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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.
1
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?
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
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Jan 17 2021
STATUS
approved