A320775 a(n) is the least exponent k greater than 1 such that prime(n)^k starts and ends in prime(n).

21, 41, 24, 33, 171, 361, 461, 471, 1281, 1091, 231, 221, 236, 61, 861, 2761, 241, 546, 3261, 1991, 6081, 421, 9541, 5731, 4461, 1621, 21501, 10381, 5051, 1301, 16301, 30051, 18601, 13601, 3171, 8991, 7561, 3201, 33501, 8701, 17351, 5601, 13551, 901, 10301, 871

Offset: 1

Author

Paolo P. Lava, Dec 03 2018

Keywords

• nonn
• base

Comments

a(n) always exists. Let p be a prime other than 2 or 5, and m its length in base 10. Let r be the multiplicative order of p mod 10^m. Then p^k ends in p if and only if k-1 is a multiple of r. p^(j*r+1) starts with p if and only if for some integer s, s + log_10(p)) <= (j*r+1)*log_10(p) < s + frac(log_10(p+1)). This is true for some j because r*log_10(p) is irrational and the fractional parts of the multiples of an irrational number are dense in [0,1]. - Robert Israel, Dec 12 2018

If all integers are considered instead of only primes, not all of them can satisfy the requirement. For instance see A075823 for two digits numbers.

Record values beyond 10^5 are: a(51) = 138801, a(74) = 193701, a(88) = 1766101. Also, a(98) = 282076 and a(100) = 438501 would be record values if not preceded by a(88). - M. F. Hasler, Dec 14 2018

Examples

			2^21 = 2097152 and 21 is the least exponent;
3^41 = 36472996377170786403 and 41 is the least exponent.
		

Links

• M. F. Hasler, Table of n, a(n) for n = 1..100
• Carlos Rivera, Puzzle 934. The prime 7499, The Prime Puzzles and Problems Connection.
• Carlos Rivera, Conjecture 81, The Prime Puzzles and Problems Connection.

Crossrefs

Cf. A000040, A075823, A182944.

Programs

Maple

f:= proc(n) local p, k,m,q,r;
  p:= ithprime(n);
  m:= ilog10(p)+1;
  q:= numtheory:-order(p,10^m);
  for k from q+1 by q do
    r:= p^k;
    if p = floor(r/10^(ilog10(r)+1-m))
      then return k
    fi;
  od
end proc:
f(1):= 21: f(3):= 24:
map(f, [$1..50]); # Robert Israel, Dec 12 2018

Mathematica

a[p_] := Module[{d=IntegerDigits[p]}, nd=Length[d];k=2; While[IntegerDigits[p^k][[1;;nd]] != d || IntegerDigits[p^k][[-nd;;-1]] != d, k++]; k]; a/@Prime@Range@10 (* Amiram Eldar, Dec 10 2018 *)

PARI

isokd(d, dpk) = {for (i=1, #d, if (dpk[i] != d[i], return (0));); return (1);}
isok(p, k) = {my(dpk=digits(p^k), d = digits(p)); if (!isokd(d, dpk), return (0)); isokd(Vecrev(d), Vecrev(dpk));}
a(n) = {my(k=2, p = prime(n)); while (!isok(p, k), k++); k;} \\ Michel Marcus, Dec 10 2018

PARI

apply( {A320775(n, d=logint(n=prime(n), 10)+1, K=if(n>5||n==3,znorder(Mod(n, 10^d)),n+18), f(x)=x\10^(logint(x, 10)+1-d))=forstep(k=1+K,oo,K, n==f(n^k)&&return(k))}, [1..20]) \\ Define A320775 & test it via apply(). - M. F. Hasler, Dec 10 2018