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a(n) = 4^n*(1-2*n).
5

%I #38 Jul 13 2024 02:45:55

%S 1,-4,-48,-320,-1792,-9216,-45056,-212992,-983040,-4456448,-19922944,

%T -88080384,-385875968,-1677721600,-7247757312,-31138512896,

%U -133143986176,-566935683072,-2405181685760,-10170482556928

%N a(n) = 4^n*(1-2*n).

%C With the n-th term of A000984 (C(2n,n)) as numerator, |a(n)| is the denominator of the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. - _Shel Kaphan_, Jan 12 2023

%H G. C. Greubel, <a href="/A144704/b144704.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-16).

%F G.f.: (1-12*x)/(1-4*x)^2.

%F From _Amiram Eldar_, Aug 05 2020: (Start)

%F Sum_{n>=0} 1/a(n) = 1 - arctanh(1/2)/2.

%F Sum_{n>=0} (-1)^(n+1)/a(n) = 1 + arctan(1/2)/2. (End)

%F E.g.f.: (1 - 8*x)*exp(4*x). - _G. C. Greubel_, Jun 16 2022

%F Sum_{n >= 1} x^(2*n-1)/a(n) = 1/4 * log((1 - x/2)/(1 + x/2)). Eldar's two summations above follow from this on setting x = 1 and x = i. - _Peter Bala_, Jul 08 2024

%t LinearRecurrence[{8,-16},{1,-4},30] (* _Harvey P. Dale_, Jun 12 2019 *)

%o (SageMath) [4^n*(1-2*n) for n in (0..30)] # _G. C. Greubel_, Jun 16 2022

%Y Hankel transform of A100320.

%Y Cf. A000302, A165747.

%K easy,sign

%O 0,2

%A _Paul Barry_, Sep 19 2008