A360641 Numbers k where A093653(k)/A000120(k) sets a new record.

1, 2, 4, 8, 12, 16, 24, 36, 66, 72, 132, 144, 264, 420, 528, 840, 1026, 1056, 1680, 2052, 4104, 8208, 16416, 32832, 65664, 73920, 84000, 110880, 118800, 131328, 133380, 237600, 263340, 266760, 526680, 533520, 1053360, 1067040, 2106720, 2134080, 3160080, 4213440

Offset

1,2

Comments

Analogous to superabundant numbers (A004394) as A175526 is analogous to abundant numbers (A005101).

The corresponding record values are 1, 2, 3, 4, 9/2, 5, 6, 15/2, 8, 10, ... .

This sequence is infinite since A093653(k)/A000120(k) is unbounded: A093653(2^m)/A000120(2^m) = m+1 for all m >= 0.

Links

Amiram Eldar, Table of n, a(n) for n = 1..58 (terms below 10^10)

Example

The values of A093653(k)/A000120(k) for k=1..10 are 1, 2, 3/2, 3, 3/2, 3, 4/3, 4, 5/2 and 3. The record values, 1, 2, 3 and 4, occur at 1, 2, 4 and 8, the first 4 terms of this sequence.

Mathematica

seq[nmax_] := Module[{s = {}, rm = 0, r}, Do[If[(r = DivisorSum[n, DigitCount[#, 2, 1] &]/DigitCount[n, 2, 1]) > rm, rm = r; AppendTo[s, n]], {n, 1, nmax}]; s]; seq[10^4]

Prog

(PARI) lista(kmax) = {my(rm = 0, r); for(k = 1, kmax, r = sumdiv(k, d, hammingweight(d))/hammingweight(k); if(r > rm, rm = r; print1(k, ", "))); }
(Python) # uses imports and definitions in A093653, A000120
from itertools import count, islice
def f(n): return A093653(n)/A000120(n)
def agen(r=0): yield from ((m, r:=fm)[0] for m in count(1) if (fm:=f(m))>r)
print(list(islice(agen(), 34))) # Michael S. Branicky, Feb 15 2023

Crossrefs

Cf. A000120, A004394, A005101, A093653, A093687, A175522, A175526, A360642.

Sequence in context: A322447 A170892 A246468 * A326807 A052184 A152768

Adjacent sequences: A360638 A360639 A360640 * A360642 A360643 A360644

Keyword

nonn,base

Author

Amiram Eldar, Feb 15 2023

Status

approved