A247698 Brady numbers: B(n) = B(n - 1) + B(n - 2) with B(1) = 2308 and B(2) = 4261.
2308, 4261, 6569, 10830, 17399, 28229, 45628, 73857, 119485, 193342, 312827, 506169, 818996, 1325165, 2144161, 3469326, 5613487, 9082813, 14696300, 23779113, 38475413, 62254526, 100729939, 162984465, 263714404, 426698869, 690413273, 1117112142, 1807525415, 2924637557, 4732162972, 7656800529
Offset: 1
Links
- Logan Cooper, Table of n, a(n) for n = 1..1000 (truncated from 9966 to 1000 terms by _M. F. Hasler_, May 10 2017)
- Brady Haran and Matt Parker, Brady Numbers, Numberphile video (2014).
- Index entries for linear recurrences with constant coefficients, signature (1,1).
Programs
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Haskell
brady = let makeSeq a b = a : makeSeq b (a + b) in makeSeq 2308 4261
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Magma
m:=50; R
:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(2308+1953*x)/(1-x-x^2))); // G. C. Greubel, Sep 07 2018 -
Maple
Brady1 := proc(n::posint) option remember, system; if n = 1 then 2308 elif n = 2 then 4261 else thisproc( n - 1 ) + thisproc( n - 2 ) end if end proc: seq( Brady1( n ), n = 1 .. 100 ); # James McCarron, Oct 05 2019 # alternate program Brady2 := ( n :: posint ) -> coeff( series(x*(2308+1953*x)/(1-x-x^2),x,n+1), x^n ): seq( Brady2( n ), n = 1 .. 100 ); # James McCarron, Oct 05 2019
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Mathematica
LinearRecurrence[{1, 1}, {2308, 4261}, n] Rest[CoefficientList[Series[x*(2308+1953*x)/(1-x-x^2), {x,0,50}], x]] (* G. C. Greubel, Sep 07 2018 *) -
PARI
Vec(-x*(1953*x+2308)/(x^2+x-1) + O(x^50)) \\ Colin Barker, Sep 23 2014
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PARI
a(n)=([1,1;1,0]^n*[1953;355])[1,1] \\ Charles R Greathouse IV, Jan 20 2016
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Python
def A247698_list(n): list = [2308, 4261] + [0] * (n - 2) for i in range(2, n): list[i] = list[i - 1] + list[i - 2] return list print(A247698_list(32)) # M. Eren Kesim, Jun 28 2021
Formula
a(n) = a(n-1) + a(n-2).
G.f.: x*(2308 + 1953*x) / (1-x-x^2). - Colin Barker, Sep 23 2014
a(n) = k*phi^n + o(1), where k = 976.5 + sqrt(354578.45) = 1571.96.... - Charles R Greathouse IV, Sep 28 2014
a(n) = 2308*A000045(n-2) + 4261*A000045(n-1) = 1953*A000045(n+1) + 355*A000045(n). - M. F. Hasler, May 10 2017
a(n) = F(n+17) - F(n+8) - 9*F(n) - F(n-14) for F(n) = A000045(n). - Greg Dresden, Jul 07 2022
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