%I #8 Apr 10 2023 02:42:02

%S 1,2,4,8,16,32,64,128,256,336,420,840,1680,3360,6720,7560,15120,30240,

%T 60480,95760,120960,176400,191520,257040,352800,383040,514080,1028160,

%U 1681680,2056320,2998800,3112200,5525520,5997600,6224400,8353800,12448800,16216200,24897600

%N Numbers k with record values of the ratio A000005(k)/A246600(k) between the total number of divisors of k and the number of divisors d of k such that the bitwise OR of k and d is equal to k.

%C This sequence is infinite since the ratio A000005(k)/A246600(k) is unbounded. For example, if k = 2^m then A000005(k)/A246600(k) = m+1.

%C All the terms except for 1 are in A355670.

%H Amiram Eldar, <a href="/A361937/b361937.txt">Table of n, a(n) for n = 1..52</a>

%e The ratios A000005(k)/A246600(k) for k = 1, 2, 3 and 4 are 1, 2, 1 and 3. The record values, 1, 2 and 3, occur at 1, 2 and 4, the first 3 terms of this sequence.

%t r[n_] := DivisorSigma[0,n]/DivisorSum[n, Boole[BitOr[#, n] == n] &];

%t seq[kmax_] := Module[{rm = 0, k = 1, s = {}, r1}, Do[r1 = r[k]; If[r1 > rm, rm = r1; AppendTo[s, k]], {k, 1 , kmax}]; s]; seq[10^6]

%o (PARI) r(n) = numdiv(n)/sumdiv(n, d, bitor(d, n) == n);

%o lista(kmax) = {my(rm = 0, r1); for(k = 1, kmax, r1 = r(k); if(r1 > rm, rm = r1; print1(k, ", "))); }

%Y Cf. A000005, A246600, A355670, A359080, A359082, A359083.

%Y Similar sequences: A307870, A335832, A361807.

%K nonn,base

%O 1,2

%A _Amiram Eldar_, Mar 31 2023