cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

Original entry on oeis.org

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647
Offset: 0

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Author

Keywords

Comments

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths: 1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
From Gary W. Adamson, Jul 25 2016: (Start)
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
1, 1, 0, 0, 0, 0, ...
2, 0, 3, 0, 0, 0, ...
2, 0, 0, 3, 0, 0, ...
2, 0, 0, 0, 3, 0, ...
2, 0, 0, 0, 0, 3, ...
...
The triangle generated from M is:
1,
1, 1,
3, 1, 3,
11, 3, 3, 9,
41, 11, 9, 9, 27,
...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). (End)
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

Examples

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...
		

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
  • 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).
  • P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

Crossrefs

Programs

  • Haskell
    a001075 n = a001075_list !! n
    a001075_list =
       1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
    -- Reinhard Zumkeller, Aug 11 2011
    
  • Magma
    I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017
  • Maple
    A001075 := proc(n)
        orthopoly[T](n,2) ;
    end proc:
    seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018
  • Mathematica
    Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
    CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
    LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
    Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
    ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
    a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)
  • PARI
    {a(n) = subst(poltchebi(abs(n)), x, 2)};
    
  • PARI
    {a(n) = real((2 + quadgen(12))^abs(n))};
    
  • PARI
    {a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};
    
  • PARI
    a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016
    
  • PARI
    my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017
    
  • SageMath
    [lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009
    
  • SageMath
    def a(n):
        Q = QuadraticField(3, 't')
        u = Q.units()[0]
        return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014
    

Formula

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

Extensions

More terms from James Sellers, Jul 10 2000
Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A052530 a(n) = 4*a(n-1) - a(n-2), with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 8, 30, 112, 418, 1560, 5822, 21728, 81090, 302632, 1129438, 4215120, 15731042, 58709048, 219105150, 817711552, 3051741058, 11389252680, 42505269662, 158631825968, 592022034210, 2209456310872, 8245803209278, 30773756526240
Offset: 0

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Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

a(n-1) and a(n+1) are the solutions for c if b = a(n) in (b^2 + c^2)/(b*c + 1) = 4 and there are no other pairs of solutions apart from consecutive pairs of terms in this sequence. Cf. A061167. - Henry Bottomley, Apr 18 2001
a(n)^2 for n >= 1 gives solutions to A007913(3*x+4) = A007913(x). - Benoit Cloitre, Apr 07 2002
For all terms k of the sequence, 3*k^2 + 4 is a perfect square. Limit_{n->oo} a(n)/a(n-1) = 2 + sqrt(3). - Gregory V. Richardson, Oct 06 2002
a(n) = the number of compositions of the integer 2*n into even parts, where each part 2*i comes in 2*i colors. (Dedrickson, Theorem 3.2.6) An example is given below. Cf. A052529, A095263. - Peter Bala, Sep 17 2013
Except for an initial 1, this is the p-INVERT of (1, 1, 1, 1, 1, ...) for p(S) = 1 - 2*S - 2*S^2; see A291000. - Clark Kimberling, Aug 24 2017
a(n+1) is the number of spanning trees of the graph P_n, where P_n is a 2 X n grid with two additional vertices, u and v, where u is adjacent to (1,1) and (2,1), and v is adjacent to (1,n) and (2,n). - Kevin Long, May 04 2018
a(n) is also the output of Tesler's formula for the number of perfect matchings of an m X n Mobius band where m is even and n is odd, specialized to m=2. (The twist is on the length-n side.) - Sarah-Marie Belcastro, Feb 15 2022
In general, values of x and y which satisfy (x^2 + y^2)/(x*y + 1) = k^2 are any two adjacent terms of a second-order recurrence with initial terms 0 and k and signature (k^2,-1). This can also be expressed as a first-order recurrence a(n+1) = (k^2*a(n) + sqrt((k^4-4)*a(n)^2 + 4*k^2))/2, n > 1. - Gary Detlefs, Feb 27 2024

Examples

			Colored compositions. a(2) = 8: There are two compositions of 4 into even parts, namely 4 and 2 + 2. Using primes to indicate the coloring of parts, the 8 colored compositions are 4, 4', 4'', 4''', 2 + 2, 2 + 2', 2' + 2 and 2' + 2'. - _Peter Bala_, Sep 17 2013
		

