%I #36 Sep 08 2022 08:46:21
%S 0,1,20,325,5000,75625,1137500,17078125,256250000,3844140625,
%T 57664062500,864970703125,12974609375000,194619384765625,
%U 2919291992187500,43789385986328125,656840820312500000,9852612457275390625,147789187622070312500,2216837818145751953125,33252567291259765625000
%N a(n) = 5^(n-1)*(3^n - 1)/2.
%C It is easy to prove that when a(n) is divisible by 7 it is also divisible by 13, but the converse does not always hold. - _Bruno Berselli_, May 22 2018
%H Stephan Ramon Garcia, Yu Xuan Hong, Florian Luca, Elena Pinsker, Carlo Sanna, Evan Schechter and Adam Starr, <a href="https://arxiv.org/abs/1607.07951">p-Adic Quotient Sets</a>, arXiv:1607.07951 [math.NT], 2018 (see Example 5.5, page 15).
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (20,-75).
%F O.g.f.: x/((1 - 5*x)*(1 - 15*x)).
%F E.g.f.: (-1 + exp(10*x))*exp(5*x)/10. - _Bruno Berselli_, May 22 2018
%F a(n) = 20*a(n-1) - 75*a(n-2), n>1.
%t Table[5^(n - 1) (3^n - 1)/2, {n, 0, 25}]
%o (Magma) [5^(n-1)*(3^n-1)/2: n in [0..20]];
%o (PARI) a(n) = 5^(n-1) * (3^n - 1) / 2 \\ _Felix Fröhlich_, May 22 2018
%Y Subsequence of A047239 (after 0).
%K nonn,easy
%O 0,3
%A _Vincenzo Librandi_, May 22 2018