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.

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A079935 a(n) = 4*a(n-1) - a(n-2) with a(1) = 1, a(2) = 3.

Original entry on oeis.org

1, 3, 11, 41, 153, 571, 2131, 7953, 29681, 110771, 413403, 1542841, 5757961, 21489003, 80198051, 299303201, 1117014753, 4168755811, 15558008491, 58063278153, 216695104121, 808717138331, 3018173449203, 11263976658481, 42037733184721, 156886956080403
Offset: 1

Views

Author

Benoit Cloitre and Paul D. Hanna, Jan 20 2003

Keywords

Comments

See A001835 for another version.
Greedy frac multiples of sqrt(3): a(1)=1, Sum_{n>0} frac(a(n)*x) < 1 at x=sqrt(3).
The n-th greedy frac multiple of x is the smallest integer that does not cause Sum_{k=1..n} frac(a(k)*x) to exceed unity; an infinite number of terms appear as the denominators of the convergents to the continued fraction of x.
Binomial transform of A002605. - Paul Barry, Sep 17 2003
In general, Sum_{k=0..n} binomial(2n-k,k)*j^(n-k) = (-1)^n* U(2n, i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
The Hankel transform of this sequence is [1,2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
From Richard Choulet, May 09 2010: (Start)
This sequence is a particular case of the following situation:
a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3) = (a(n+2)*a(n+1)+q)/a(n)
where q is given in Z to have Q = (a*b^2 + q*b + a + q)/(a*b) itself in Z.
The g.f is f: f(z) = (1 + a*z + (b-Q)*z^2 + (a*b + q - a*Q)*z^3)/(1 - Q*z^2 + z^4);
so we have the linear recurrence: a(n+4) = Q*a(n+2) - a(n).
The general form of a(n) is given by:
a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (b-Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p) and
a(2*m+1) = a*Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (a*b+q-a*Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p).
(End)
x-values in the solution to 3*x^2 - 2 = y^2. - Sture Sjöstedt, Nov 25 2011
From Wolfdieter Lang, Oct 12 2020: (Start)
[X(n) = S(n, 4) - S(n-1, 4), Y(n) = X(n-1)] gives all positive solutions of X^2 + Y^2 - 4*X*Y = -2, for n = -oo..+oo, where the Chebyshev S-polynomials are given in A049310, with S(-1, 0) = 0, and S(-|n|, x) = - S(|n|-2, x), for |n| >= 2.
This binary indefinite quadratic form has discriminant D = +12. There is only this family representing -2 properly with X and Y positive, and there are no improper solutions.
See also the preceding comment by Sture Sjöstedt.
See the formula for a(n) = X(n-1), for n >= 1, in terms of S-polynomials below.
This comment is inspired by a paper by Robert K. Moniot (private communication). See his Oct 04 2020 comment in A027941 related to the case of x^2 + y^2 - 3*x*y = -1 (special Markov solutions). (End)
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 and n are both even, specialized to m=2. (The twist is on the length-n side.) - Sarah-Marie Belcastro, Feb 15 2022

Examples

			a(4) = 41 since frac(1*x) + frac(3*x) + frac(11*x) + frac(41*x) < 1, while frac(1*x) + frac(3*x) + frac(11*x) + frac(k*x) > 1 for all k > 11 and k < 41.
		

Crossrefs

Cf. A002530 (denominators of convergents to sqrt(3)), A079934, A079936, A001353.
Cf. A001835 (same except for the first term).
Row 4 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • Haskell
    a079935 n = a079935_list !! (n-1)
    a079935_list =
       1 : 3 : zipWith (-) (map (4 *) $ tail a079935_list) a079935_list
    -- Reinhard Zumkeller, Aug 14 2011
    
  • Magma
    I:=[1,3]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 06 2015
    