Crossrefs

Programs

  • Haskell
    a052530 n = a052530_list !! n
    a052530_list =
       0 : 2 : zipWith (-) (map (* 4) $ tail a052530_list) a052530_list
    -- Reinhard Zumkeller, Sep 29 2011
    
  • Magma
    I:=[0,2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Feb 25 2019
    
  • Maple
    spec := [S,{S=Sequence(Prod(Union(Z,Z),Sequence(Z),Sequence(Z)))},unlabeled]: seq(combstruct[count](spec, size=n), n=0..20);
    s := sqrt(3): a := n -> ((2-s)^n-(s+2)^n)/(s*(s-2)*(s+2)):
    seq(simplify(a(n)), n=0..24); # Peter Luschny, Apr 28 2020
  • Mathematica
    p=1; c=2; a[0]=0; a[1]=c; a[n_]:=a[n]=p*c^2*a[n-1]-a[n-2]; Table[a[n], {n, 0, 20}]
    NestList[2 # + Sqrt[4 + 3 #^2]&, 0, 200] (* Zak Seidov, Mar 31 2011 *)
    LinearRecurrence[{4, -1}, {0, 2}, 25] (* T. D. Noe, Jan 09 2012 *)
    CoefficientList[Series[2x/(1-4x+x^2),{x,0,30}],x] (* Harvey P. Dale, May 31 2023 *)
  • PARI
    { polya002(p,c,m) = local(v,w,j,a); w=0; print1(w,", "); v=c; print1(v,", "); j=1; while(j<=m,a=p*c^2*v-w; print1(a,", "); w=v; v=a; j++) };
    polya002(1,2,25)
    
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(2*x/(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019
    
  • PARI
    first(n) = n = max(n, 2); my(res = vector(n)); res[1] = 0; res[2] = 2; for(i = 3, n, res[i] = 4 * res[i-1] - res[i-2]); res \\ David A. Corneth, Apr 28 2020
    
  • Sage
    (2*x/(1-4*x+x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

Formula

G.f.: 2*x/(1 - 4*x + x^2).
Invert transform of even numbers: a(n) = 2*Sum_{k=1..n} k*a(n-k). - Vladeta Jovovic, Apr 27 2001
From Gregory V. Richardson, Oct 06 2002: (Start)
a(n) = Sum_{alpha} -(1/3)*(-1 + 2*alpha)*alpha^(-1 - n), alpha = root of (1 - 4*Z + Z^2).
a(n) = (((2+sqrt(3))^(n+1) - (2-sqrt(3))^(n+1)) - ((2+sqrt(3))^n - (2-sqrt(3))^n) + ((2+sqrt(3))^(n-1) - (2-sqrt(3))^(n-1)))/(3*sqrt(3)). (End)
a(n) = A071954(n) - 2. - N. J. A. Sloane, Feb 20 2005
a(n) = (2*sinh(2n*arcsinh(1/sqrt(2))))/sqrt(3). - Herbert Kociemba, Apr 24 2008
a(n) = 2*A001353(n). - R. J. Mathar, Oct 26 2009
a(n) = ((3 - 2*sqrt(3))/3)*(2 - sqrt(3))^(n - 1) + ((3 + 2*sqrt(3))/3)*(2 + sqrt(3))^(n - 1). - Vincenzo Librandi, Nov 20 2010
a(n) = floor((2 + sqrt(3))^n/sqrt(3)). - Zak Seidov, Mar 31 2011
a(n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/sqrt(3). (See Horadam for construction.) - Johannes Boot, Jan 08 2012
a(n) = A217233(n) + A217233(n-1) with A217233(-1) = -1. - Bruno Berselli, Oct 01 2012
a(n) = A001835(n+1) - A001835(n). - Kevin Long, May 04 2018
E.g.f.: (exp((2 + sqrt(3))*x) - exp((2 - sqrt(3))*x))/sqrt(3). - Franck Maminirina Ramaharo, Nov 12 2018
a(n+1) = 2*a(n) + sqrt(3*a(n)^2 + 4), n > 1. - Gary Detlefs, Feb 27 2024

