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a(n) = (2*n+1)*25^n.
2

%I #21 Feb 26 2022 09:19:32

%S 1,75,3125,109375,3515625,107421875,3173828125,91552734375,

%T 2593994140625,72479248046875,2002716064453125,54836273193359375,

%U 1490116119384765625,40233135223388671875,1080334186553955078125,28870999813079833984375,768341124057769775390625

%N a(n) = (2*n+1)*25^n.

%H Vincenzo Librandi, <a href="/A166725/b166725.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (50,-625).

%F G.f.: (1+25*x)/(1-25*x)^2.

%F Sum_{k>=0} 1/a(k) = (5/2)*log(3/2).

%F E.g.f.: (50*x + 1)*exp(25*x). - _G. C. Greubel_, May 24 2016

%F Sum_{n>=0} (-1)^n/a(n) = 5*arctan(1/5). - _Amiram Eldar_, Feb 26 2022

%t Table[5^(2*n) *(2*n + 1), {n,0,10}] (* _G. C. Greubel_, May 24 2016 *)

%t LinearRecurrence[{50,-625},{1,75},30] (* _Harvey P. Dale_, Mar 02 2018 *)

%o (PARI) a(n)=(2*n+1)*25^n

%o (Magma) [(2*n+1)*25^n: n in [0..20]]; // _Vincenzo Librandi_, Jun 08 2011

%Y Cf. A058962 ((2n+1)*4^n), A155988 ((2n+1)*9^n), A016578 (log(3/2)).

%K nonn

%O 0,2

%A _Jaume Oliver Lafont_, Oct 20 2009