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A165225 a(0)=1, a(1)=5, a(n) = 10*a(n-1) - 5*a(n-2) for n > 1.

%I A165225 #38 May 13 2022 07:02:09
%S A165225 1,5,45,425,4025,38125,361125,3420625,32400625,306903125,2907028125,
%T A165225 27535765625,260822515625,2470546328125,23401350703125,
%U A165225 221660775390625,2099601000390625,19887706126953125,188379056267578125
%N A165225 a(0)=1, a(1)=5, a(n) = 10*a(n-1) - 5*a(n-2) for n > 1.
%C A165225 Sum_{k=1..(m-1)/2} tan^(2n) (k*Pi/m) is an integer when m >= 3 is an odd integer (see AMM and Crux Mathematicorum links); twice this sequence is the particular case m = 5. - _Bernard Schott_, Apr 25 2022
%H A165225 Harvey P. Dale, <a href="/A165225/b165225.txt">Table of n, a(n) for n = 0..1000</a>.
%H A165225 Michel Bataille and Li Zhou, <a href="https://doi.org/10.2307/30037561">A Combinatorial Sum Goes on Tangent</a>, The American Mathematical Monthly, Vol. 112, No. 7 (Aug. - Sep., 2005), Problem 11044, pp. 657-659.
%H A165225 D. J. Smeenk, <a href="https://cms.math.ca/publications/crux/issue?volume=35&amp;issue=5">Problem 3452</a>, Crux Mathematicorum, Vol. 35, No. 5 (2009), pp. 325 and 327; <a href="https://cms.math.ca/publications/crux/issue?volume=36&amp;issue=5">Solution to Problem 3452</a> by Roy Barbara, ibid., Vol. 36, No. 5 (2010), pp. 341-342.
%H A165225 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-5).
%F A165225 Limit_{n->oo} a(n+1)/a(n) = 5 + 2*sqrt(5) = 9.47213595...
%F A165225 G.f.: (1-5x)/(1-10x+5x^2).
%F A165225 a(n) = ((5 - 2*sqrt(5))^n + (5 + 2*sqrt(5))^n)/2. - _Klaus Brockhaus_, Sep 25 2009
%F A165225 a(n) = (tan(Pi/5)^(2*n) + tan(2*Pi/5)^(2*n))/2 (Smeenk, 2009). - _Amiram Eldar_, Apr 03 2022
%t A165225 LinearRecurrence[{10,-5},{1,5},30] (* _Harvey P. Dale_, Dec 23 2019 *)
%Y A165225 Cf. A019934, A019970.
%Y A165225 Similar with: A000244 (m=3), 2*this sequence (m=5), A108716 (m=7), A353410 (m=9), A275546 (m=11), A353411 (m=13).
%K A165225 easy,nonn
%O A165225 0,2
%A A165225 _Philippe Deléham_, Sep 09 2009
%E A165225 More terms from _Klaus Brockhaus_, Sep 25 2009