If the binary representations of the first several primes are reversed, the resulting number is also irreducible (i.e., either a prime or 1). This was pointed out by John Rickert, who cited the following reflective pairs of primes reverse prime prime (or unit) 2 10 01 1 (or 010 to 010) 3 11 11 3 5 101 101 5 7 111 111 7 11 1011 1101 13 13 1101 1011 11 17 10001 10001 17 The first prime whose binary reversal has a proper factorization is 19 = 10011, which gives 11001=25. Still, of the primes less than 2048, more than half of them have prime reflective partners. I suspect the proportion of such primes becomes arbitrarily small as we go to larger and larger numbers. To illustrate, here's a little table showing the number of such primes less than x for various values of x: number of primes < x total number that give primes under x of primes < x binary reversal ratio -------- ------------- --------------------- ------- 100 25 21 0.8400 1000 168 102 0.6071 10000 1229 509 0.4141 100000 9592 3054 0.3184 1000000 78498 20054 0.2555 10000000 664579 141772 0.2133 I would expect that the fraction of primes whose reflective pairs are also primes is roughly what would be expected based on a random selection of an odd number with the correct number of binary digits for each prime. Thus it becomes increasingly less likely as the density of primes (slowly) decreases. Also, there are presumably infinitely many such prime pairs, although I don't know how to prove it. One interesting approach to this problem might be based on interpreting the binary representations as polynomials with reciprocal roots, as discussed in The Fundamental Theorem For Palindromic Polynomials. By the way, here's some C code for counting reflective pairs: #include <stdio.h> #include <stdlib.h> #include <math.h> unsigned long ic,d,nmax, nn, n, i, imax, mm, kk, aa[30]; unsigned long ictot,ptot; main () { nmax=1000; for(nn=3;nn<=nmax;nn=nn+2) { ic=0;for(d=3;d<sqrt(nn+1);d=d+2)if(nn%d==0)ic=1; if(ic==0) { for(i=0;i<31;i++)aa[i]=0; n=nn;i=0; while(n>0){aa[i]=n%2;n=(n-aa[i])/2;i++;}imax=i-1; mm=0;for(i=0;i<=imax;i++)mm=2*mm+aa[i]; ic=0;for(d=3;d<sqrt(mm+1);d=d+2)if(mm%d==0)ic=1; ictot=ictot+ic; ptot=ptot+1; printf(" %lu %lu %lu %lu %lu \n",ptot,nn,mm,ic,ictot); } } } On a slightly related point, we can also consider "complete cyclic sets of primes", i.e., primes whose base-b expansions are primes under all the rotations of their digits. Obviously there are none in the base 2 other than repunits, i.e., primes of the form 2^p - 1. However, for any higher base there exist complete cyclical prime sets. Focusing on the base 10, some trivial examples of two-digit cyclic pairs of primes are 11 13 37 17 79 11 31 73 71 97 For slightly less trivial examples, we have the three-digit cyclical prime sets 113 197 199 337 311 719 919 733 131 971 991 373 Among four-digit decimal numbers the only cyclic sets of primes are 1193 3779 3119 9377 9311 7937 1931 7793 Likewise among the five-digit decimal numbers we have only the sets 11939 19937 91193 71993 39119 37199 93911 93719 19391 99371 Finally, among six-digit decimal numbers the only cyclic sets of primes are 193939 199933 919393 319993 391939 331999 939193 933199 393919 993319 939391 999331 There are no complete cyclic sets of decimal primes with 7 or 8 digits. The next known cycle set is the repunit with 19 digits, then the repunits with 23, 317, 1031 digits, and so on. If we exclude repunits, it is not known if there are any complete cyclic sets of primes in the base 10 with more than 6 digits. Of course, we can also find cyclic sets of primes in other bases. For example, in the base 7 we have the set 11515 51151 15115 51511 15151 Also, whenever a repunit is a prime, it gives a degenerate cyclic set, such as the number 1111111 in the base 3 (which equals the prime 1093 in decimal).

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