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A346729
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Maximum number of divisors among n-bit numbers.
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1
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1, 2, 4, 6, 8, 12, 16, 20, 24, 32, 40, 48, 64, 80, 96, 120, 144, 168, 200, 240, 288, 360, 432, 504, 600, 720, 864, 1008, 1152, 1344, 1600, 1920, 2304, 2688, 3072, 3584, 4096, 4800, 5760, 6720, 7680, 8640, 10080, 11520, 13824, 16128, 18432, 20736, 23040, 27648
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OFFSET
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1,2
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COMMENTS
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a(n) is the maximum value of tau(k)=A000005(k) for k in the interval [2^(n-1), 2^n - 1]. For n >= 3, that smallest k at which tau(k) is maximized in that interval is A036484(n).
No term is repeated: for n >= 1, if k is the number in [2^(n-1), 2^n - 1] at which tau(k) is maximized (i.e., tau(k) = a(n)), then 2k, which will be a number in [2^n, 2^(n+1) - 1], will have more divisors than k has, so a(n+1) >= tau(2k) > tau(k) = a(n).
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LINKS
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EXAMPLE
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There are four 3-bit numbers: 4 = 100_2, 5 = 101_2 = 5, 6 = 110_2, 7 = 111_2. 5 and 7 are both prime, so each has 2 divisors; 4 = 2^2 has 3 divisors (1, 2, and 4), and 6 = 2*3 has 4 divisors (1, 2, 3, and 6). Thus, the maximum number of divisors among 3-bit numbers is A000005(6) = 4, so a(3)=4.
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MATHEMATICA
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a[n_]:=Max[Table[DivisorSigma[0, k], {k, 2^(n-1), 2^n-1}]]; Table[a[n], {n, 23}] (* Stefano Spezia, Aug 02 2021 *)
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PROG
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(Python)
from sympy import divisors
def a(n): return max(len(divisors(n)) for n in range(2**(n-1), 2**n))
(PARI) a(n) = vecmax(apply(numdiv, [2^(n-1)..2^n-1])); \\ Michel Marcus, Aug 03 2021
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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