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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A386394 a(n) = (5^(2*n) + 3*2^(5*n-2))/7.

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%I A386394 #12 Jul 23 2025 01:01:29
%S A386394 7,199,5743,168151,4990207,149920999,4553331343,139603087351,
%T A386394 4314710904607,134256051681799,4200826222176943,132042253318646551,
%U A386394 4165747461421299007,131813802646096802599,4180788781690478542543,132853918439479834845751,4228042325697967752173407
%N A386394 a(n) = (5^(2*n) + 3*2^(5*n-2))/7.
%C A386394 a(n) is integer for n > 0.
%C A386394 Proof: It is sufficient to prove that 5^(2*n) + 3*2^(5*n-2) is divisible by 7. Since from 5 == -2 (mod 7) follows that 5^(2*n) == 2^(2*n) (mod 7) and from 2^5 == 2^2 (mod 7) follows 2^(5*n-2) == 2^(2*(n-1)) (mod 7), one gets that 5^(2*n) + 3*2^(5*n-2) == 2^(2*n) + 3*2^(2*(n-1)) (mod 7)  and  2^(2*n) + 3*2^(2*(n-1)) = 2^(2*(n-1))*(4 + 3) == 0 (mod 7). QED
%D A386394 James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 5.1.24 on page 158.
%H A386394 Vincenzo Librandi, <a href="/A386394/b386394.txt">Table of n, a(n) for n = 1..200</a>
%H A386394 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (57,-800).
%F A386394 a(n) = 57*a(n-1) - 800*a(n-2) for n > 2.
%F A386394 G.f.: x*(7 - 200*x)/((1 - 25*x)*(1 - 32*x)).
%F A386394 E.g.f.: (3*exp(32*x) + 4*exp(25*x) - 7)/28.
%F A386394 a(n) = (A009969(n) + 3*A013824(n-1))/7.
%t A386394 a[n_]:=(5^(2n)+3*2^(5n-2))/7; Array[a,17]
%o A386394 (Magma) [(5^(2*n)+3*2^(5*n-2))/7 : n in [1..20]]; // _Vincenzo Librandi_, Jul 21 2025
%Y A386394 Cf. A009969, A013824.
%K A386394 nonn,easy
%O A386394 1,1
%A A386394 _Stefano Spezia_, Jul 20 2025