login
a(n) = 2^(4*n+1).
11

%I #73 Apr 17 2022 09:22:51

%S 2,32,512,8192,131072,2097152,33554432,536870912,8589934592,

%T 137438953472,2199023255552,35184372088832,562949953421312,

%U 9007199254740992,144115188075855872,2305843009213693952

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

%C a(n) ~ -Pi*E(2*n)/B(2*n), E(n) Euler number, B(n) Bernoulli number. - _Peter Luschny_, Oct 28 2012

%C Equivalently, powers of 2 with final digit 2. - _Muniru A Asiru_, Mar 15 2019

%C As phi(a(n)) = (2^n)^4 is a perfect biquadrate (where phi is the Euler totient A000010), this is a subsequence of A078164 and A307690. - _Bernard Schott_, Mar 28 2022

%H Vincenzo Librandi, <a href="/A013776/b013776.txt">Table of n, a(n) for n = 0..200</a>

%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>

%H <a href="/index/Di#divseq">Index to divisibility sequences</a>

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (16).

%F From _Philippe Deléham_, Nov 23 2008: (Start)

%F a(n) = 16*a(n-1), n > 0, a(0) = 2.

%F G.f.: 2/(1 - 16*x). (End)

%F From _Peter Bala_, Nov 29 2015: (Start)

%F a(n) = Sum_{k = 0..n} binomial(2*k,k)*binomial(4*n + 2 - 2*k, 2*n + 1 - k).

%F Bisection of A264960. (End)

%F a(n) = A000079(A016813(n)). - _Michel Marcus_, Nov 30 2015

%F a(n) = Sum_{k = 0..2*n} binomial(4*n + 2, 2*k + 1) = A004171(2*n). - _Peter Bala_, Nov 25 2016

%F E.g.f.: 2*exp(16*x). - _G. C. Greubel_, Jun 30 2019

%F From _Bernard Schott_, Apr 15 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 8/15.

%F Sum_{n>=0} (-1)^n/a(n) = 8/17. (End)

%e G.f. = 2 + 32*x + 512*x^2 + 8192*x^3 + 131072*x^4 + 2097152*x^5 + ...

%p [2^(4*n+1)$n=0..20]; # _Muniru A Asiru_, Apr 10 2019

%t 2^(4*Range[0,20]+1) (* _G. C. Greubel_, Mar 15 2019 *)

%t NestList[16#&,2,20] (* _Harvey P. Dale_, Jul 28 2019 *)

%o (Magma) [2^(4*n+1): n in [0..20]]; // _Vincenzo Librandi_, Jun 27 2011

%o (PARI) a(n)=2<<(4*n) \\ _Charles R Greathouse IV_, Apr 07 2012

%o (GAP) List([0..20],n->2^(4*n+1)); # _Muniru A Asiru_, Mar 15 2019

%o (Sage) [2^(4*n+1) for n in (0..20)] # _G. C. Greubel_, Mar 15 2019

%Y Subsequence of A307690.

%Y Cf. A000010, A004171, A016813, A264960, A307690.

%Y Intersection of A000079 and A078164.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_

%E Wrong comment deleted by _Kevin Ryde_, Apr 16 2022