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A020988 a(n) = (2/3)*(4^n-1). 57

%I

%S 0,2,10,42,170,682,2730,10922,43690,174762,699050,2796202,11184810,

%T 44739242,178956970,715827882,2863311530,11453246122,45812984490,

%U 183251937962,733007751850,2932031007402,11728124029610,46912496118442

%N a(n) = (2/3)*(4^n-1).

%C Numbers whose binary representation is 10, n times (see A163662(n) for n >= 1). - _Alexandre Wajnberg_, May 31 2005

%C Numbers whose base-4 representation consists entirely of 2's; twice base-4 repunits. - _Franklin T. Adams-Watters_, Mar 29 2006

%C Expected time to finish a random Tower of Hanoi problem with 2n disks using optimal moves, so (since 2n is even and A010684(2n) = 1) a(n) = A060590(2n). - _Henry Bottomley_, Apr 05 2001

%C a(n) is the number of derangements of [2n + 3] with runs consisting of consecutive integers. E.g., a(1) = 10 because the derangements of {1, 2, 3, 4, 5} with runs consisting of consecutive integers are 5|1234, 45|123, 345|12, 2345|1, 5|4|123, 5|34|12, 45|23|1, 345|2|1, 5|4|23|1, 5|34|2|1 (the bars delimit the runs). - _Emeric Deutsch_, May 26 2003

%C For n > 0, also smallest numbers having in binary representation exactly n + 1 maximal groups of consecutive zeros: A087120(n) = a(n-1), see A087116. - _Reinhard Zumkeller_, Aug 14 2003

%C Number of walks of length 2n + 3 between any two diametrically opposite vertices of the cycle graph C_6. Example: a(0) = 2 because in the cycle ABCDEF we have two walks of length 3 between A and D: ABCD and AFED. - _Emeric Deutsch_, Apr 01 2004

%C From _Paul Barry_, May 18 2003: (Start)

%C Row sums of triangle using cumulative sums of odd-indexed rows of Pascal's triangle (start with zeros for completeness):

%C 0 0

%C 1 1

%C 1 4 4 1

%C 1 6 14 14 6 1

%C 1 8 27 49 49 27 8 1 (End)

%C a(n) gives the position of the n-th zero in A173732, i.e., A173732(a(n)) = 0 for all n and this gives all the zeros in A173732. - _Howard A. Landman_, Mar 14 2010

%C Smallest number having alternating bit sum -n. Cf. A065359. For n = 0, 1, ..., the last digit of a(n) is 0, 2, 0, 2, ... . - _Washington Bomfim_, Jan 22 2011

%C Number of toothpicks minus 1 in the toothpick structure of A139250 after 2^n stages. - _Omar E. Pol_, Mar 15 2012

%C For n > 0 also partial sums of the odd powers of 2 (A004171). - _K. G. Stier_, Nov 04 2013

%C Values of m such that binomial(4*m + 2, m) is odd. Cf. A002450. - _Peter Bala_, Oct 06 2015

%C For a(n) > 2, values of m such that m is two steps away from a power of 2 under the Collatz iteration. - _Roderick MacPhee_, Nov 10 2016

%H G. C. Greubel, <a href="/A020988/b020988.txt">Table of n, a(n) for n = 0..1000</a> (terms 0..170 from Vincenzo Librandi)

%H Andrei Asinowski, Cyril Banderier, Benjamin Hackl, <a href="https://benjamin-hackl.at/downloads/2019_ABH_popstack-extremal.pdf">On extremal cases of pop-stack sorting</a>, Permutation Patterns (Z├╝rich, Switzerland, 2019).

%H Peter Bala, <a href="/A002450/a002450.txt">A characterization of A002450, A020988 and A080674.</a>

%H John Brillhart and Peter Morton, <a href="http://www.maa.org/programs/maa-awards/writing-awards/a-case-study-in-mathematical-research-the-golay-rudin-shapiro-sequence">A case study in mathematical research: the Golay-Rudin-Shapiro sequence</a>, Amer. Math. Monthly, 103 (1996) 854-869.

