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%I #51 Nov 10 2023 21:16:42
%S 30,96,192,210,384,420,480,768,960,1080,1440,1536,1920,2160,2310,2880,
%T 3072,3360,3840,4320,4620,5760,6144,6300,6480,6720,7680,8640,9240,
%U 11520,12288,12600,12960,13440,13860,15360,17280,18480,23040,24576,25920,26880,30030
%N Numbers (including primorials) that satisfy "three rules that highly composite numbers must have" (see Comments below) but are not highly composite numbers.
%C The three rules that the highly composite numbers (and this sequence too) must have are:
%C - The primes in the product have to be consecutive and start with 2,
%C - The exponents of the primes must be decreasing,
%C - The final exponent of the primes must be 1 (except 4 = 2^2 and 36 = 2^2 * 3^2, but those are highly composite numbers and are excluded).
%C Except for numbers of the form 2^n * 3, the terms are divisible by 10, because a(n) has the form of 2^n * 3^m * 5^j * c = (2 * 5) * 2^(n - 1) * 3^m * 5^(j - 1) * c.
%C Except for terms that are primorials, all others are divisible by 12, because a(n) has the form 2^n * 3^m * c = (2^2 * 3) * 2^{n - 2} * 3^(m - 1) * c.
%C All terms are divisible by 6, because a(n) has the form 2^n * 3^m * c = (2 * 3) * 2^(n - 1) * 3^(m - 1) * c.
%C It seems that eulerphi(a(n)) is always divisible by 8.
%H Michael De Vlieger, <a href="/A308508/b308508.txt">Table of n, a(n) for n = 1..16384</a> (first 137 terms from Giovanni Resta).
%t limit = 2*10^5; hc = {1}; r=1; Do[t = DivisorSigma[0, n]; If[t > r, r=t; AppendTo[hc, n]], {n, 2, limit, 2}]; ok[n_] := Block[{f = FactorInteger[n]}, ! MemberQ[hc, n] && f[[-1, 2]] == 1 && Max[ Differences[Last /@ f]] <= 0 && Union[ Differences[ PrimePi[ First /@ f]]] == {1}]; Select[Range[2, limit, 2], ok] (* _Giovanni Resta_, Jun 10 2019 *)
%Y Contains A002110(n) (primorials).
%Y Does not contain A002182(n) (highly composite numbers).
%K nonn
%O 1,1
%A _Pham G. Hoang_, Jun 02 2019
%E a(40)-a(43) from _Giovanni Resta_, Jun 04 2019
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