A262988 Number of distinct primes, including n if prime, obtained by cyclically shifting the digits of n.
0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 2, 1, 0, 1, 2, 0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 2, 1, 0, 1, 1, 0, 2, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 2, 0, 2, 1, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0
Offset: 1
Examples
a(1013) = 2, because of the four cyclic permutations of the digits of 1013 (1013, 131, 1310, 3101) two, namely 1013 and 131, are prime and those two primes are distinct.
Links
- Felix Fröhlich, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
f[n_] := Block[{len = IntegerLength@ n, s = {n}}, Do[AppendTo[s, FromDigits@ RotateRight@ IntegerDigits@ s[[k - 1]]], {k, 2, len}]; DeleteDuplicates@ Select[s, PrimeQ]]; Array[ Length@ f@ # &, {87}] (* Michael De Vlieger, Oct 07 2015 *)
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PARI
rot(n) = if(#Str(n)==1, v=vector(1), v=vector(#n-1)); for(i=2, #n, v[i-1]=n[i]); u=vector(#n); for(i=1, #n, u[i]=n[i]); v=concat(v, u[1]); v eva(n) = x=0; for(k=1, #n, x=x+(n[k]*10^(#n-k))); x a(n) = i=0; r=rot(digits(n)); while(r!=digits(n), if(ispseudoprime(eva(r)), i++); r=rot(r)); if(ispseudoprime(eva(r)), i++); i
Comments