OFFSET
1,2
COMMENTS
Numbers k such that A002378(k) = k*(k+1) is deficient.
"In 1700, Charles de Neuveglise claimed the product of two consecutive integers n(n+1) with n>=3 is abundant." - Tattersall, p. 144. In other words, de Neuveglise claimed that all oblong numbers greater than 6 are abundant. In fact, up to A002378(1100), 17.6% of the oblong numbers are deficient. The per-100 count of deficient oblong numbers from A002378(1) to A002378(1100) is 16, 19, 19, 16, 17, 20, 18, 17, 17, 15, 20. For most deficient oblong numbers A002378(k) in this range, either k or k+1 is prime, but this is not always the case, explaining why the density of deficient oblong numbers does not decrease in line with the primes.
All the terms are congruent to 1 or 4 mod 6, and there are no terms that are congruent to 0, 4, 15, or 19 mod 20. Therefore, the asymptotic density of this sequence is less than 4/15. The numbers of terms that do not exceed 10^k, for k = 1, 2, ..., are 2, 16, 174, 1831, 18237, 182432, 1824453, 18241059, 182414767, 1824169736, ... . Apparently, the asymptotic density of this sequence equals 0.18241... . - Amiram Eldar, Mar 15 2024
REFERENCES
James J. Tattersall, Elementary Number Theory in Nine Chapters, Second Edition, Cambridge University Press, 2005.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
EXAMPLE
The third deficient oblong number is A002378(13) = 13*14 = 182: sigma(182) = 336 < 364 = 2*182.
MATHEMATICA
Select[Range[400], DivisorSigma[1, o = # (# + 1)] < 2 o &] (* Amiram Eldar, Jun 21 2019 *)
PROG
(PARI) for(n=1, 400, o=n*(n+1); if(sigma(o)<2*o, print1(n, ", ")))
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Chris Fry, Jun 22 2011
STATUS
approved