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%I #67 Jul 08 2022 08:21:18
%S 2308,4261,6569,10830,17399,28229,45628,73857,119485,193342,312827,
%T 506169,818996,1325165,2144161,3469326,5613487,9082813,14696300,
%U 23779113,38475413,62254526,100729939,162984465,263714404,426698869,690413273,1117112142,1807525415,2924637557,4732162972,7656800529
%N Brady numbers: B(n) = B(n - 1) + B(n - 2) with B(1) = 2308 and B(2) = 4261.
%C B(n) / B(n - 1) approaches the golden ratio as n approaches infinity.
%H Logan Cooper, <a href="/A247698/b247698.txt">Table of n, a(n) for n = 1..1000</a> (truncated from 9966 to 1000 terms by _M. F. Hasler_, May 10 2017)
%H Brady Haran and Matt Parker, <a href="https://www.youtube.com/watch?v=D8ntDpBm6Ok">Brady Numbers</a>, Numberphile video (2014).
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,1).
%F a(n) = a(n-1) + a(n-2).
%F G.f.: x*(2308 + 1953*x) / (1-x-x^2). - _Colin Barker_, Sep 23 2014
%F a(n) = k*phi^n + o(1), where k = 976.5 + sqrt(354578.45) = 1571.96.... - _Charles R Greathouse IV_, Sep 28 2014
%F a(n) = 2308*A000045(n-2) + 4261*A000045(n-1) = 1953*A000045(n+1) + 355*A000045(n). - _M. F. Hasler_, May 10 2017
%F a(n) = F(n+17) - F(n+8) - 9*F(n) - F(n-14) for F(n) = A000045(n). - _Greg Dresden_, Jul 07 2022
%p Brady1 := proc(n::posint)
%p option remember, system;
%p if n = 1 then
%p 2308
%p elif n = 2 then
%p 4261
%p else
%p thisproc( n - 1 ) + thisproc( n - 2 )
%p end if
%p end proc:
%p seq( Brady1( n ), n = 1 .. 100 );
%p # _James McCarron_, Oct 05 2019
%p # alternate program
%p Brady2 := ( n :: posint ) -> coeff( series(x*(2308+1953*x)/(1-x-x^2),x,n+1), x^n ):
%p seq( Brady2( n ), n = 1 .. 100 );
%p # _James McCarron_, Oct 05 2019
%t LinearRecurrence[{1, 1}, {2308, 4261}, n]
%t Rest[CoefficientList[Series[x*(2308+1953*x)/(1-x-x^2), {x,0,50}], x]] (* _G. C. Greubel_, Sep 07 2018 *)
%o (Haskell) brady = let makeSeq a b = a : makeSeq b (a + b) in makeSeq 2308 4261
%o (PARI) Vec(-x*(1953*x+2308)/(x^2+x-1) + O(x^50)) \\ _Colin Barker_, Sep 23 2014
%o (PARI) a(n)=([1,1;1,0]^n*[1953;355])[1,1] \\ _Charles R Greathouse IV_, Jan 20 2016
%o (Magma) m:=50; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(2308+1953*x)/(1-x-x^2))); // _G. C. Greubel_, Sep 07 2018
%o (Python)
%o def A247698_list(n):
%o list = [2308, 4261] + [0] * (n - 2)
%o for i in range(2, n):
%o list[i] = list[i - 1] + list[i - 2]
%o return list
%o print(A247698_list(32)) # _M. Eren Kesim_, Jun 28 2021
%K nonn,easy
%O 1,1
%A _Sebastian Zimmer_, Sep 22 2014
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