Extensions

More terms from James Sellers, Jun 06 2000
Edited by N. J. A. Sloane, Nov 11 2006
a(0) changed to 0 and entry revised accordingly by Max Alekseyev, Nov 15 2007
Signs in definition corrected by John W. Layman, Nov 20 2007

A001571 a(n) = 4*a(n-1) - a(n-2) + 1, with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 9, 35, 132, 494, 1845, 6887, 25704, 95930, 358017, 1336139, 4986540, 18610022, 69453549, 259204175, 967363152, 3610248434, 13473630585, 50284273907, 187663465044, 700369586270, 2613814880037, 9754889933879, 36405744855480, 135868089488042
Offset: 0

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Author

Keywords

Comments

Second member of the Diophantine pair (m,k) that satisfies 3(m^2 + m) = k^2 + k: a(n) = k. - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002

References

  • 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).

Crossrefs

Programs

  • Magma
    I:=[0,2]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2)+1: n in [1..30]]; // Vincenzo Librandi, Jun 07 2015
    
  • Magma
    [(Evaluate(ChebyshevU(n+1), 2) + Evaluate(ChebyshevU(n), 2) - 1)/2 : n in [0..30]]; // G. C. Greubel, Feb 02 2022
    
  • Maple
    f := gfun:-rectoproc({a(0) = 0, a(1) = 2, a(n) = 4*a(n - 1) - a(n - 2) + 1}, a(n), remember): map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 25 2020
  • Mathematica
    a[0]=0; a[1]=2; a[n_]:= a[n]= 4a[n-1] -a[n-2] +1; Table[a[n], {n, 0, 24}] (* Robert G. Wilson v, Apr 24 2004 *)
    Table[(ChebyshevU[n,2] +ChebyshevU[n-1,2] -1)/2, {n,0,30}] (* G. C. Greubel, Feb 02 2022 *)
  • Sage
    [(chebyshev_U(n,2) + chebyshev_U(n-1,2) - 1)/2 for n in (0..30)] # G. C. Greubel, Feb 02 2022

Formula

a(n) = (A001834(n) - 1)/2.
G.f.: x*(2-x)/( (1-x)*(1-4*x+x^2) ). - Simon Plouffe in his 1992 dissertation.
a(n) = sqrt((-2 + (2 - sqrt(3))^n + (2 + sqrt(3))^n)*(2 + (2 - sqrt(3))^(1 + n) + (2 + sqrt(3))^(1 + n)))/(2*sqrt(2)). - Gerry Martens, Jun 05 2015
E.g.f.: (exp(2*x)*(cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) - exp(x))/2. - Franck Maminirina Ramaharo, Nov 12 2018
a(n) = 2*A061278(n) - A061278(n-1). - R. J. Mathar, Feb 06 2020
a(n) = ((1+sqrt(3))*(2+sqrt(3))^n + (1-sqrt(3))*(2-sqrt(3))^n)/4 - (1/2). - Vladimir Pletser, Jan 15 2021
a(n) = (ChebyshevU(n, 2) + ChebyshevU(n-1, 2) - 1)/2. - G. C. Greubel, Feb 02 2022

Extensions

Better description from Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002
More terms and new description from Robert G. Wilson v, Apr 24 2004

A003699 Number of Hamiltonian cycles in C_4 X P_n.

Original entry on oeis.org

1, 6, 22, 82, 306, 1142, 4262, 15906, 59362, 221542, 826806, 3085682, 11515922, 42978006, 160396102, 598606402, 2234029506, 8337511622, 31116016982, 116126556306, 433390208242, 1617434276662, 6036346898406, 22527953316962, 84075466369442, 313773912160806
Offset: 1

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Author

Keywords

Comments

a(n) is the number of generalized compositions of n when there are i^2+i-1 different types of i, (i = 1, 2, ...). - Milan Janjic, Sep 24 2010
Is this the same as the sequence visible in Table 5 of Pettersson, 2014? - N. J. A. Sloane, Jun 05 2015

References

  • F. Faase, On the number of specific spanning subgraphs of the graphs G X P_n, Ars Combin. 49 (1998), 129-154.