  • Maple
    f:= gfun:-rectoproc({a(n) = 4*a(n-1) - a(n-2),a(1)=1,a(2)=3}, a(n), remember):
    seq(f(n),n=1..30); # Robert Israel, Jun 05 2015
  • Mathematica
    a[n_] := (MatrixPower[{{1, 2}, {1, 3}}, n].{{1}, {1}})[[1, 1]]; Table[ a[n], {n, 0, 23}] (* Robert G. Wilson v, Jan 13 2005 *)
    LinearRecurrence[{4,-1},{1,3},30] (* or *) CoefficientList[Series[ (1-x)/(1-4x+x^2),{x,0,30}],x]  (* Harvey P. Dale, Apr 26 2011 *)
    a[n_] := Sqrt[2/3] Cosh[(-1 - 2 n) ArcCsch[Sqrt[2]]];
    Table[Simplify[a[n-1]], {n, 1, 12}] (* Peter Luschny, Oct 13 2020 *)
  • PARI
    a(n)=([0,1; -1,4]^(n-1)*[1;3])[1,1] \\ Charles R Greathouse IV, Mar 18 2017
    
  • PARI
    my(x='x+O('x^30)); Vec((1-x)/(1-4*x+x^2)) \\ G. C. Greubel, Feb 25 2019
  • Sage
    [lucas_number1(n,4,1)-lucas_number1(n-1,4,1) for n in range(1, 25)] # Zerinvary Lajos, Apr 29 2009
    

Formula

For n > 0, a(n) = ceiling( (2+sqrt(3))^n/(3+sqrt(3)) ).
From Paul Barry, Sep 17 2003: (Start)
G.f.: (1-x)/(1-4*x+x^2).
E.g.f.: exp(2*x)*(sinh(sqrt(3)*x)/sqrt(3) + cosh(sqrt(3)*x)).
a(n) = ( (3+sqrt(3))*(2+sqrt(3))^n + (3-sqrt(3))*(2-sqrt(3))^n )/6 (offset 0). (End)
a(n) = Sum_{k=0..n} binomial(2*n-k, k)*2^(n-k). - Paul Barry, Jan 22 2005 [offset 0]
a(n) = (-1)^n*U(2*n, i*sqrt(2)/2), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005 [offset 0]
a(n) = Jacobi_P(n,-1/2,1/2,2)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006 [offset 0]
a(n) = sqrt(2+(2-sqrt(3))^(2*n-1) + (2+sqrt(3))^(2*n-1))/sqrt(6). - Gerry Martens, Jun 05 2015
a(n) = (1/2 + sqrt(3)/6)*(2-sqrt(3))^n + (1/2 - sqrt(3)/6)*(2+sqrt(3))^n. - Robert Israel, Jun 05 2015
a(n) = S(n-1,4) - S(n-2,4) = (-1)^(n-1)*S(2*(n-1), i*sqrt(2)), with Chebyshev S-polynomials (A049310), the imaginary unit i, S(-1, x) = 0, for n >= 1. See also the formula above by Paul Barry (with offset 0). - Wolfdieter Lang, Oct 12 2020
a(n) = sqrt(2/3)*cosh((-1 - 2*n) arccsch(sqrt(2))), where arccsch is the inverse hyperbolic cosecant function (with offset 0). - Peter Luschny, Oct 13 2020
From Peter Bala, May 04 2025: (Start)
a(n) = (1/sqrt(3)) * sqrt(1 - T(2*n-1, -2)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(n) divides a(3*n-1); a(n) divides a(5*n-2); in general, for k >= 0, a(n) divides a((2*k+1)*n - k).
The aerated sequence [b(n)]n>=1 = [1, 0, 3, 0, 11, 0, 41, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -6, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 2} 1/(a(n) - 1/a(n)) = 1/2 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A052530(n-1) - 1/A052530(n).) (End)

A134451 Ternary digital root of n.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 27 2007

