
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