A273255 Isolated deficient numbers that are divisible by 3.

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

Adjacent sequences: A273252 A273253 A273254 * A273256 A273257 A273258

Keyword

nonn

Author

Timothy L. Tiffin, Aug 28 2016

Status

approved