Keywords

Comments

Continued fraction expansion of sqrt(3) - 1. - N. J. A. Sloane, Dec 17 2007. Cf. A040001, A048878/A002530.
Minimum number of terms required to express n as a sum of odd numbers.
Shadow transform of even numbers A005843. - Michel Marcus, Jun 06 2013
From Jianing Song, Nov 01 2022: (Start)
For n > 0, a(n) is the minimal gap of distinct numbers coprime to n. Proof: denote the minimal gap by b(n). For odd n we have A058026(n) > 0, hence b(n) = 1. For even n, since 1 and -1 are both coprime to n we have b(n) <= 2, and that b(n) >= 2 is obvious.
The maximal gap is given by A048669. (End)

Examples

			n=42: A007089(42) = '1120', A053735(42) = 1+1+2+0 = 4,
A007089(4)='11', A053735(4)=1+1=2: therefore a(42) = 2.
0.732050807568877293527446341... = 0 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ...)))). - _Harry J. Smith_, May 31 2009
		

Crossrefs

Cf. A000010, A055034, A134452, A160390 (decimal expansion).
Apart from a(0) the same as A040001.
Related base-3 sequences: A053735, A134451, A230641, A230642, A230643, A230853, A230854, A230855, A230856, A230639, A230640, A010063 (trajectory of 1).

Programs

Formula

a(n) = n if n <= 2, otherwise a(A053735(n)).
a(A005408(n)) = 1; a(A005843(n)) = 2 for n>0;
a(n) = 0 if n=0, otherwise A000034(n-1).
a(n) = ((n+1) mod 2) + 2*sign(n) - 1. - Wesley Ivan Hurt, Dec 06 2013
Multiplicative with a(2^e) = 2, a(p^e) = 1 for odd prime p. - Andrew Howroyd, Aug 06 2018
a(0) = A055034(1) / A000010(1), a(n) = A000010(n+1) / A055034(n+1), n>1. - Torlach Rush, Oct 29 2019
Dirichlet g.f.: zeta(s)*(1+1/2^s). - Amiram Eldar, Jan 01 2023

A080599 Expansion of e.g.f.: 2/(2-2*x-x^2).

Original entry on oeis.org

1, 1, 3, 12, 66, 450, 3690, 35280, 385560, 4740120, 64751400, 972972000, 15949256400, 283232149200, 5416632421200, 110988861984000, 2425817682288000, 56333385828720000, 1385151050307024000, 35950878932544576000, 982196278209226080000, 28175806418228108640000
Offset: 0

Views

Author

Detlef Pauly (dettodet(AT)yahoo.de), Feb 24 2003

Keywords

Comments

Number of ordered partitions of {1,..,n} with at most 2 elements per block. - Bob Proctor, Apr 18 2005
In other words, number of preferential arrangements of n things (see A000670) in which each clump has size 1 or 2. - N. J. A. Sloane, Apr 13 2014
Recurrences (of the hypergeometric type of the Jovovic formula) mean: multiplying the sequence vector from the left with the associated matrix of the recurrence coefficients (here: an infinite lower triangular matrix with the natural numbers in the main diagonal and the triangular series in the subdiagonal) recovers the sequence up to an index shift. In that sense, this sequence here and many other sequences of the OEIS are eigensequences. - Gary W. Adamson, Feb 14 2011
Number of intervals in the weak (Bruhat) order of S_n that are Boolean algebras. - Richard Stanley, May 09 2011
a(n) = D^n(1/(1-x)) evaluated at x = 0, where D is the operator sqrt(1+2*x)*d/dx. Cf. A000085, A005442 and A052585. - Peter Bala, Dec 07 2011
From Gus Wiseman, Jul 04 2020: (Start)
Also the number of (1,1,1)-avoiding or cubefree sequences of length n covering an initial interval of positive integers. For example, the a(0) = 1 through a(3) = 12 sequences are:
() (1) (11) (112)
(12) (121)
(21) (122)
(123)
(132)
(211)
(212)
(213)
(221)
(231)
(312)
(321)
(End)

