OFFSET
1,1
COMMENTS
Someone called James Davis found that 13532385396179 = 13 * 53^2 * 3853 * 96179, showing that a composite number can be equal the concatenation of the primes and exponents in its canonical prime factorization. In general, if a composite number is equal the concatenation in base b of the primes and exponents in its prime factorization, then let's call it a Davis number to base b.
Conjecture: a composite number can be a Davis number to at most one base.
Let (d_1,...,d_r) be the ordered tuple of prime factors and exponents > 1 in the prime factorization of n (e.g., 4617 = 3^5 * 19 -> (3,5,19), 13532385396179 = 13 * 53^2 * 3853 * 96179 -> (13,53,2,3853,96179)), then n is a Davis number to base b if and only if n = d_1*b^{s_1} + ... + d_{r-1}*b^{s_{r-1}} + d_r, where s_i = (Sum_{j=i+1..r} floor(log_b(d_j))) + r-i. In particular, we must have b dividing n - d_r.
Suppose that p^e is a Davis number to some base b, with e >= 2. We have p^e = p*b^(floor(log_b(e))+1) + e in base b, hence e is divisible by p. If b <= e, then we have p^e <= p*b^(log_b(e)+1) + e <= p*e^2 + e, which is impossible, and so we must have b > e. Conversely, when e is divisible by p and p^e > 4, p^e is a Davis number to base (p^e-e)/p > e.
No term can be squarefree: for primes p_1 < ... < p_r, the concatenation of p_1, ..., p_r in base b is p_1 * b^(Sum_{i=2..r} (floor(log_b(p_i))+1)) + ... >= p_1*...*p_r + ... > p_1*...*p_r.
Here are some examples that are near-miss of being Davis numbers to base 10. Each is equal to the concatenation of the factors and exponents in its generalized factorization (we call n = (q_1)^(e_1) * ... * (q_k)^(e_k) a generalized factorization of n, where 1 < q_1 < ... < q_k, (q_1,...,q_k) are pairwise coprime but are not necessarily primes, and exponents 1 are omitted; the number of such factorizations is A327399(n)):
2592 = 2^5 * 9^2;
34425 = 3^4 * 425;
312325 = 31^2 * 325;
492205 = 49^2 * 205;
36233196159122085048010973936921313644799483579440006455257 = 3^6 * 2331961591220850480109739369 * 21313644799483579440006455257. (Note that in the last four examples, we can add as many trailing zeros as we want).
EXAMPLE
In base 6: 24 = 2^4 (in decimal: 16 = 2^4);
In base 8: 33 = 3^3 (in decimal: 27 = 3^3);
In base 29: 26 = 2^6 (in decimal: 64 = 2^6);
In base 124: 28 = 2^8 (in decimal: 256 = 2^8);
In base 241: 36 = 3^6 (in decimal: 729 = 3^6);
In base 507: 2A = 2^A (in decimal: 1024 = 2^10);
In base 624: 55 = 5^5 (in decimal: 3125 = 5^5);
In base 2042: 2C = 2^C (in decimal: 4096 = 2^12);
In base 11: 3518 = 3^5 * 18 (in decimal: 4617 = 3^5 * 19).
See A384540 for more nontrivial examples.
PROG
(PARI) F(n, b) = my(f=factor(n), d=[]); for(i=1, #f~, d=concat(d, digits(f[i, 1], b)); if(f[i, 2]>1, d=concat(d, digits(f[i, 2], b)))); fromdigits(d, b)
isA384537(n) = {
if(issquarefree(n), return(0)); my(f=factor(n), dr);
if(#f~ == 1, return(n > 4 && f[1, 2] % f[1, 1] == 0));
dr = if(f[#f~, 2] == 1, f[#f~, 1], f[#f~, 2]);
fordiv(n - dr, b, if(b>=2 && F(n, b)==n, return(b))); return(0);
} \\ returns the (smallest) base to which n is a Davis number whenever possible
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Jianing Song, Jun 02 2025
STATUS
approved