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%I #26 Dec 31 2016 01:44:23
%S 351,2001,2211,2751,2991,3009,3249,3711,3849,4509,4731,5169,6231,7071,
%T 7209,7911,8889,9351,9591,9729,11409,13749,14211,14769,17151,17991,
%U 18129,18591,18831,18849,19551,20151,20481,21489,22191,22989,23169,23451,24051,25689
%N Isolated deficient numbers that are divisible by 3.
%C 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.
%C 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.
%C 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.
%H Timothy L. Tiffin, <a href="/A273255/b273255.txt">Table of n, a(n) for n = 1..151725</a> (terms < 10^8)
%F a(n) = 3*A273125(n).
%e a(1) = 351, since 351 = A274849(27) = A276050(10) and 351 = 3*117 = 3*A273125(1).
%Y Cf. A273125, subsequence of A274849 and A276050.
%Y Subsequence of A016945.
%K nonn
%O 1,1
%A _Timothy L. Tiffin_, Aug 28 2016