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A201455 a(n) = 3*a(n-1) + 4*a(n-2) for n>1, a(0)=2, a(1)=3. 5

%I #27 Sep 08 2022 08:46:01

%S 2,3,17,63,257,1023,4097,16383,65537,262143,1048577,4194303,16777217,

%T 67108863,268435457,1073741823,4294967297,17179869183,68719476737,

%U 274877906943,1099511627777,4398046511103,17592186044417,70368744177663,281474976710657

%N a(n) = 3*a(n-1) + 4*a(n-2) for n>1, a(0)=2, a(1)=3.

%C This is the Lucas sequence V(3,-4).

%C Inverse binomial transform of this sequence is A087451.

%H Bruno Berselli, <a href="/A201455/b201455.txt">Table of n, a(n) for n = 0..200</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas_sequence#Specific_names">Lucas sequence: Specific names</a>.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,4).

%F G.f.: (2-3*x)/((1+x)*(1-4*x)).

%F a(n) = 4^n+(-1)^n.

%F a(n) = A086341(A047524(n)) for n>0, a(0)=2.

%F a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 25*x^2))/2 )^n for n >= 1. - _Peter Bala_, Jun 23 2015

%F a(n) = (2/4^n) * Sum_{k = 0..n} binomial(4*n+1, 4*k). - _Peter Bala_, Feb 06 2019

%t RecurrenceTable[{a[n] == 3 a[n - 1] + 4 a[n - 2], a[0] == 2, a[1] == 3}, a[n], {n, 25}]

%o (Magma) [n le 1 select n+2 else 3*Self(n)+4*Self(n-1): n in [0..25]];

%o (Maxima) a[0]:2$ a[1]:3$ a[n]:=3*a[n-1]+4*a[n-2]$ makelist(a[n], n, 0, 25);

%o (PARI) Vec((2-3*x)/((1+x)*(1-4*x)) + O(x^30)) \\ _Michel Marcus_, Jun 26 2015

%Y Cf. for the same recurrence with initial values (i,i+1): A015521 (Lucas sequence U(3,-4); i=0), A122117 (i=1), A189738 (i=3).

%Y Cf. for similar closed form: A014551 (2^n+(-1)^n), A102345 (3^n+(-1)^n), A087404 (5^n+(-1)^n).

%Y Cf. A052539, A024036.

%K nonn,easy

%O 0,1

%A _Bruno Berselli_, Jan 09 2013

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)