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.

6, 14, 32, 90, 140, 200, 294, 1832, 1070, 888, 1130, 2180, 2478, 2972, 4298, 5592, 1328, 9552, 30594, 19334, 16142, 15684, 81464, 28230, 31908, 19610, 35618, 82074, 44294, 43332, 34062, 89690, 162144, 134514, 173360, 31398, 404598, 212702, 188030, 542604, 265622

Offset

1,1

Comments

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.

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:

.

n A045881(n) a(n) 3-divisor number

-- ---------- ---- ----------------

1 4 6 4 = 2^2

2 8 14 9 = 3^2

3 24 32 25 = 5^2

7 114 294 121 = 11^2

9 524 1070 529 = 23^2

12 1670 2180 1681 = 41^2

.

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?

Links

Table of n, a(n) for n=1..41.

Example

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.
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.
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.

Crossrefs

Cf. A045881.

Sequence in context: A225972 A332724 A078836 * A142875 A074981 A066510

Adjacent sequences: A340732 A340733 A340734 * A340736 A340737 A340738

Keyword

nonn

Author

Jon E. Schoenfield, Jan 17 2021

Status

approved