OFFSET
1,1
COMMENTS
Numbers q.r such that q.r = 3*q*r, when q and r have the same number of digits, "." means concatenation, r = 2q and r may not begin with 0.
We must solve the Diophantine equation q.r = q*10^m+r = 3 * q*r where m = length(q) = length(r).
The number of solutions is infinite with (r, q) = ((10^n+2)/3, (10^n+2)/6) and n >= 1.
Note that 15 satisfies also q.r = 3*q*r, 15 = 3*1*5 with here r = 5*q.
For further information about the general equation q.r = k * q*r, see A347541.
Problem proposed on the French website Diophante (see link).
LINKS
Diophante, A1945 - Concaténations en tous genres (in French).
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).
FORMULA
a(n) = (10^n+2)^2 / 6.
a(n) = A133384(n-1)^2/6.
G.f.: 6*x*(4-155*x+250*x^2)/((1-x)*(1-10*x)*(1-100*x)). - Stefano Spezia, Oct 25 2021
a(n) = 3*A102807(n)/2. - Hugo Pfoertner, Oct 30 2021
EXAMPLE
a(1) = 12^2 / 6 = 24 and 2.4 = 3 * 2*4.
a(2) = 102^2 / 6 = 1734 and 17.34 = 3 * 17*34.
MAPLE
seq((10^n+2)^2 / 6, n=1..14);
MATHEMATICA
Table[(10^n + 2)^2/6, {n, 1, 14}] (* Amiram Eldar, Oct 24 2021 *)
PROG
(Python)
def a(n): return (10**n+2)**2//6
print([a(n) for n in range(1, 15)]) # Michael S. Branicky, Oct 24 2021
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Bernard Schott, Oct 24 2021
STATUS
approved