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Search: id:A052918
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| A052918 |
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a(0)=1, a(1)=5, a(n+1) = 5*a(n) + a(n-1). |
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+0 14
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| 1, 5, 26, 135, 701, 3640, 18901, 98145, 509626, 2646275, 13741001, 71351280, 370497401, 1923838285, 9989688826, 51872282415, 269351100901, 1398627786920, 7262490035501, 37711077964425, 195817879857626
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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[A085448(n)]^2 - 29*[a(n-1)]^2 = 4*(-1)^n. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 01 2003
a(p) == 29^((p-1)/2)) mod p, for odd primes p. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Feb 22 2009]
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LINKS
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Index entries for sequences related to linear recurrences with constant coefficients
Tanya Khovanova, Recursive Sequences
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 901
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
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G.f.: -1/(-1+5*x+x^2)
a(3n)=A041047(5n), a(3n+1)=A041047(5n+3), a(3n+2)=2*A041047(5n+4) - Henry Bottomley (se16(AT)btinternet.com), May 10 2000.
Sum(1/29*(5+2*_alpha)*_alpha^(-1-n), _alpha=RootOf(-1+5*_Z+_Z^2))
a(n-1) = [[(5 + sqrt 29)/2]^n - [(5 - sqrt 29)/2)^n]] / (sqrt 29). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 01 2003
a(n)= U(n, 5*I/2)*(-I)^n with I^2=-1 and Chebyshev's U(n, x/2)=S(n, x) polynomials. See triangle A049310.
Let M = {{0, 1}, {1, 5}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a = v[n][[1]]. - Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 29 2005 - T. D. Noe (noe(AT)sspectra.com), Jan 19 2006
a(n)=F(n, 5), the n-th Fibonacci polynomial evaluated at x=5. - T. D. Noe (noe(AT)sspectra.com), Jan 19 2006
a(n), n>0 = denominator of n-th convergent to [1, 4, 5, 5, 5,...]. Continued fraction [1, 4, 5, 5, 5,...] = .807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701,... - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 21 2007
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MAPLE
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spec := [S, {S=Sequence(Union(Z, Z, Z, Z, Z, Prod(Z, Z)))}, unlabeled]: seq(combstruct[count](spec, size=n), n=0..20);
a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 26 2006
with(combinat):a:=n->fibonacci(n, 5):seq(a(n), n=1..22); # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Dec 07 2008]
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MATHEMATICA
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a=0; lst={}; s=0; Do[a=s-(a-1); AppendTo[lst, a]; s+=a*5, {n, 3*4!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Oct 27 2009]
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PROGRAM
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(Other) sage: [lucas_number1(n, 5, -1) for n in xrange(1, 22)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 24 2009]
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CROSSREFS
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Cf. A000045, A000129, A006190, A001076, A005668, A085448, A099365 (squares), A100237.
Cf. A100237.
Sequence in context: A047755 A047768 A022032 this_sequence A018903 A083331 A076025
Adjacent sequences: A052915 A052916 A052917 this_sequence A052919 A052920 A052921
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KEYWORD
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easy,nonn,new
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AUTHOR
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encyclopedia(AT)pommard.inria.fr, Jan 25 2000
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EXTENSIONS
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Corrected formula: [A087130(n)]^2 - 29*a(n-1)^2 = 4*(-1)^n. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Oct 07 2008
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