Examples

			From _Gus Wiseman_, Jul 04 2020: (Start)
The a(0) = 1 through a(3) = 12 ordered set partitions with block sizes <= 2 are:
  {}  {{1}}  {{1,2}}    {{1},{2,3}}
             {{1},{2}}  {{1,2},{3}}
             {{2},{1}}  {{1,3},{2}}
                        {{2},{1,3}}
                        {{2,3},{1}}
                        {{3},{1,2}}
                        {{1},{2},{3}}
                        {{1},{3},{2}}
                        {{2},{1},{3}}
                        {{2},{3},{1}}
                        {{3},{1},{2}}
                        {{3},{2},{1}}
(End)
		

Crossrefs

Column k=2 of A276921.
Cubefree numbers are A004709.
(1,1)-avoiding patterns are A000142.
(1,1,1)-avoiding compositions are A232432.
(1,1,1)-matching patterns are A335508.
(1,1,1)-avoiding permutations of prime indices are A335511.
(1,1,1)-avoiding compositions are ranked by A335513.
(1,1,1,1)-avoiding patterns are A189886.

Programs

  • Magma
    [n le 2 select 1 else (n-1)*Self(n-1) + Binomial(n-1,2)*Self(n-2): n in [1..31]]; // G. C. Greubel, Jan 31 2023
    
  • Maple
    a:= n-> n! *(Matrix([[1,1], [1/2,0]])^n)[1,1]:
    seq(a(n), n=0..40);  # Alois P. Heinz, Jun 01 2009
    a:= gfun:-rectoproc({a(n) = n*a(n-1)+(n*(n-1)/2)*a(n-2),a(0)=1,a(1)=1},a(n),remember):
    seq(a(n), n=0..40); # Robert Israel, Nov 01 2015
  • Mathematica
    Table[n!*SeriesCoefficient[-2/(-2+2*x+x^2),{x,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 13 2012 *)
    Round@Table[n! ((1+Sqrt[3])^(n+1) - (1-Sqrt[3])^(n+1))/(2^(n+1) Sqrt[3]), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 31 2015 *)
  • PARI
    Vec(serlaplace((2/(2-2*x-x^2) + O(x^30)))) \\ Michel Marcus, Nov 02 2015
    
  • SageMath
    A002605=BinaryRecurrenceSequence(2,2,0,1)
    def A080599(n): return factorial(n)*A002605(n+1)/2^n
    [A080599(n) for n in range(41)] # G. C. Greubel, Jan 31 2023

Formula

a(n) = n*a(n-1) + (n*(n-1)/2)*a(n-2). - Vladeta Jovovic, Aug 22 2003
E.g.f.: 1/(1-x-x^2/2). - Richard Stanley, May 09 2011
a(n) ~ n!*((1+sqrt(3))/2)^(n+1)/sqrt(3). - Vaclav Kotesovec, Oct 13 2012
a(n) = n!*((1+sqrt(3))^(n+1) - (1-sqrt(3))^(n+1))/(2^(n+1)*sqrt(3)). - Vladimir Reshetnikov, Oct 31 2015
a(n) = A090932(n) * A002530(n+1). - Robert Israel, Nov 01 2015

A048788 a(2n+1) = a(2n) + a(2n-1), a(2n) = 2*a(2n-1) + a(2n-2); a(n) = n for n = 0, 1.

Original entry on oeis.org

0, 1, 2, 3, 8, 11, 30, 41, 112, 153, 418, 571, 1560, 2131, 5822, 7953, 21728, 29681, 81090, 110771, 302632, 413403, 1129438, 1542841, 4215120, 5757961, 15731042, 21489003, 58709048, 80198051, 219105150, 299303201, 817711552, 1117014753
Offset: 0

