OFFSET
1,2
COMMENTS
Let S be the prime signature that occurs most frequently among the numbers in the interval I_n = [2^(n-1), 2^n - 1]; then a(n) is the smallest number in I_n whose prime signature is S.
For each n in 1..34, there is only one prime signature in I_n that occurs with the maximum frequency.
Conjecture: for all n >= 1, there is only one prime signature in I_n that occurs with the maximum frequency.
For each n in 1..34, the most common prime signature in I_n is squarefree: 1 for n=1, p for n=2,3,5; p*q for n=4,6..19; p*q*r for 20..34. From limited sampling of portions of the interval I_n for a number of values of n > 34, it appears that p*q*r continues to be the most common prime signature among n-bit numbers for values of n up to roughly 60; for values larger than about 60, p*q*r*s becomes more common than p*q*r.
Is the most common prime signature in I_n squarefree for every n?
EXAMPLE
For n=1, the interval I_1 consists only of the single integer 1, so a(1)=1. (We can disregard 0 even though it is a 1-bit number because 0 has no prime signature.)
For n=2, the integers in I_2 = [2, 3] have the same prime signature (both are primes), so a(2)=1.
For n=3, the integers in I_3 = [4, 7] are 4=2^2, 5 (prime), 6=2*3, and 7 (prime), so their prime signatures are p^2, p, p*q, and p, respectively, where p and q are distinct primes; the prime signature that occurs most frequently is p, which is occurs at 5 and 7, so a(3)=5.
For n=4, I_4 = [8, 15]; the integers are 8=2^3, 9=3^2, 10=2*5, 11 (prime), 12=2^2*3, 13 (prime), 14=2*7, and 15=3*5, so the prime signatures are p^3, p^2, p*q, p, p^2*q, p, p*q, and p*q, of which p*q occurs most frequently (at 10, 14, and 15), so a(4)=10.
For n=20, of the 524288 integers in the interval [524288, 1048575], 109245 (about 21%) have the prime signature p*q*r; this is the most common prime signature in the interval, and the smallest number having that prime signature in the interval is 524291 = 29*101*179, so a(20)=524291.
PROG
(Python)
from sympy import factorint
from collections import Counter
def a(n):
c, d = Counter(), dict()
for i in range(2**(n-1), 2**n):
t = tuple(sorted(factorint(i).values()))
if t not in d: d[t] = i
c.update([t])
return d[c.most_common(1)[0][0]]
print([a(n) for n in range(1, 19)]) # Michael S. Branicky, Mar 27 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Mar 26 2021
EXTENSIONS
a(26)-a(29) from Jinyuan Wang, Mar 27 2021
a(30)-a(34) from Jon E. Schoenfield, Mar 28 2021
STATUS
approved