login
A360641
Numbers k where A093653(k)/A000120(k) sets a new record.
2
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
KEYWORD
nonn,base
AUTHOR
Amiram Eldar, Feb 15 2023
STATUS
approved