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a(n) = 4^n*(4^n+1)/2.
7

%I #45 Sep 08 2022 08:44:49

%S 1,10,136,2080,32896,524800,8390656,134225920,2147516416,34359869440,

%T 549756338176,8796095119360,140737496743936,2251799847239680,

%U 36028797153181696,576460752840294400,9223372039002259456,147573952598266347520,2361183241469182345216

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

%H Vincenzo Librandi, <a href="/A026244/b026244.txt">Table of n, a(n) for n = 0..200</a>

%H P. J. Szablowski, <a href="http://arxiv.org/abs/1403.0386">On moments of Cantor and related distributions</a>, arXiv preprint arXiv:1403.0386 [math.PR], 2014.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (20,-64)

%F From _Paul Barry_, Mar 11 2004: (Start)

%F With interpolated zeros 0, 1, 0, 10, ... has a(n) = (4^n - (-4)^n + 2*2^n - 2*(-2)^n)/16 and counts walks of length n between adjacent vertices of the 4-cube.

%F G.f.: (1 - 10*x)/((1 - 4*x)*(1 - 16*x)). (End)

%F From _Philippe Deléham_, Sep 08 2009: (Start)

%F a(n) = Sum_{k = 0..n} 9*k*binomial(2n, 2k) = Sum_{k = 0..n} 9^k*A086645(n, k);

%F a(n) = 8^n*T(n,5/4) where T is the Chebyshev polynomial of first kind;

%F e.g.f.: exp(10*x)*cosh(6*x). (End)

%F a(n) = (2*(n+1))! * [x^(2*(n+1))] (cosh(x)^4-1)/4. - _Vladimir Kruchinin_, Oct 19 2016

%F a(n) = 64^n * a(-n) for all n in Z. - _Michael Somos_, Jul 02 2017

%p seq(binomial(-4^n, 2), n=0..18); # _Zerinvary Lajos_, Feb 22 2008

%t Table[4^n (4^n + 1)/2, {n, 0, 19}] (* _Alonso del Arte_, Jun 18 2019 *)

%o (Magma) [4^n*(4^n+1)/2: n in [0..30]]; // _Vincenzo Librandi_, May 01 2011

%o (PARI) a(n)=4^n*(4^n+1)/2 \\ _Charles R Greathouse IV_, Oct 07 2015

%o (Scala) ((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)(_ * _)).map((n: BigInt) => n * (n + 1)/2) // _Alonso del Arte_, Jun 22 2019

%Y Cf. A052539.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_