Views

Author

Robin Trew (trew(AT)hcs.harvard.edu), Dec 11 1999

Keywords

Comments

Numerators of continued fraction convergents to sqrt(3) - 1 (A160390). See A002530 for denominators. - N. J. A. Sloane, Dec 17 2007
Convergents are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, ... - Clark Kimberling, Sep 21 2013
A strong divisibility sequence, that is gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. - Peter Bala, Jun 06 2014
From Sarah-Marie Belcastro, Feb 15 2022: (Start)
a(n) is also the number of perfect matchings of an edge-labeled 2 X (n-1) Mobius band grid graph, or equivalently the number of domino tilings of a 2 X (n-1) Mobius band grid. (The twist is on the length-n side.)
a(n) is also the output of Lu and Wu's formula for the number of perfect matchings of an m X n Mobius band grid, specialized to m = 2 with the twist on the length-n side.
2*a(n) is the number of perfect matchings of an edge-labeled 2 X (n-1) projective planar grid graph, or equivalently the number of domino tilings of a 2 X (n-1) projective planar grid. (End)

References

  • Russell Lyons, A bird's-eye view of uniform spanning trees and forests, in Microsurveys in Discrete Probability, AMS, 1998.

Crossrefs

Bisections are A001835 and A052530.

Programs

  • GAP
    a:=[0,1,2,3];; for n in [5..40] do a[n]:=4a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
  • Magma
    I:=[0,1,2,3]; [n le 4 select I[n] else 4*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Dec 10 2013
    
  • Maple
    seq( simplify( `if`(`mod`(n,2)=0, 2*ChebyshevU((n-2)/2, 2), ChebyshevU((n-1)/2, 2) - ChebyshevU((n-3)/2, 2)) ), n=0..40); # G. C. Greubel, Dec 23 2019
  • Mathematica
    Numerator[NestList[(2/(2 + #))&, 0, 40]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
    CoefficientList[Series[x(1+2x-x^2)/(1-4x^2+x^4), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 10 2013 *)
    a0[n_]:= ((3+Sqrt[3])*(2-Sqrt[3])^n-((-3+Sqrt[3])*(2+Sqrt[3])^n))/6 // Simplify
    a1[n_]:= 2*Sum[a0[i], {i, 1, n}]
    Flatten[MapIndexed[{a1[#-1],a0[#]}&,Range[20]]] (* Gerry Martens, Jul 10 2015 *)
    Round@Table[With[{r=1+Sqrt[2], s=1+Sqrt[3]}, ((r + (-1)^n/r) s^n/2^(n/2) - (1/r + (-1)^n r) 2^(n/2)/s^n) Sqrt[6]/12], {n, 0, 20}] (* or *) LinearRecurrence[ {0,4,0,-1}, {0,1,2,3}, 40] (* Vladimir Reshetnikov, May 11 2016 *)
    Table[If[EvenQ[n], 2*ChebyshevU[(n-2)/2, 2], ChebyshevU[(n-1)/2, 2] - ChebyshevU[(n-3)/2, 2]], {n, 0, 40}] (* G. C. Greubel, Dec 23 2019 *)
  • PARI
    main(size)=v=vector(size); v[1]=0;v[2]=1;v[3]=2;v[4]=3;for(i=5, size, v[i]=4*v[i-2] - v[i-4]); v; \\ Anders Hellström, Jul 11 2015
    
  • PARI
    a=vector(50); a[1]=1; a[2]=2; for(n=3, #a, if(n%2==1, a[n]=a[n-1]+a[n-2], a[n]=2*a[n-1]+a[n-2])); concat(0, a) \\ Colin Barker, Jan 30 2016
    
  • PARI
    a(n)=([0,1,0,0;0,0,1,0;0,0,0,1;-1,0,4,0]^n*[0;1;2;3])[1,1] \\ Charles R Greathouse IV, Mar 16 2017
    
  • PARI
    apply( {A048788(n)=imag((2+quadgen(12))^(n\/2)*if(bittest(n, 0), quadgen(12)-1, 2))}, [0..30]) \\ M. F. Hasler, Nov 04 2019
    
