A011900 a(n) = 6*a(n-1) - a(n-2) - 2 with a(0) = 1, a(1) = 3.
1, 3, 15, 85, 493, 2871, 16731, 97513, 568345, 3312555, 19306983, 112529341, 655869061, 3822685023, 22280241075, 129858761425, 756872327473, 4411375203411, 25711378892991, 149856898154533, 873430010034205, 5090723162050695, 29670908962269963, 172934730611569081
Offset: 0
Examples
G.f. = 1 + 3*x + 15x^2 + 85*x^3 + 493*x^4 + 2871*x^5 + 16731*x^6 + ... - _Michael Somos_, Feb 23 2019
References
- Mario Velucchi "The Pell's equation ... an amusing application" in Mathematics and Informatics Quarterly, to appear 1997.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
- Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
- S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
- S. N. Perepechko, Number of perfect matchings on triangular lattices of fixed width, DIMA'2015 slides. [see: page 12]
- Index entries for linear recurrences with constant coefficients, signature (7,-7,1).
Crossrefs
Programs
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Magma
I:=[1,3]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2) - 2: n in [1..40]]; // Vincenzo Librandi, Dec 05 2015
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Maple
f:= gfun:-rectoproc({a(n)=6*a(n-1)-a(n-2)-2,a(0)=1,a(1)=3},a(n),remember): seq(f(n),n=0..40); # Robert Israel, Dec 16 2015
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Mathematica
a[0] = 1; a[1] = 3; a[n_]:= a[n]= 6 a[n-1] -a[n-2] -2; Table[a@ n, {n,0,40}] (* Michael De Vlieger, Dec 05 2015 *) Table[(Fibonacci[2n + 1, 2] + 1)/2, {n, 0, 40}] (* Vladimir Reshetnikov, Sep 16 2016 *) LinearRecurrence[{7,-7,1},{1,3,15},40] (* Harvey P. Dale, Feb 16 2017 *) a[ n_] := (4 + ChebyshevT[n, 3] + ChebyshevT[n + 1, 3])/8; (* Michael Somos, Feb 23 2019 *)
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PARI
Vec((1-4*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Altug Alkan, Dec 06 2015
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SageMath
[(1+lucas_number1(2*n+1,2,-1))//2 for n in range(41)] # G. C. Greubel, Oct 17 2024
Formula
a(n) = (A001653(n+1) + 1)/2.
a(n) = (((1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1))/sqrt(8) + 1)/2.
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3); a(1) = 1, a(2) = 3, a(3) = 15. Also a(n) = 1/2 + ( (1-sqrt(2))/(-4*sqrt(2)) )*(3-2*sqrt(2))^n + ( (1+sqrt(2))/(4*sqrt(2)) )*(3+2*sqrt(2))^n. - Antonio Alberto Olivares, Dec 23 2003
Sqrt(2) = Sum_{n>=0} 1/a(n); a(n) = a(n-1) + floor(1/(sqrt(2) - Sum_{k=0..n-1} 1/a(k))) (n>0) with a(0)=1. - Paul D. Hanna, Jan 25 2004
For n>k, a(n+k) = A001541(n)*A001653(k) - A053141(n-k-1); e.g., 493 = 99*5 - 2. For n<=k, a(n+k)=A001541(n)*A001653(k) - A053141(k-n); e.g., 85 = 3*29 - 2. - Charlie Marion, Oct 18 2004
a(n+1) = 3*a(n) - 1 + sqrt(8*a(n)^2 - 8*a(n) + 1), a(1)=1. - Richard Choulet, Sep 18 2007
a(n+1) = a(n) * (a(n) + 2) / a(n-1) for n>=1 with a(0)=1 and a(1)=3. - Paul D. Hanna, Apr 08 2012
G.f.: (1 - 4*x + x^2)/((1-x)*(1 - 6*x + x^2)). - R. J. Mathar, Oct 26 2009
Sum_{k=a(n)..A001109(n+1)} k = a(n)*A001109(n+1) = A011906(n+1). Example n=2, 3+4+5+6=18, 3*6=18. - Paul Cleary, Dec 05 2015
a(n) = a(-1-n) for all n in Z. - Michael Somos, Feb 23 2019
E.g.f.: (2*exp(x) + exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/4. - Stefano Spezia, Mar 16 2024
Extensions
More terms and comments from Wolfdieter Lang
Comments