A085324 a(n) is the least exponent so that reverse(n^a(n)) is a prime number. a(n)=0 if no such exponent exists, namely when e.g., n = 3k or n = 11k, k > 1.
0, 1, 1, 2, 1, 0, 1, 8, 0, 0, 1, 0, 1, 1, 0, 1, 1, 0, 2, 1, 0, 0, 8, 0, 13, 47, 0, 2, 7, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 2, 2, 0, 5, 0, 0, 22, 15, 0, 6, 1, 0, 3, 10, 0, 0, 143, 0, 88, 12, 0, 4, 2, 0, 4, 8, 0, 39, 83, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 8, 0, 6, 11, 0, 2, 28, 0, 0, 2, 0, 1, 1, 0, 292, 1, 0, 1, 1
Offset: 1
Examples
For n=46, a(46)=22 means that reversion of 46^22 gives a prime: 6100744433653913942689966672393877083.
Links
- Robert Israel, Table of n, a(n) for n = 1..202
Programs
-
Maple
Rev:= proc(n) local L; L:= convert(n,base,10); add(L[-i]*10^(i-1),i=1..nops(L)) end proc: f:= proc(n) local k; if igcd(n,33) <> 1 then return 0 fi; if n mod 10 = 0 then return procname(n/10) fi; for k from 1 do if isprime(Rev(n^k)) then return k fi od: end proc: f(1):= 0: f(3):= 1; f(11):= 1; map(f, [$1..100]); # Robert Israel, Apr 09 2018
-
Mathematica
nd[x_, y_] := 10*x+y; tn[x_] := Fold[nd, 0, x]; bac[x_] := tn[Reverse[IntegerDigits[x]]] t={list without 3k and 11k numbers}; le=Length[t]; Table[f=1; Do[s=bac[Part[t, n]^k]; If[PrimeQ[s]&&Equal[f, 1], Print[{k, Part[t, n], s}]; f=0], {k, 1, 300}], {n, 1, le}]
Formula
a(3k) = a(11k) = 0 for k > 1 because reversion does not make a prime from any of their powers.
Comments