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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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23
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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; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| [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]
For more information about this type of recurrence follow the Khovanova link and see A054413, A086902 and A178765. [Johannes W. Meijer, Jun 12 2010] Binomial transform of A015523. [Johannes W. Meijer, Aug 01 2010]
For positive n, a(n) equals the permanent of the nXn tridiagonal matrix with 5's along the main diagonal and 1's along the superdiagonal and the subdiagonal. [From John M. Campbell, Jul 08 2011]
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LINKS
| Tanya Khovanova, Recursive Sequences
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 901
Index entries for sequences related to Chebyshev polynomials.
Index to sequences with linear recurrences with constant coefficients, signature (5,1).
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FORMULA
| 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
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 12 2010: (Start)
a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2.
Limit(A087130(n)/A052918(n-1), n=infinity) = sqrt(29).
(End)
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Aug 01 2010: (Start)
Limit(a(n+k)/a(k),k=infinity) = (A087130(n)+A052918(n-1)*sqrt(29))/2
(End)
Contribution by L. Edson Jeffery, Jan 07 2012 (Start):
Define the 2 X 2 matrix A=[1,1; 5,4]. Then:
a(n) = (1/5)*[A^(n+2)-A^(n+1)]_(1,1);
a(n) = [A^(n+1)]_(1,2);
a(n) = (1/5)*[A^(n+1)]_(2,1);
a(n) = [sum[k=0..n, A^k]]_(2,2). (End)
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MAPLE
| 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
| 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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PROG
| (Other) sage: [lucas_number1(n, 5, -1) for n in xrange(1, 22)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 24 2009]
(PARI) Vec(-1/(-1+5*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Nov 20 2011
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CROSSREFS
| Cf. A000045, A000129, A006190, A001076, A005668, A085448, A099365 (squares), A100237, A100237.
Sequence in context: A047755 A047768 A022032 * A018903 A083331 A076025
Adjacent sequences: A052915 A052916 A052917 * A052919 A052920 A052921
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KEYWORD
| easy,nonn
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AUTHOR
| encyclopedia(AT)pommard.inria.fr, Jan 25 2000
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EXTENSIONS
| 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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