cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A383365 Numbers k with a record number of proper divisors, where all of them have binary weights that are different from the binary weight of k.

This page as a plain text file.
%I A383365 #13 Apr 25 2025 03:10:22
%S A383365 1,3,15,63,231,405,495,1485,2475,4095,14175,21735,24255,31185,79695,
%T A383365 190575,218295,239085,294525,904365,1276275,2789325,3586275,4937625,
%U A383365 6912675,10072755,17342325,17972955,26801775,46621575,80405325,192567375,326351025,333107775,654729075
%N A383365 Numbers k with a record number of proper divisors, where all of them have binary weights that are different from the binary weight of k.
%C A383365 The corresponding record values are 0, 1, 3, 5, 7, 9, 11, 15, 17, ... (see the link for more values).
%C A383365 All terms are odd as an even number k has proper divisor k/2 with the same binary weight. - _David A. Corneth_, Apr 24 2025
%H A383365 David A. Corneth, <a href="/A383365/b383365.txt">Table of n, a(n) for n = 1..91</a> (first 46 terms from Amiram Eldar)
%H A383365 Amiram Eldar, <a href="/A383365/a383365.txt">Table of n, a(n), A032741(a(n)) for n = 1..46</a>.
%e A383365 a(1) = 1 since 1 has no proper divisors.
%e A383365 a(2) = 3 since 3 has one proper divisor, 1, and A000120(1) = 1 != A000120(3) = 2, while 2 also has one proper divisor, 1, but A000120(2) = A000120(1) = 1.
%e A383365 a(3) = 15 since 15 has 3 proper divisors, 1, 3 and 5, and A000120(1) = 1 and A000120(3) = A000120(5) = 2 are different from A000120(15) = 4. All the numbers below 15 have fewer proper divisors with this property.
%t A383365 q[k_] := DivisorSum[k, 1 &, DigitCount[#, 2, 1] == DigitCount[k, 2, 1] &] == 1; seq[kmax_] := Module[{s = {}, d, dm = 0}, Do[d = DivisorSigma[0, k]; If[d > dm && q[k], dm = d; AppendTo[s, k]], {k, 1, kmax}]; s]; seq[10^5]
%o A383365 (PARI) is1(k) = {my(h = hammingweight(k)); sumdiv(k, d, hammingweight(d) == h) ==  1};
%o A383365 list(kmax) = {my(d, dm = 0); for(k = 1, kmax, d = numdiv(k); if(d > dm && is1(k), dm = d; print1(k, ", ")));}
%Y A383365 Cf. A000120, A032741, A380844, A383363, A383364.
%K A383365 nonn,base
%O A383365 1,2
%A A383365 _Amiram Eldar_, Apr 24 2025