|
Search: id:A117608
|
|
|
| A117608 |
|
Let p be an element of A110028, the set of all primes that have prime digits, a prime number of digits, and whose digits sum to a prime. 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). |
|
+0
1
|
|
| 23, 223, 227, 337, 353, 557, 577, 773, 22573, 23327, 25253, 27527, 32233, 33353, 33377, 35353, 35537, 53777, 57557, 75577, 77377
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
EXAMPLE
|
a(4)=337 since 337, 373, and 733 all have the same sorted list of digits [3,3,7].
|
|
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;
|
|
CROSSREFS
|
Cf. A000040, A062088, A110028.
Sequence in context: A112747 A027481 A110028 this_sequence A038514 A125386 A091628
Adjacent sequences: A117605 A117606 A117607 this_sequence A117609 A117610 A117611
|
|
KEYWORD
|
base,easy,nonn
|
|
AUTHOR
|
Walter Kehowski (wkehowski(AT)cox.net), Apr 06 2006
|
|
|
Search completed in 0.002 seconds
|
