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A001541
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a(0) = 1, a(1) = 3; for n > 1, a(n) = 6a(n-1) - a(n-2).
(Formerly M3037 N1231)
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80
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1, 3, 17, 99, 577, 3363, 19601, 114243, 665857, 3880899, 22619537, 131836323, 768398401, 4478554083, 26102926097, 152139002499, 886731088897, 5168247530883, 30122754096401, 175568277047523, 1023286908188737, 5964153172084899, 34761632124320657
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OFFSET
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0,2
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COMMENTS
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Chebyshev polynomials of the first kind evaluated at 3.
a(n) solves for x in x^2 - 8*y^2 = 1, the corresponding y being A001109(n). For n>0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy, Sep 09 2003
Formula: ((-1+sqrt(2))^n+(1+sqrt(2))^n+(1-sqrt(2))^n+(-1-sqrt(2))^n)/4 (with interpolated zeros) E.g.f. cosh(x)cosh(sqrt(2)x) (with interpolated zeros). - Paul Barry, Sep 18 2003
Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(8) - Benoit Cloitre, Feb 14 2004
Appears to give all solutions >1 to the equation : x^2=ceiling(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 24, 2004
a(n+1) - A001542(n+1) = A090390(n+1) - A046729(n) = A001653(n); a(n+1) - 4*A079291(n+1) = (-1)^(n+1). Formula generated by the floretion - .5'i + .5'j - .5i' + .5j' - 'ii' + 'jj' - 2'kk' + 'ij' + .5'ik' + 'ji' + .5'jk' + .5'ki' + .5'kj' + e - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Nov 16 2004
This sequence give numbers n such that (n-1)*(n+1)/2 is a perfect square. Remark : (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i=sqrt(-1) so i is also in the sequence. - Pierre CAMI, Apr 20 2005
a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is divisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk, Nov 24 2006
a(n)=A001333(2*n) [From Ctibor O. Zizka, Aug 13 2008]
The upper principal convergents to 2^(1/2), beginning with 3/2, 17/12, 99/70, 577/408, comprise a strictly decreasing sequence; essentially, numerators=A001541 and denominators=A001542. - Clark Kimberling, Aug 26 2008
Also index of sequence A082532 for which A082532=1 [From Carmine Suriano, Sep 07 2010]
Numbers n such that sigma(n-1) and sigma(n+1) are both odd numbers. [From Juri-Stepan Gerasimov, Mar 28 2011]
Also, numbers such that floor[a(n)^2/2] is a square: base 2 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001075. - M. F. Hasler, Jan 15 2012
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REFERENCES
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I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.
Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
H. Brocard, Notes e'le'mentaires sur le proble`me de Peel, Nouvelles Correspondance Math\'{e}matique, 4 (1878), 161-169.
John M. Campbell, An Integral Representation of Kekule' Numbers, and Double Integrals Related to Smarandache Sequences, Arxiv preprint arXiv:1105.3399, 2011.
J. W. L. Glaisher, On Eulerian numbers (formulae, residues, end-figures), with the values of the first twenty-seven, Quarterly Journal of Mathematics, vol. 45, 1914, pp. 1-51.
D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
Problem E1976, Amer. Math. Monthly, 75 (1968), 683-684.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..200
Tanya Khovanova, Recursive Sequences
_Simon 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.
_Simon Plouffe_, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
Index entries for two-way infinite sequences
Index entries for sequences related to linear recurrences with constant coefficients, signature (6,-1)
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
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G.f.: (1-3*x)/(1-6*x+x^2) - Barry Williams and Wolfdieter Lang, May 05 2000.
E.g.f.: exp(3x)cosh(2sqrt(2)x). Binomial transform of A084128. - Paul Barry, May 16 2003
a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2.
a(n) = Cosh[2n*ArcSinh[1]] - Herbert Kociemba, Apr 24 2008
a(n) = sqrt{8*[(A001109(n))^2] + 1} = T(n, 3), with Chebyshev's T-polynomials A053120.
a(n) ~ (1/2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
For all elements x of the sequence, 2*x^2 - 2 is a square. Lim. as n -> inf. of a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 10 2002. [Corrected by Peter Pein, Mar 09 2009]
a(n) = 3*A001109(n)-A001109(n-1), n >= 1. - Barry Williams and Wolfdieter Lang, May 05 2000.
