A346730 -id:A346730 - cp's OEIS frontend

A153876 a(n) = Sum_{i=2^(n-1)..2^n-1} sigma_0(i), sigma_0(i) number of divisors of n, n positive integer.

1, 4, 11, 29, 68, 160, 364, 820, 1813, 3981, 8674, 18782, 40387, 86443, 184232, 391188, 827787, 1746443, 3674573, 7712561, 16151933, 33757505, 70422235, 146659055, 304947023, 633152157, 1312820598, 2718674046, 5623413203, 11618957217, 23982175093, 49452872529

Offset: 1

Ctibor O. Zizka, Jan 03 2009

This sequence tells how many binary numbers with n digits are there in the multiplication matrix [1,...,2^n -1]x[1,...,2^n -1]. In general, counting how many base-B numbers of length n are there in the multiplication matrix [1,...,B^n -1]x[1,...,B^n -1] gives a(n)= sum_{i=B^(n-1),(B^n)-1} sigma_0(i). Besides this motivation it is interesting to see the behavior of partial sums of sigma_0(i) on growing intervals : a(n)= sum_{i=f(n-1),f(n)} sigma_0(i).

Chai Wah Wu, Table of n, a(n) for n = 1..76

Cf. A000005, A006218, A095256, A140480, A346730.

a(n) = sum(i=2^(n-1), 2^n-1, numdiv(i)); \\ Michel Marcus, Oct 10 2021

from math import isqrt
def A153876(n): return ((t:=isqrt(b:=(1<Chai Wah Wu, Oct 23 2023

a(n) = A085831(n) - A085831(n-1)-1. - R. J. Mathar, Jan 05 2009

a(n) = Sum_{k>=1} k * A346730(n,k). - Alois P. Heinz, Aug 01 2021

a(14)-a(32) from Alois P. Heinz, Aug 01 2021

A300509 a(n) is the number of numbers in the interval [2^(n-1), 2^n-1] that have exactly n divisors.

Original entry on oeis.org

1, 2, 1, 4, 1, 6, 1, 25, 3, 10, 1, 212, 1, 27, 8, 3625, 1, 1291, 1, 7687, 18, 265, 1, 629369, 4, 885, 695, 365370, 1, 685360, 1, 178723829, 131, 10782, 12, 311470930, 1, 38692, 413, 6162245368, 1, 381481569, 1, 1067082439, 139407, 513855, 1

Offset: 1

Jon E. Schoenfield, May 25 2018

Number of n-digit binary numbers with exactly n divisors.
If p is an odd prime, then the only p-digit binary number having exactly p divisors is 2^(p-1), so a(p) = 1.
Only squares have an odd number of divisors, so for odd values of n, a(n) is the number of numbers in the interval [ceiling(sqrt(2^(n-1))), floor(sqrt(2^n-1))] whose squares have exactly n divisors. The next few odd-indexed terms are a(41) = 1, a(43) = 1, a(45) = 139407, a(47) = 1, and a(49) = 8. - Jon E. Schoenfield, May 26 2018

			a(1) = 1 because the only number in the interval [2^(1-1), 2^1 - 1] = [1, 1] having exactly 1 divisor is 1.
a(2) = 2 because each of the two numbers in the interval [2^(2-1), 2^2 - 1] = [2, 3] has exactly 2 divisors.
a(8) = 25 because the numbers in the interval [2^(8-1), 2^8 - 1] = [128, 255] having exactly 8 divisors are the 1 number of the form p^7 {i.e., 2^7 = 128}, the 8 numbers of the form p^3 * q {135, 136, 152, 184, 189, 232, 248, 250}, and the 16 numbers of the form p*q*r {130, 138, 154, 165, 170, 174, 182, 186, 190, 195, 222, 230, 231, 238, 246, 255}; 1 + 8 + 16 = 25.

Cf. A000005, A284398.

Main diagonal of A346730.

a(n) = sum(k=2^(n-1), 2^n-1, numdiv(k)==n); \\ Michel Marcus, May 26 2018

a(26)-a(38) from Giovanni Resta, May 26 2018
a(39) from Jon E. Schoenfield, May 26 2018
a(40)-a(41) from Giovanni Resta, May 27 2018
a(42)-a(47) from Jon E. Schoenfield, May 27 2018

A346729 Maximum number of divisors among n-bit numbers.

Original entry on oeis.org

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

Offset: 1

Jon E. Schoenfield, Jul 30 2021

nonn

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).

			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.

Cf. A000005, A002183, A036484, A346730.

a[n_]:=Max[Table[DivisorSigma[0,k],{k,2^(n-1),2^n-1}]]; Table[a[n],{n,23}] (* Stefano Spezia, Aug 02 2021 *)

a(n) = vecmax(apply(numdiv, [2^(n-1)..2^n-1])); \\ Michel Marcus, Aug 03 2021

from sympy import divisors
def a(n): return max(len(divisors(n)) for n in range(2**(n-1), 2**n))
print([a(n) for n in range(1, 18)]) # Michael S. Branicky, Aug 02 2021

A346695 Numbers with more divisors than digits in their binary representation.

Original entry on oeis.org

6, 12, 18, 20, 24, 28, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 105, 108, 110, 112, 114, 120, 126, 132, 140, 144, 150, 156, 160, 162, 168, 176, 180, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 225, 228, 234, 240, 252, 260

Offset: 1

Alex Meiburg, Jul 29 2021

Not all terms are perfect or abundant, with 105 being the first deficient term.
There are no primes in the sequence, and 6 is the only semiprime.
By the same comments as those at A175495, this sequence is infinite.
This sequence is a subsequence of A175495.
It is natural to conjecture that this sequence has asymptotic density 0. However, after the first three terms where a(n)/n = 6 -- a function which would increase to infinity if the asymptotic density were zero -- it drops, and it seems to take a long time to get that high again. The first time it gets above 5.0 is at a(30243)=151216. Even as high as a(2188516)=10000000, the density is only ~1/4.57.
The number of terms with m binary digits is Sum_{k>m} A346730(m,k). - Jon E. Schoenfield, Jul 31 2021

			12 has 6 divisors: {1,2,3,4,6,12}. 12 is written in binary as 1100, which has 4 digits. Since 6 > 4, 12 is in the sequence.

Cf. A000005, A070939.
Cf. A135772 (equal number rather than more).
Cf. A175495 (where "binary digits in n" is replaced by "log_2(n)").

Select[Range[1000], (DivisorSigma[0, #] > Floor[1 + Log2[#]]) &]

isok(m) = numdiv(m) > #binary(m); \\ Michel Marcus, Jul 29 2021

from sympy import divisor_count
def ok(n): return divisor_count(n) > n.bit_length()
print(list(filter(ok, range(1, 261)))) # Michael S. Branicky, Jul 29 2021

cp's OEIS Frontend

A153876 a(n) = Sum_{i=2^(n-1)..2^n-1} sigma_0(i), sigma_0(i) number of divisors of n, n positive integer.

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A300509 a(n) is the number of numbers in the interval [2^(n-1), 2^n-1] that have exactly n divisors.

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A346729 Maximum number of divisors among n-bit numbers.

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A346695 Numbers with more divisors than digits in their binary representation.

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