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a(n) = (2*n+1)*5^(2*n+1).
5

%I #17 Sep 08 2022 08:45:06

%S 5,375,15625,546875,17578125,537109375,15869140625,457763671875,

%T 12969970703125,362396240234375,10013580322265625,274181365966796875,

%U 7450580596923828125,201165676116943359375,5401670932769775390625,144354999065399169921875

%N a(n) = (2*n+1)*5^(2*n+1).

%C J. Machin (died 1751) used Pi/4 = 4*Sum_{n=0..inf} (-1)^n/((2*n+1)*5^(2*n+1)) - Sum_{n=0..inf} (-1)^n/((2*n+1)*239^(2*n+1)) to calculate Pi to 100 decimal places.

%D H. Doerrie, 100 Great Problems of Elementary Mathematics, Dover, NY, 1965, p. 73

%H Colin Barker, <a href="/A072172/b072172.txt">Table of n, a(n) for n = 0..700</a>

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

%F From _Colin Barker_, Aug 25 2016: (Start)

%F a(n) = 50*a(n-1) - 625*a(n-2) for n>1.

%F G.f.: 5*(1+25*x)/(1-25*x)^2.

%F (End)

%F From _Ilya Gutkovskiy_, Aug 25 2016: (Start)

%F E.g.f.: 5*(1 + 50*x)*exp(25*x).

%F Sum_{n>=0} 1/a(n) = arctanh(1/5) = 0.2027325540540821...

%F Sum_{n>=0} (-1)^n/a(n) = arctan(1/5) = A105532 (End)

%p seq((2*n+1)*5^(2*n+1), n=0..20); # _G. C. Greubel_, Aug 26 2019

%t Table[(2*n+1)*5^(2*n+1), {n,0,20}] (* _G. C. Greubel_, Aug 26 2019 *)

%o (PARI) Vec(5*(1+25*x)/(1-25*x)^2 + O(x^20)) \\ _Colin Barker_, Aug 25 2016

%o (PARI) vector(20, n, (2*n-1)*5^(2*n-1) ) \\ _G. C. Greubel_, Aug 26 2019

%o (Magma) [(2*n+1)*5^(2*n+1): n in [0..20]]; // _G. C. Greubel_, Aug 26 2019

%o (Sage) [(2*n+1)*5^(2*n+1) for n in (0..20)] # _G. C. Greubel_, Aug 26 2019

%o (GAP) List([0..20], n-> (2*n+1)*5^(2*n+1)); # _G. C. Greubel_, Aug 26 2019

%Y Cf. A072173.

%Y Cf. A157332. - _Jaume Oliver Lafont_, Mar 03 2009

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_, Jun 30 2002