Crossrefs

Column k=4 of A359855.
First differences of A052530 and A071954.

Programs

  • GAP
    a:=[6,22];; for n in [3..20] do a[n]:=4a[n-1]-a[n-2]; od; Concatenation([1], a); # G. C. Greubel, Dec 23 2019
  • Magma
    I:=[1,6,22]; [n le 3 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 13 2018
    
  • Maple
    seq( simplify( `if`(n=1, 1, 2*(ChebyshevU(n-1,2) - ChebyshevU(n-2,2))) ), n=1..30); # G. C. Greubel, Dec 23 2019
  • Mathematica
    Join[{1},LinearRecurrence[{4,-1},{6,22},30]] (* Harvey P. Dale, Jul 19 2011 *)
    Table[If[n<2, n, 2*(ChebyshevU[n-1, 2] - ChebyshevU[n-2, 2])], {n,30}] (* G. C. Greubel, Dec 23 2019 *)
  • Maxima
    (a[1] : 1, a[2] : 6, a[3] : 22, a[n] := 4*a[n - 1] - a[n - 2], makelist(a[n], n, 1, 50)); /* Franck Maminirina Ramaharo, Nov 12 2018 */
    
  • PARI
    vector(30, n, if(n==1, 1, 2*(polchebyshev(n-1, 2, 2) - polchebyshev(n-2, 2, 2))) ) \\ G. C. Greubel, Dec 23 2019
    
  • Sage
    [1]+[2*(chebyshev_U(n-1,2) - chebyshev_U(n-2,2)) for n in (2..30)] # G. C. Greubel, Dec 23 2019
    

Formula

a(n) = 2 * A001835(n), n > 1.
From Benoit Cloitre, Mar 28 2003: (Start)
a(n) = ceiling((1 - sqrt(1/3))*(2 + sqrt(3))^n) for n > 1.
a(1) = 1, a(2) = 6, a(3) = 22 and for n > 3, a(n) = 4*a(n-1) - a(n-2). (End)
O.g.f.: x*(1 + 2*x - x^2)/(1-4*x+x^2) = -2 - x + 2*(1 - 3*x)/(1-4*x+x^2). - R. J. Mathar, Nov 23 2007
From Franck Maminirina Ramaharo, Nov 12 2018: (Start)
a(n) = ((1 + sqrt(3))*(2 - sqrt(3))^n - (1 - sqrt(3))*(2 + sqrt(3))^n)/sqrt(3), n > 1.
E.g.f.: ((1 + sqrt(3))*exp((2 - sqrt(3))*x) - (1 - sqrt(3))*exp((2 + sqrt(3))*x) - (2 + x)*sqrt(3))/sqrt(3). (End)
a(n) = 2*(ChebyshevU(n-1, 2) - ChebyshevU(n-2, 2)) for n >1, with a(1)=1. - G. C. Greubel, Dec 23 2019

A228946 Numbers m such that m^3 - k^3 is a square for some k < m, k > 0.

Original entry on oeis.org

8, 10, 14, 28, 32, 33, 40, 56, 57, 65, 71, 72, 74, 78, 90, 105, 112, 114, 126, 128, 130, 132, 140, 148, 154, 155, 160, 176, 193, 200, 217, 218, 224, 228, 250, 252, 260, 266, 273, 280, 284, 288, 296, 297, 305, 312, 329, 336, 344, 349, 350, 360, 392
Offset: 1

Views

Author

M. F. Hasler, Oct 05 2013

Keywords

Comments

See A038596 = A038597^2 for the possible values of n^3-k^3.

Crossrefs

Programs

Formula

a(n) = A066648(n)^(1/3). - Amiram Eldar, Mar 20 2025

A082840 a(n) = 4*a(n-1) - a(n-2) + 3, with a(0) = -1, a(1) = 1.

