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A006190
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a(n) = 3*a(n-1) + a(n-2).
(Formerly M2844)
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68
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0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930, 72171863277, 238367471761, 787274278560, 2600190307441
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Denominators of continued fraction convergents to (3+Sqrt[13])/2. - Benoit Cloitre, Jun 14 2003
A006190(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393)...such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003
Form the 4-node graph with matrix A=[1,1,1,1;1,1,1,0;1,1,0,1;1,0,1,1]. Then A006190 counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004
a(n+1) is the binomial transform of A006138. - Paul Barry, May 21 2006
a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006
Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007
a(p) == 13^((p-1)/2)) mod p, for odd primes p. [From Gary W. Adamson, Feb 22 2009]
Contribution from Gary W. Adamson, Aug 06 2010: (Start)
Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1)
dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33) (End)
For n >=2, a(n) equals the permanent of the (n-1)X(n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. [From John M. Campbell, Jul 08 2011]
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REFERENCES
| H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.
Ayoub B. Ayoub, "Fibonacci-like sequences and Pell equations", The College Mathematics Journal, Vol. 38 (2007), pp. 49-53.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 128.
A. F. Horadam, Generating identities for generalized Fibonacci and Lucas triples, Fib. Quart., 15 (1977), 289-292.
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1997), 318-328.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
| T. D. Noe, Table of n, a(n) for n=0..200
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 158
Tanya Khovanova, Recursive Sequences
S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
Index entries for sequences related to Chebyshev polynomials.
Index entries for sequences related to linear recurrences with constant coefficients
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FORMULA
| G.f.: x/(1-3*x-x^2).
a(3n)=2*A041019(5n-1), a(3n+1)=A041019(5n), a(3n+2)=A041019(5n+3).
n>=1 a(2n)=3*A004190(n-1); a(3n)=10*A041613(n-1) - Benoit Cloitre, Jun 14 2003
a(n-1) + a(n+1) = A006497(n); [A006497(n)]^2 - 13[a(n)]^2 = 4(-1)^n - Gary W. Adamson, Jun 15 2003
a(n)=U(n-1, (3/2)i)(-i)^(n-1), i^2=-1 - Paul Barry, Nov 19 2003
a(n)=sum{k=0..n-1, binomial(n-k-1, k)3^(n-2k-1)} - Paul Barry, Oct 02 2004
a(n)=F(n, 3), the n-th Fibonacci polynomial evaluated at x=3.
Let M = {{0, 1}, {1, 3}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = Abs[v[n][[1]]]. - Roger L. Bagula, May 29 2005
a(n+1)=sum{k=0..n, sum{j=0..n-k, C(k,j)C(n-j,k)*2^(k-j)}}=sum{k=0..n, sum{j=0..n-k, C(k,j)C(n-j,k)*2^(n-j-k)}}; a(n+1)=sum{k=0..floor(n/2), comb(n-k,k)3^(n-2k)}=sum{k=0..n, comb(k,n-k)3^(2k-n)}; - Paul Barry, May 21 2006
E.g.f.: exp(3x/2)*sinh(sqrt(13)x/2)/(sqrt(13)/2); - Paul Barry, Jun 03 2006
a(n)=(ap^n-am^n)/(ap-am), with ap := (3+sqrt(13))/2, am := (3-sqrt(13))/2.
Let C = (3+sqrt(13))/2 = exp ArcSinh(3/2) = 3.3027756377... Then C^n, n>0 = a(n)*(1/C) + a(n+1). Example: C^3 = 36.02775637... = a(3)*(1/C) + a(4) = 10*(.302775637...) + 33. Let X = the 2 X 2 matrix [0, 1; 1, 3]. Then X^n = [a(n-1), a(n); a(n), a(n+1)]. - Gary W. Adamson, Dec 21 2007
1/3 = 3/(1*10) + 3/(3*33) + 3/(10*109) + 3/(33*360) + 3/(109*1189) + ... - Gary W. Adamson, Mar 16 2008
a(n)=((3 + Sqrt[3^2 + 4])/2)^(n) - ((3 - Sqrt[3^2 + 4])/2)^(n))/(Sqrt[3^2 + 4]) [From Artur Jasinski, Oct 07 2008]
a(n)=((3+sqrt(13))^n-(3-sqrt(13))^n)/(2^n*sqrt(13)). [From Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009]
Contribution from Johannes W. Meijer, Jun 12 2010: (Start)
Limit(a(n+k)/a(k), k=infinity) = (A006497(n) + A006190(n)*sqrt(13))/2.
Limit(A006497(n)/A006190(n), n=infinity) = sqrt(13).
(End)
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MAPLE
| a[0]:=0: a[1]:=1: for n from 2 to 25 do a[n]:= 3*a[n-1]+a[n-2] end do: seq(a[n], n=0..25); - Emeric Deutsch, Sep 03 2007
A006190:=-1/(-1+3*z+z**2); [S. Plouffe in his 1992 dissertation, without the leading 0.]
seq(combinat[fibonacci](n, 3), n=0..20) ; # R. J. Mathar, Dec 07 2011
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MATHEMATICA
| a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, -1, 24}] (from Robert G. Wilson v Jan 13 2005)
m = 3; Table[Simplify[(((m + Sqrt[m^2 + 4])/2)^(n) - ((m - Sqrt[m^2 + 4])/2)^(n))/(Sqrt[m^2 + 4])], {n, 0, 20}] [From Artur Jasinski, Oct 07 2008]
a=0; lst={a}; s=0; Do[a=s-(a-1); AppendTo[lst, a]; s+=a*3, {n, 3*4!}]; lst [From Vladimir Orlovsky, Oct 27 2009]
LinearRecurrence[{3, 1}, {0, 1}, 40] (* or *) CoefficientList[Series[x/ (1-3x-x^2), {x, 0, 39}], x] (* From Harvey P. Dale, Apr 20 2011 *)
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PROG
| (PARI) a(n)=if(n<1, 0, contfracpnqn(vector(n, i, 2+(i>1)))[2, 1])
(Sage) [lucas_number1(n, 3, -1) for n in xrange(0, 26)] # [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Apr 22 2009]
(MAGMA)[ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+Self(n-2): n in [1..40] ]; // Vincenzo Librandi, Aug 19 2011
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CROSSREFS
| Row sums of Pascal's rhombus (A059317). Also row sums of triangle A054456(n, m). Cf. A000045, A000129, A001076.
Cf. A006497, A052906.
Cf. A000129 [From Gary W. Adamson, Aug 06 2010]
Sequence in context: A126184 A060557 A018920 * A020704 A113299 A126931
Adjacent sequences: A006187 A006188 A006189 * A006191 A006192 A006193
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KEYWORD
| easy,nonn,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| More terms from Larry Reeves (larryr(AT)acm.org), Apr 17 2000
Second formula corrected by Johannes W. Meijer (meijgia(AT)hotmail.com), Jun 02 2010
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