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5th binomial transform of (1,0,1,0,1,...), A059841.
5

%I #42 Jan 13 2024 20:32:29

%S 1,5,26,140,776,4400,25376,148160,872576,5169920,30757376,183495680,

%T 1096779776,6563901440,39316299776,235629363200,1412702437376,

%U 8471919656960,50814338072576,304817308958720,1828628975845376

%N 5th binomial transform of (1,0,1,0,1,...), A059841.

%C Binomial transform of A081187.

%C a(n) is also the number of words of length n over an alphabet of six letters, of which a chosen one appears an even number of times. See a comment in A007582, also for the crossrefs, for the 1- to 11-letter word cases. - _Wolfdieter Lang_, Jul 17 2017

%H Vincenzo Librandi, <a href="/A081187/b081187.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-24).

%F a(n) = 10*a(n-1) - 24*a(n-2) with n > 1, a(0)=1, a(1)=5.

%F G.f.: (1-5*x)/((1-4*x)*(1-6*x)).

%F E.g.f.: exp(5*x)*cosh(x).

%F a(n) = (4^n + 6^n)/2.

%F a(n) = Sum_{k=0..floor(n/2)} C(n, 2k)*5^(n-2k).

%F E.g.f.: exp(5*x)*cosh(x) = (1/2)*E(0), where E(k) = 1 + 2^k/(3^k - 6*x*9^k/(6*x*3^k + (k+1)*2^k/E(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, Nov 21 2011

%F a(n) = A074612(n)/2. - _G. C. Greubel_, Jan 13 2024

%p seq( (4^n + 6^n)/2, n=0..25); # _G. C. Greubel_, Dec 26 2019

%t CoefficientList[Series[(1-5x)/((1-4x)(1-6x)), {x, 0, 25}], x] (* _Vincenzo Librandi_, Aug 07 2013 *)

%t LinearRecurrence[{10,-24}, {1,5}, 26] (* _G. C. Greubel_, Dec 26 2019 *)

%o (Magma) [4^n/2 + 6^n/2: n in [0..25]]; // _Vincenzo Librandi_, Aug 07 2013

%o (PARI) vector(26, n, (4^(n-1) + 6^(n-1))/2) \\ _G. C. Greubel_, Dec 26 2019

%o (Sage) [(4^n + 6^n)/2 for n in (0..25)] # _G. C. Greubel_, Dec 26 2019

%o (GAP) List([0..25], n-> (4^n + 6^n)/2); # _G. C. Greubel_, Dec 26 2019

%Y Cf. A007582, A074612, A081189.

%K nonn,easy

%O 0,2

%A _Paul Barry_, Mar 11 2003