For n>=1, a(n) = A001652(n) - A001652(n-1) - Charlie Marion, Jul 01 2003
For n>0, a(n)^2 +1=2*A001653(n-1)*A001653(n). - Charlie Marion, Dec 21 2003
a(n)^2+a(n+1)^2=2*(A001653(2n+1)-A001652(2n)). - Charlie Marion, Mar 17 2003
a(n) = Sum_{k>=0} binomial(2*n, 2*k)*2^k = Sum_{k>=0} A086645(n, k)*2^k . - Philippe Deléham, Feb 29 2004
a(n)*A002315(n+k)=A001652(2n+k)+A001652(k)+1; for k>0, a(n+k)*A002315(n) = A001652(2n+k)-A001652(k-1). - Charlie Marion, Mar 17 2003
For n>k, a(n)*A001653(k)=A011900(n+k)+A053141(n-k-1). For n<=k, a(n)*A001653(k)=A011900(n+k)+A053141(k-n). - Charlie Marion, Oct 18 2004
A053141(n+1) + A055997(n+1) = a(n+1) + A001109(n+1). - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Sep 16 2004
a(n) = Sqrt[ A055997(2n) ]. - Alexander Adamchuk, Nov 24 2006
a(2n) = A056771(n). a(2n+1) = 3*A077420(n). - Alexander Adamchuk, Feb 01 2007
(A000129(n)^2)*4+(-1)^n - Vim Wenders, Mar 28 2007
2*a(k)*A001653(n)*A001653(n+k)=A001653(n)^2+A0001653(n+k)^2+A001542(k)^2. - Charlie Marion, Oct 12 2007
A028982(a(n)-1)+2=A028982(a(n)+1). [Juri-Stepan Gerasimov, Mar 28 2011]
a(n) = 2*A001108(n)+1. -Paul Weisenhorn, Dec 17 2011
a(n) = sqrt(2*x^2+1) with x being A001542(n). - Zak Seidov, Jan 30 2013
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MAPLE
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a[0]:=1: a[1]:=3: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 26 2006
A001541:=-(-1+3*z)/(1-6*z+z**2); [Simon Plouffe in his 1992 dissertation.]
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MATHEMATICA
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Table[Round[N[(1/2) (3 + 2 Sqrt[2])^n + (1/2) (3 - 2 Sqrt[2])^n]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
a[ n_] := If[n == 0, 1, With[{m = Abs @ n}, m Sum[4^i Binomial[m + i, 2 i]/(m + i), {i, 0, m}]]] (* Michael Somos Jul 11 2011 *)
a[ n_] := ChebyshevT[ n, 3] (* Michael Somos Jul 11 2011 *)
LinearRecurrence[{6, -1}, {1, 3}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
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PROG
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(PARI) {a(n) = real((3 + quadgen( 32))^n)}
(PARI) {a(n) = subst( poltchebi( abs(n)), x, 3)}
(PARI) {a(n) = if( n<0, a(-n), polsym(1 - 6*x + x^2 , n) [n+1] / 2)}
(PARI) {a(n) = polchebyshev( n, 1, 3)} /* Michael Somos Jul 11 2011 */
(PARI) a(n)=([1, 2, 2; 2, 1, 2; 2, 2, 3]^n)[3, 3] [Vim Wenders, Mar 28 2007]
(Magma)[n: n in [1..10000000] |IsSquare(8*(n^2-1))] [V. Librandi Nov 18 2010]
(Haskell)
a001541 n = a001541_list !! (n-1)
a001541_list =
1 : 3 : zipWith (-) (map (* 6) $ tail a001541_list) a001541_list
-- Reinhard Zumkeller, Oct 06 2011
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CROSSREFS
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Bisection of A001333. A003499(n)=2a(n).
Cf. A046090, A001109, A053142.
Cf. A084130.
Cf. A055997 = numbers n such that n(n-1)/2 is a square. Cf. A001601.
Cf. A056771, A077420.
Cf. A005319.
Cf. A082532.
Row 1 of array A188645.
Cf. A001542.
Sequence in context: A142988 A056660 A155610 * A161940 A074565 A054365
Adjacent sequences: A001538 A001539 A001540 * A001542 A001543 A001544
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KEYWORD
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nonn,easy,nice
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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More terms from Clark Kimberling, Aug 26 2008
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STATUS
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approved
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