%H Nobushige Kurokawa, <a href="http://cage.ugent.be/~kthas/Fun/library/Kurokawa.pdf">Zeta functions over F_1</a>, Proc. Japan Acad., 81, Ser. A (2005), 180-184. See Theorem 3 (3).

%H Andrei K. Svinin, <a href="http://arxiv.org/abs/1603.05748">Tuenter polynomials and a Catalan triangle</a>, arXiv:1603.05748 [math.CO], 2016. See p.3.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-4).

%F a(n) = 4*a(n-1) + 2, a(0) = 0.

%F a(n) = A026644(2*n).

%F a(n) = A007583(n) - 1 = A039301(n+1) - 2 = A083584(n-1) + 1.

%F E.g.f. : (2/3)*(exp(4*x)-exp(x)). - _Paul Barry_, May 18 2003

%F a(n) = A007583(n+1) - 1 = A039301(n+2) - 2 = A083584(n) + 1. - _Ralf Stephan_, Jun 14 2003

%F G.f.: 2*x/((1-x)*(1-4*x)). - _R. J. Mathar_, Sep 17 2008

%F a(n) = a(n-1) + 2^(2n-1), a(0) = 0. - _Washington Bomfim_, Jan 22 2011

%F a(n) = A193652(2*n). - _Reinhard Zumkeller_, Aug 08 2011

%F a(n) = 5*a(n-1) - 4*a(n-2) (n > 1), a(0) = 0, a(1) = 2. - _L. Edson Jeffery_, Mar 02 2012

%F a(n) = (2/3)*A024036(n). - _Omar E. Pol_, Mar 15 2012

%F a(n) = 2*A002450(n). - _Yosu Yurramendi_, Jan 24 2017

%F From _Seiichi Manyama_, Nov 24 2017: (Start)

%F Zeta_{GL(2)/F_1}(s) = Product_{k = 1..4} (s-k)^(-b(2,k)), where Sum b(2,k)*t^k = t*(t-1)*(t^2-1). That is Zeta_{GL(2)/F_1}(s) = (s-3)*(s-2)/((s-4)*(s-1)).

%F Zeta_{GL(2)/F_1}(s) = Product_{n > 0} (1 - (1/s)^n)^(-A295521(n)) = Product_{n > 0} (1 - x^n)^(-A295521(n)) = (1-3*x)*(1-2*x)/((1-4*x)*(1-x)) = 1 + Sum_{k > 0} a(k-1)*x^k (x=1/s). (End)

%p A020988 := proc(n)

%p 2*(4^n-1)/3 ;

%p end proc: # _R. J. Mathar_, Feb 19 2015

%t Table[FromDigits[Flatten[Table[{1, 0}, {i, n}]], 2], {n, 0, 23}] (* _Robert G. Wilson v_, Jun 01 2005 *)

%t (2(4^Range[0, 30] - 1))/3 (* or *) LinearRecurrence[{5, -4}, {0, 2}, 30] (* _Harvey P. Dale_, Sep 25 2013 *)

%o (PARI) an=0;print1(an,", ");for(n=1,23,an+=2^(2*n-1);print1(an,", ")) \\ _Washington Bomfim_, Jan 22 2011

%o (MAGMA) [(2/3)*(4^n-1): n in [0..40] ]; // _Vincenzo Librandi_, Apr 28 2011

%o (PARI) vector(100, n, n--; (2/3)*(4^n-1)) \\ _Altug Alkan_, Oct 06 2015

%o (PARI) Vec(2*z/((1-z)*(1-4*z)) + O(z^30)) \\ _Altug Alkan_, Oct 11 2015

%o (Scala) (((List.fill(20)(4: BigInt)).scanLeft(1: BigInt)(_ * _)).map(2 * _)).scanLeft(0: BigInt)(_ + _) // _Alonso del Arte_, Sep 12 2019

%Y Cf. A020989, A295521.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E Edited by _N. J. A. Sloane_, Sep 06 2006

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Last modified November 14 17:24 EST 2019. Contains 329126 sequences. (Running on oeis4.)