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a(n) = (6^(n+1) - (-5)^(n+1))/11.
4

%I #45 Jan 05 2025 19:51:36

%S 1,1,31,61,991,2821,32551,117181,1093711,4609141,37420471,175694701,

%T 1298308831,6569149861,45518414791,242592910621,1608145354351,

%U 8885932672981,57130293303511,323708273492941,2037617072598271

%N a(n) = (6^(n+1) - (-5)^(n+1))/11.

%C Hankel transform is := 1,30,0,0,0,0,0,0,0,0,0,0,... - _Philippe Deléham_, Nov 02 2008

%C The ratio a(n+1)/a(n) converges to 6 as n approaches infinity. - _Felix P. Muga II_, Mar 10 2014

%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

%D F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.

%H Vincenzo Librandi, <a href="/A053430/b053430.txt">Table of n, a(n) for n = 0..1000</a>

%H A. K. Whitford, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/15-1/whitford-a.pdf">Binet's Formula Generalized</a>, Fibonacci Quarterly, Vol. 15, No. 1, 1979, pp. 21, 24, 29.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (1,30).

%F G.f.: -1/(5*x+1)/(6*x-1). - _R. J. Mathar_, Nov 16 2007

%F a(0)=1, a(1)=1, a(n) = a(n-1) + 30*a(n-2). - _Harvey P. Dale_, May 09 2012

%p A053430:=n->( 6^(n+1)-(-5)^(n+1) )/11; seq(A053430(n), n=0..20); # _Wesley Ivan Hurt_, Mar 11 2014

%t Join[{a=1,b=1},Table[c=b+30*a;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 01 2011 *)

%t Table[(6^(n+1)-(-5)^(n+1))/11,{n,0,20}] (* _Harvey P. Dale_, May 09 2012 *)

%t LinearRecurrence[{1,30},{1,1},21] (* _Harvey P. Dale_, May 09 2012 *)

%t CoefficientList[Series[-1/(5 x + 1)/(6 x - 1), {x, 0, 40}], x] (* _Vincenzo Librandi_, Mar 11 2014 *)

%o (PARI) a(n) = ( 6^(n+1)-(-5)^(n+1) )/11; \\ _Joerg Arndt_, Mar 10 2014

%o (Magma) [(6^(n+1)-(-5)^(n+1))/11: n in [0..30]]; // _Vincenzo Librandi_, Mar 11 2014

%Y Cf. A001045, A015441, A053404.

%K easy,nonn

%O 0,3

%A _Barry E. Williams_, Jan 10 2000

%E More terms from _James A. Sellers_, Feb 02 2000