  • PARI
    {a(n) = my(s=1,m=n); if(n<0,s=-(-1)^n; m=-n); polcoeff(x*(1+2*x-x^2)/(1-4*x^2+x^4) + x*O(x^m), m)*s}; /* Michael Somos, Sep 17 2020 */
    
  • Sage
    @CachedFunction
    def a(n):
        if (mod(n,2)==0): return 2*chebyshev_U((n-2)/2, 2)
        else: return chebyshev_U((n-1)/2, 2) - chebyshev_U((n-3)/2, 2)
    [a(n) for n in (0..40)] # G. C. Greubel, Dec 23 2019
    

Formula

G.f.: x*(1+2*x-x^2)/(1-4*x^2+x^4). - Paul Barry, Sep 18 2009
a(n) = 4*a(n-2) - a(n-4). - Vincenzo Librandi, Dec 10 2013
a(2*n-1) = A001835(n); a(2*n) = 2*A001353(n). - Peter Bala, Jun 06 2014
From Gerry Martens, Jul 11 2015: (Start)
Interspersion of 2 sequences [a1(n-1),a0(n)] for n>0:
a0(n) = ((3+sqrt(3))*(2-sqrt(3))^n-((-3+sqrt(3))*(2+sqrt(3))^n))/6.
a1(n) = 2*Sum_{i=1..n} a0(i). (End)
a(n) = ((r + (-1)^n/r)*s^n/2^(n/2) - (1/r + (-1)^n*r)*2^(n/2)/s^n)*sqrt(6)/12, where r = 1 + sqrt(2), s = 1 + sqrt(3). - Vladimir Reshetnikov, May 11 2016
a(n) = 2*ChebyshevU(n-1, 2) if n is even and ChebyshevU(n, 2) - ChebyshevU(n-1, 2) if n in odd. - G. C. Greubel, Dec 23 2019
a(n) = -(-1)^n*a(-n) for all n in Z. - Michael Somos, Sep 17 2020

Extensions

Denominator of g.f. corrected by Paul Barry, Sep 18 2009
Incorrect g.f. deleted by Colin Barker, Aug 10 2012

A152063 Triangle read by rows. Coefficients of the Fibonacci product polynomials F(n) = Product_{k=1..(n - 1)/2} (1 + 4*cos^2(k*Pi/n)).

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 5, 5, 1, 6, 8, 1, 8, 19, 13, 1, 9, 25, 21, 1, 11, 42, 65, 34, 1, 12, 51, 90, 55, 1, 14, 74, 183, 210, 89, 1, 15, 86, 234, 300, 144, 1, 17, 115, 394, 717, 654, 233, 6, 18, 130, 480, 951, 954, 377, 1, 20, 165, 725, 1825, 2622, 1985, 610, 1, 21, 183, 855
Offset: 1

Views

Author

Gary W. Adamson and Roger L. Bagula, Nov 22 2008

Keywords

Comments

The triangle A125076 is formed by reading upward sloping diagonals. - Gary W. Adamson, Nov 26 2008
Bisection of the triangle: odd-indexed rows are reversals of the rows of A126124, even-indexed rows are the reversals of the rows of A123965. - Gary W. Adamson, Aug 15 2010

Examples

			First few rows of the triangle are:
1;
1;
1, 2;
1, 3;
1, 5, 5;
1, 6, 8;
1, 8, 19, 13;
1, 9, 25, 21;
1, 11, 42, 65, 34;
1, 12, 51, 90, 55;
1, 14, 74, 183, 210, 89;
1, 15, 86, 234, 300, 144;
1, 17, 115, 394, 717, 654, 233;
1, 18, 130, 480, 951, 954, 377;
1, 20, 165, 725, 1825, 2622, 1985, 610;
1, 21, 183, 855, 2305, 3573, 2939, 987;
...
By row, alternate signs (+,-,+,-,...) with descending exponents. Rows with n terms have exponents (n-1), (n-2), (n-3),...;
Example: There are two rows with 4 terms corresponding to the polynomials
x^3 - 8x^2 + 19x - 13 (roots associated with the heptagon); and
x^3 - 9x^2 + 25x - 21 (roots associated with the 9-gon (nonagon)).
		

