cp's OEIS Frontend

A340735 a(n) is the smallest positive integer that begins a run of exactly 2*n-1 consecutive integers having at least 4 divisors each.

%I A340735 #6 Jan 19 2021 21:18:03
%S A340735 6,14,32,90,140,200,294,1832,1070,888,1130,2180,2478,2972,4298,5592,
%T A340735 1328,9552,30594,19334,16142,15684,81464,28230,31908,19610,35618,
%U A340735 82074,44294,43332,34062,89690,162144,134514,173360,31398,404598,212702,188030,542604,265622
%N A340735 a(n) is the smallest positive integer that begins a run of exactly 2*n-1 consecutive integers having at least 4 divisors each.
%C A340735 If "integers having at least 4 divisors each" in this sequence's definition were replaced with "integers having at least 3 divisors each" (i.e., composite numbers), the resulting sequence would be A045881.
%C A340735 A045881(n) = a(n) except when the run of 2*n-1 consecutive composite numbers beginning with A045881(n) includes a number with exactly 3 divisors (i.e., the square of a prime). The first six such exceptions are as follows:
%C A340735 .
%C A340735          n  A045881(n)  a(n)  3-divisor number
%C A340735         --  ----------  ----  ----------------
%C A340735          1        4        6        4 =  2^2
%C A340735          2        8       14        9 =  3^2
%C A340735          3       24       32       25 =  5^2
%C A340735          7      114      294      121 = 11^2
%C A340735          9      524     1070      529 = 23^2
%C A340735         12     1670     2180     1681 = 41^2
%C A340735 .
%C A340735 There are no other exceptions among the first 672 terms of A045881 (see the b-file there). Can it be proved that there are no other exceptions?
%e A340735 a(1)=6 because 6=2*3 (which has 4 divisors, {1,2,3,6}) is the first isolated number that has at least 4 divisors.
%e A340735 a(2)=14 because 14 is the first number that begins a run of exactly 2*2-1=3 consecutive integers having at least 4 divisors each: tau(14)=tau(2*7)=4; tau(15)=tau(3*5)=4; tau(16)=tau(2^4)=5.
%e A340735 a(3)=32 because 32 is the first number that begins a run of exactly 2*3-1=5 consecutive integers having at least 4 divisors each: tau(32)=tau(2^5)=6; tau(33)=tau(3*11)=4; tau(34)=tau(2*17)=4; tau(35)=tau(5*7)=4; tau(36)=tau(2^2*3^2)=9.
%Y A340735 Cf. A045881.
%K A340735 nonn
%O A340735 1,1
%A A340735 _Jon E. Schoenfield_, Jan 17 2021