A348589 a(n) = (10^n+2)^2 / 6.

24, 1734, 167334, 16673334, 1666733334, 166667333334, 16666673333334, 1666666733333334, 166666667333333334, 16666666673333333334, 1666666666733333333334, 166666666667333333333334, 16666666666673333333333334, 1666666666666733333333333334

Offset: 1

Author

Bernard Schott, Oct 24 2021

Keywords

• nonn
• base
• easy

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).

Examples

			a(1) = 12^2 / 6 = 24 and 2.4 = 3 * 2*4.
a(2) = 102^2 / 6 = 1734 and 17.34 = 3 * 17*34.
		

Links

• Diophante, A1945 - Concaténations en tous genres (in French).
• Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Crossrefs

Cf. A102807, A133384, A147553.

Subsequence of A347541.

Programs

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 *)

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

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