A334904 a(n) is the least integer b such that the fractions (b^0)/p, (b^1)/p, ..., (b^(r-1))/p where p is the n-th prime, produce the A006556(n) distinct cycles.

1, 2, 2, 1, 2, 2, 1, 1, 1, 1, 3, 2, 3, 2, 1, 2, 1, 1, 2, 7, 2, 3, 2, 3, 1, 2, 2, 2, 1, 1, 3, 1, 2, 2, 1, 3, 2, 2, 1, 2, 1, 1, 7, 1, 2, 3, 2, 1, 2, 1, 1, 7, 7, 3, 1, 1, 1, 6, 2, 3, 2, 2, 2, 11, 1, 2, 2, 1, 2, 2, 2, 7, 1, 2, 1, 1, 1, 2, 3, 7, 1, 2, 7, 1, 3, 2, 3, 3, 1, 2, 2, 13

Offset: 1

Author

George Plousos, May 15 2020

Keywords

• nonn
• base

Comments

With the exception of the prime numbers 2 and 5, the values of r mentioned above form the sequence A006556.

Detection of all different cycles of digits in the decimal expansions of 1/p, 2/p, ..., (p-1)/p where p=n-th prime. If for the n-th prime p the number of different cycles of digits is equal to r, then there will be the smallest integer b in the interval 0 < b < p with the following property: The fractions (b^0)/p, (b^1)/p, ..., (b^(r-1))/p will produce r different cycles of digits. In this case the term a(n) of the sequence becomes equal to b.

Examples

			For n=13, prime(13)=41, there are A006556(13)=8 cycles.
With b=3, we get (normally, these fractions should be in the form (b^k mod p)/p):
  frac(3^0 / 41) = 0.02439 (1)
  frac(3^1 / 41) = 0.07317 (2)
  frac(3^2 / 41) = 0.21951 (3)
  frac(3^3 / 41) = 0.65853 (4)
  frac(3^4 / 41) = 0.97560 (5)
  frac(3^5 / 41) = 0.92682 (6)
  frac(3^6 / 41) = 0.78048 (7)
  frac(3^7 / 41) = 0.34146 (8=r)
So a(13) = 3.
		

Crossrefs

Cf. A006556.

Programs

PARI

\\ default(realprecision, 1000)
nbc(p) = (p-1)/znorder(Mod(10, p));
len(p) = znorder(Mod(10, p));
pad(x, sz) = {while(#digits(x) != sz, x*=10); x;}
cmpc(x,y) = {if (x==y, return (0)); my(dx=digits(x), dy=digits(y), v=dx); for (k=1, #dx, v=vector(#v, k, if (k==#v, v[1], v[k+1])); if (v == dy, return (0));); return (1);}
decimals(x, sz) = pad(floor(1.0*10^sz*x), sz);
a(n) = {my(p=prime(n)); if ((p==2), return (1)); if ((p==5), return (2)); my(sz=len(p), nb=nbc(p), m=1); while (#vecsort(vector(f(p), k, decimals((m^(k-1) % p)/p, sz)),cmpc,8) != nb, m++); m;} \\ Michel Marcus, May 29 2020