A348589 - OEIS

%I #36 Nov 02 2021 14:08:31

%S 24,1734,167334,16673334,1666733334,166667333334,16666673333334,

%T 1666666733333334,166666667333333334,16666666673333333334,

%U 1666666666733333333334,166666666667333333333334,16666666666673333333333334,1666666666666733333333333334

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

%C 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.

%C We must solve the Diophantine equation q.r = q*10^m+r = 3 * q*r where m = length(q) = length(r).

%C The number of solutions is infinite with (r, q) = ((10^n+2)/3, (10^n+2)/6) and n >= 1.

%C Note that 15 satisfies also q.r = 3*q*r, 15 = 3*1*5 with here r = 5*q.

%C For further information about the general equation q.r = k * q*r, see A347541.

%C Problem proposed on the French website Diophante (see link).

%H Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/1024-a1945-concatenations-en-tous-genres">A1945 - Concaténations en tous genres</a> (in French).

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-1110,1000).

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

%F a(n) = A133384(n-1)^2/6.

%F G.f.: 6*x*(4-155*x+250*x^2)/((1-x)*(1-10*x)*(1-100*x)). - _Stefano Spezia_, Oct 25 2021

%F a(n) = 3*A102807(n)/2. - _Hugo Pfoertner_, Oct 30 2021

%e a(1) = 12^2 / 6 = 24 and 2.4 = 3 * 2*4.

%e a(2) = 102^2 / 6 = 1734 and 17.34 = 3 * 17*34.

%p seq((10^n+2)^2 / 6, n=1..14);

%t Table[(10^n + 2)^2/6, {n, 1, 14}] (* _Amiram Eldar_, Oct 24 2021 *)

%o (Python)

%o def a(n): return (10**n+2)**2//6

%o print([a(n) for n in range(1, 15)]) # _Michael S. Branicky_, Oct 24 2021

%Y Cf. A102807, A133384, A147553.

%Y Subsequence of A347541.

%K nonn,base,easy

%O 1,1

%A _Bernard Schott_, Oct 24 2021