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a(n) = 16*a(n-1) - 56*a(n-2) for n>1, with a(0)=1, a(1)=16.
2

%I #29 Sep 08 2022 08:45:40

%S 1,16,200,2304,25664,281600,3068416,33325056,361369600,3915710464,

%T 42414669824,459354931200,4974457389056,53867442077696,

%U 583309459456000,6316374594945024,68396663789584384,740629643316428800

%N a(n) = 16*a(n-1) - 56*a(n-2) for n>1, with a(0)=1, a(1)=16.

%C Third binomial transform of A164609, fourth binomial transform of A164608, fifth binomial transform of A054490, sixth binomial transform of A164607, seventh binomial transform of A083100, eighth binomial transform of A164683.

%C lim_{n -> infinity} a(n)/a(n-1) = 8 + 2*sqrt(2) = 10.8284271247....

%H R. J. Mathar, <a href="/A154348/b154348.txt">Table of n, a(n) for n = 0..100</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (16,-56).

%F a(n) = 16*a(n-1) - 56*a(n-2) for n>1. - _Philippe Deléham_, Jan 12 2009

%F a(n) = ( (8 + 2*sqrt(2))^n - (8 - 2*sqrt(2))^n )/(4*sqrt(2)).

%F G.f.: 1/(1 - 16*x + 56*x^2). - _Klaus Brockhaus_, Jan 12 2009; corrected Oct 08 2009

%F E.g.f.: (1/(2*sqrt(2)))*exp(8*x)*sinh(2*sqrt(2)*x). - _G. C. Greubel_, Sep 13 2016

%t Join[{a=1,b=16},Table[c=16*b-56*a;a=b;b=c,{n,40}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 08 2011*)

%t LinearRecurrence[{16,-56},{1,16},30] (* _Harvey P. Dale_, Aug 31 2016 *)

%o (Magma) Z<x>:=PolynomialRing(Integers()); N<r>:=NumberField(x^2-2); S:=[ ((8+2*r)^n-(8-2*r)^n)/(4*r): n in [1..18] ]; [ Integers()!S[j]: j in [1..#S] ]; // _Klaus Brockhaus_, Jan 12 2009

%Y Cf. A002193 (decimal expansion of sqrt(2)), A164609, A164608, A054490, A164607, A083100, A164683.

%K nonn,easy

%O 0,2

%A Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009

%E Extended beyond a(7) by _Klaus Brockhaus_, Jan 12 2009

%E Edited by _Klaus Brockhaus_, Oct 08 2009

%E Offset corrected. - _R. J. Mathar_, Jun 19 2021