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A273255
Isolated deficient numbers that are divisible by 3.
1
351, 2001, 2211, 2751, 2991, 3009, 3249, 3711, 3849, 4509, 4731, 5169, 6231, 7071, 7209, 7911, 8889, 9351, 9591, 9729, 11409, 13749, 14211, 14769, 17151, 17991, 18129, 18591, 18831, 18849, 19551, 20151, 20481, 21489, 22191, 22989, 23169, 23451, 24051, 25689
OFFSET
1,1
COMMENTS
Each term a(n) will be an odd number, since it must be an odd multiple of 3. [Proof: If a(n) was an even multiple of 3, then a(n) = 3*2k = 6k, which indicates that it will either be a perfect number (when k = 1) or an abundant number (when k > 1). So, for a(n) to be a deficient number, it must be an odd multiple of 3.] Those odd multiples of 3 are given in 3*A273125.
a(n) will be part of a longer string of three or more consecutive isolated deficient numbers, provided that a(n)-2, a(n)+2, a(n)-4, a(n)+4, ... are also deficient. This is because a(n)-3 and a(n)+3 are both multiples of 6, and hence abundant.
The vast majority of terms (probably around 98.6%) end in either 1 or 9, with a(1) = 351 and a(6) = 3009 being the first instances of each. The first instances of the other digits are: a(91) = 58785, a(187) = 119967, a(213) = 138753. Of the first 151725 terms (those less than 10^8), 74769 end in 1, 670 end in 3, 701 end in 5, 685 end in 7, and 74900 end in 9.
LINKS
Timothy L. Tiffin, Table of n, a(n) for n = 1..151725 (terms < 10^8)
FORMULA
a(n) = 3*A273125(n).
EXAMPLE
a(1) = 351, since 351 = A274849(27) = A276050(10) and 351 = 3*117 = 3*A273125(1).
CROSSREFS
Cf. A273125, subsequence of A274849 and A276050.
Subsequence of A016945.
Sequence in context: A267939 A355973 A092374 * A264426 A231707 A220333
KEYWORD
nonn
AUTHOR
Timothy L. Tiffin, Aug 28 2016
STATUS
approved