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4th binomial transform of (1,0,1,0,1,...), A059841.
14

%I #63 Aug 23 2024 20:39:19

%S 1,4,17,76,353,1684,8177,40156,198593,986404,4912337,24502636,

%T 122336033,611148724,3054149297,15265963516,76315468673,381534296644,

%U 1907542343057,9537324294796,47685459212513,238423809278164,1192108586037617,5960511549128476

%N 4th binomial transform of (1,0,1,0,1,...), A059841.

%C Binomial transform of A007582.

%C a(n) is a companion to A005059(n): a(n) + A005059(n) = 5^n; e.g. a(4) = A005059(4) = 353 + 272 = 625 = 5^4. - _Gary W. Adamson_, Jun 30 2006

%C Number of words of length n from an alphabet of 5 letters in which a chosen letter appears an even number of times. - _James Mahoney_, Feb 03 2012 [See a comment in A007582, also for crossrefs. for the 1- to 11-letter word cases. - _Wolfdieter Lang_, Jul 17 2017]

%C The sequence of fractions x(n) = a(n+1)/a(n) satisfies a simple recurrence x(n+1) = 108 - (815 - 1500 / x(n-1)) / x(n) known as Muller's recurrence. It is used for the demonstration of an unexpected failure of floating-point computations. - _Andrey Zabolotskiy_, Sep 17 2019

%H Vincenzo Librandi, <a href="/A081186/b081186.txt">Table of n, a(n) for n = 0..200</a>

%H Christian Hill, <a href="https://scipython.com/blog/mullers-recurrence/">Muller's Recurrence</a>, 2017.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-15).

%F a(n) = 8*a(n-1) - 15*a(n-2) with n>1, a(0)=1, a(1)=4.

%F G.f.: (1-4*x)/((1-3*x)*(1-5*x)).

%F a(n) = (3^n + 5^n)/2.

%F a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k)*4^(n-2*k).

%F E.g.f.: exp(4*x) * cosh(x). - _Paul Barry_, Oct 06 2004

%e Say the alphabet is {a,b,c,d,e} and we want to know how many words of length one and two contain c an even number of times. a(1) = 4, which we can see by the four words {(a),(b),(d),(e)} and a(2) = 17, which we can see by the seventeen words {(a,a), (a,b), (a,d), (a,e), (b,a), (b,b), (b,d), (b,e), (c,c), (d,a), (d,b), (d,d), (d,e), (e,a), (e,b), (e,d), (e,e)}. - _James Mahoney_, Feb 03 2012

%p seq( (3^n + 5^n)/2, n=0..30); # _G. C. Greubel_, Dec 26 2019

%t CoefficientList[Series[(1-4x)/((1-3x)(1-5x)), {x,0,25}], x] (* _Vincenzo Librandi_, Aug 07 2013 *)

%t LinearRecurrence[{8,-15},{1,4},30] (* _Harvey P. Dale_, Apr 13 2019 *)

%o (Magma) [3^n/2+5^n/2: n in [0..25]]; // _Vincenzo Librandi_, Aug 07 2013

%o (PARI) vector(31, n,(3^(n-1) + 5^(n-1))/2 ) \\ _G. C. Greubel_, Dec 26 2019

%o (Sage) [(3^n + 5^n)/2 for n in (0..25)] # _G. C. Greubel_, Dec 26 2019

%o (GAP) List([0..25], n-> (3^n + 5^n)/2); # _G. C. Greubel_, Dec 26 2019

%Y Cf. A005059, A007582.

%K nonn,easy

%O 0,2

%A _Paul Barry_, Mar 11 2003