A117608 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).
23, 223, 227, 337, 353, 557, 577, 773, 22573, 23327, 25253, 27527, 32233, 33353, 33377, 35353, 35537, 53777, 57557, 75577, 77377, 2222333, 2222533, 2222537, 2227727, 2233337, 2233757, 2235353, 2235557, 2237773, 2277553, 2332333, 2525557, 2572777, 3333773, 3335537, 3335737
Offset: 1
Examples
a(4)=337 since 337, 373 and 733 all have the same sorted list of digits [3,3,7].
Links
- David A. Corneth, Table of n, a(n) for n = 1..10871 (terms <= 10^43)
- David A. Corneth, PARI program
Programs
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Maple
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;
Comments