OFFSET
1,1
COMMENTS
Inspired by the problem 141 of Project Euler (see the link).
The terms of this sequence are oblong numbers m = k*(k+1) with k in A024619.
When q < r < d are consecutive terms of a geometric progression of constant b = p/s noninteger, with b>1, s>=2, p>s, it is necessary that q is a multiple of s^2, so q = q' * s^2 with q' >= 1; the Euclidean division of a term m by q becomes
p*s*q' * (1+p*s*q') = (p^2*q') * (s^2*q') + p*s*q' with k = p*s*q',
so (q, r, d) = (s^2*q', p*s*q', p^2*q') is solution. (see examples).
But, as these terms are oblong, there exists also another division where the constant ratio is the integer psq' and (q,r,d) = (1, p*s*q', (p*s*q')^2) are in geometric progression.
LINKS
FORMULA
EXAMPLE
Examples for 42, 110 and 156 with consecutive ratios 3/2, 5/2, 4/3:
42 | 9 110 | 25 156 | 16
----- ----- -----
6 | 4 , 10 | 4 , 12 | 9 ,
then with consecutive ratios 2, 10 and 12:
42 | 12 110 | 100 156 | 144
----- ----- ------
6 | 3 , 10 | 1 , 12 | 1.
MATHEMATICA
Table[n*(n + 1), {n, Select[Range[80], PrimeNu[#] > 1 &]}] (* Amiram Eldar, May 23 2020 *)
PROG
(PARI) apply(x->x*(x+1), select(x->!isprimepower(x), [2..80])) \\ Michel Marcus, May 23 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Bernard Schott, May 22 2020
STATUS
approved