Original entry on oeis.org

-1, 1, 8, 34, 131, 493, 1844, 6886, 25703, 95929, 358016, 1336138, 4986539, 18610021, 69453548, 259204174, 967363151, 3610248433, 13473630584, 50284273906, 187663465043, 700369586269, 2613814880036, 9754889933878, 36405744855479, 135868089488041
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 14 2003

Keywords

Comments

Apart from the initial -1, these are the numbers k such that the triangular number k*(k + 1)/2 is the sum of three consecutive triangular numbers - see A129803. - Brian Nowell, Nov 03 2009

Crossrefs

Programs

  • GAP
    a:=[-1,1,8];; for n in [4..30] do a[n]:=5*a[n-1]-5*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Feb 25 2019
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( -(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2)) )); // G. C. Greubel, Feb 25 2019
    
  • Mathematica
    CoefficientList[Series[(-1+6x-2x^2)/((1-x)(1-4x+x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, Apr 15 2014 *)
    LinearRecurrence[{5,-5,1}, {-1,1,8}, 30] (* G. C. Greubel, Feb 25 2019 *)
  • PARI
    is(n)=ispolygonal(3/2*n*(n+1)+4,3) || n==-1 \\ Charles R Greathouse IV, Apr 14 2014
    
  • PARI
    my(x='x+O('x^30)); Vec(-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019
    
  • Sage
    (-(1-6*x+2*x^2)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019
    

Formula

a(n) = A001571(n) - 1. - N. J. A. Sloane, Nov 03 2009
G.f.: -(1 -6*x +2*x^2)/((1 - x)*(1 - 4*x + x^2)).
a(n) = -3/2 + (1/12)*( (a -2*b +5)*a^n + (b -2*a +5)*b^n ), with a = 2 + sqrt(3), b = 2 - sqrt(3):.
a(n) = -3/2 + (3/4)*A003500(n) - (1/4)*A003500(n-1).
a(n) = (1/2)*(A001834(n) - 3).
E.g.f.: ((1 + sqrt(3))*exp((2 + sqrt(3))*x) + (1 - sqrt(3))*exp((2 - sqrt(3))*x) - 6*exp(x))/4. - Franck Maminirina Ramaharo, Nov 12 2018

A072110 a(n) = 4*a(n-1) - a(n-2) - 2, with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 5, 16, 57, 210, 781, 2912, 10865, 40546, 151317, 564720, 2107561, 7865522, 29354525, 109552576, 408855777, 1525870530, 5694626341, 21252634832, 79315912985, 296011017106, 1104728155437, 4122901604640, 15386878263121
Offset: 0

Views

Author

Robert G. Wilson v, Jul 30 2002

Keywords

Crossrefs

Programs

  • GAP
    a:=[1,2,5];; for n in [4..30] do a[n]:=5*a[n-1]-5*a[n-2]+a[n-3]; od; a; # G. C. Greubel, Feb 25 2019
  • Magma
    I:=[1,2,5]; [n le 3 select I[n] else 5*Self(n-1) -5*Self(n-2) + Self(n-3): n in [1..30]]; // G. C. Greubel, Feb 25 2019
    
  • Mathematica
    a[0]=1; a[1]=2; a[n_]:=a[n] =4*a[n-1]-a[n-2] -2; Table[a[n], {n, 0, 25}]
    LinearRecurrence[{5,-5,1}, {1,2,5}, 30] (* G. C. Greubel, Feb 25 2019 *)
  • PARI
    my(x='x+O('x^30)); Vec((1-3*x)/((1-x)*(1-4*x+x^2))) \\ G. C. Greubel, Feb 25 2019
    
  • Sage
    [lucas_number1(n,4,1)+1 for n in range(26)] # Zerinvary Lajos, Jul 06 2008
    
  • Sage
    ((1-3*x)/((1-x)*(1-4*x+x^2))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019
    

Formula

a(n) = A071954(n)/2 = A001353(n) + 1.
From G. C. Greubel, Feb 25 2019: (Start)
G.f.: (1-3*x)/((1-x)*(1-4*x+x^2))
a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3).
a(n) = 1 - (-i)^(n+1)*F(n, 4*i), where i = sqrt(-1) and F(n,x) is the Fibonacci polynomial. (End)
Showing 1-7 of 7 results.