%I #43 Jun 30 2024 02:48:33
%S 42,110,156,210,240,342,420,462,506,600,702,812,930,1122,1190,1260,
%T 1332,1482,1560,1640,1806,1980,2070,2162,2352,2550,2652,2756,2970,
%U 3080,3192,3306,3422,3660,3906,4032,4290,4422,4692,4830,4970,5256,5550,5700,5852,6006,6162
%N Let m = d*q + r be the Euclidean division of m by d. The terms m of this sequence satisfy that q, r, d are consecutive positive integer terms in a geometric progression with a noninteger common ratio > 1.
%C Inspired by the problem 141 of Project Euler (see the link).
%C The terms of this sequence are oblong numbers m = k*(k+1) with k in A024619.
%C 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
%C p*s*q' * (1+p*s*q') = (p^2*q') * (s^2*q') + p*s*q' with k = p*s*q',
%C so (q, r, d) = (s^2*q', p*s*q', p^2*q') is solution. (see examples).
%C 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.
%H Project Euler, <a href="https://projecteuler.net/problem=141">Problem 141: Investigating progressive numbers, n, which are also square</a>
%F a(n) = A024619(n))* (1+A024619(n)).
%F a(n) = A002378(A024619(n)). - _Michel Marcus_, May 23 2020
%e Examples for 42, 110 and 156 with consecutive ratios 3/2, 5/2, 4/3:
%e 42 | 9 110 | 25 156 | 16
%e ----- ----- -----
%e 6 | 4 , 10 | 4 , 12 | 9 ,
%e then with consecutive ratios 2, 10 and 12:
%e 42 | 12 110 | 100 156 | 144
%e ----- ----- ------
%e 6 | 3 , 10 | 1 , 12 | 1.
%t Table[n*(n + 1), {n, Select[Range[80], PrimeNu[#] > 1 &]}] (* _Amiram Eldar_, May 23 2020 *)
%o (PARI) apply(x->x*(x+1), select(x->!isprimepower(x), [2..80])) \\ _Michel Marcus_, May 23 2020
%Y Cf. A024619, A127629, A334185, A334186.
%Y Subsequence of A002378 and of A335065.
%K nonn
%O 1,1
%A _Bernard Schott_, May 22 2020