Crossrefs

Programs

  • Maple
    P := proc(n) option remember; if n < 5 then return
    ifelse(n < 3, 1, ifelse(n = 3, 1 + 2*q, 1 + 3*q)) fi;
    (1 + 3*q)*P(n - 2) - q^2*P(n - 4) end:
    T := n -> local k; seq(coeff(P(n), q, k), k = 0..(n-1)/2):
    for n from 1 to 12 do T(n) od;  # (after F. Chapoton)  Peter Luschny, May 27 2024
    # Alternative:
    P := n -> local k; add(binomial(n-k,k)*(1+x)^(floor(n/2)-k)*x^k, k=0..floor(n/2)):
    T := n -> local k; seq(coeff(P(n), x, k), k = 0..n/2):
    for n from 0 to 12 do T(n) od; # (after F. Chapoton) Peter Luschny, May 28 2024

Formula

Recurrence (as monic polynomials) P(n+4) = (1 + 3*q)*P(n+2) - q^2*P(n). - F. Chapoton, May 27 2024
As monic polynomials, these are the numerators of the polynomials from A011973 evaluated at 1/(1+q). - F. Chapoton, May 28 2024

A189006 Array A(m,n) read by antidiagonals: number of domino tilings of the m X n grid with upper left corner removed iff m*n is odd, (m>=0, n>=0).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 5, 4, 5, 1, 1, 1, 1, 8, 11, 11, 8, 1, 1, 1, 1, 13, 15, 36, 15, 13, 1, 1, 1, 1, 21, 41, 95, 95, 41, 21, 1, 1, 1, 1, 34, 56, 281, 192, 281, 56, 34, 1, 1, 1, 1, 55, 153, 781, 1183, 1183, 781, 153, 55, 1, 1, 1, 1, 89, 209, 2245, 2415, 6728, 2415, 2245, 209, 89, 1, 1
Offset: 0

Views

Author

Alois P. Heinz, Apr 15 2011

Keywords

Examples

			A(3,3) = 4, because there are 4 domino tilings of the 3 X 3 grid with upper left corner removed:
  . .___. . .___. . .___. . .___.
  ._|___| ._|___| ._| | | ._|___|
  | |___| | | | | | |_|_| |___| |
  |_|___| |_|_|_| |_|___| |___|_|
Array begins:
  1, 1,  1,  1,   1,    1,    1, ...
  1, 1,  1,  1,   1,    1,    1, ...
  1, 1,  2,  3,   5,    8,   13, ...
  1, 1,  3,  4,  11,   15,   41, ...
  1, 1,  5, 11,  36,   95,  281, ...
  1, 1,  8, 15,  95,  192, 1183, ...
  1, 1, 13, 41, 281, 1183, 6728, ...
		

Crossrefs

Rows m=0+1, 2-12 give: A000012, A000045(n+1), A002530(n+1), A005178(n+1), A189003, A028468, A189004, A028470, A189005, A028472, A210724, A028474.
Main diagonal gives: A189002.

Programs

  • Maple
    with(LinearAlgebra):
    A:= proc(m, n) option remember; local i, j, s, t, M;
          if m=0 or n=0 then 1
        elif m1 or j>1 or s=0 then
                   if j
    				
  • Mathematica
    A[1, 1] = 1; A[m_, n_] := A[m, n] = Module[{i, j, s, t, M}, Which[m == 0 || n == 0, 1, m < n, A[n, m], True, s = Mod[n*m, 2];M[i_, j_] /; j < i := -M[j, i]; M[, ] = 0; For[i = 1, i <= n, i++, For[j = 1, j <= m, j++, t = (i-1)*m+j-s; If[i > 1 || j > 1 || s == 0, If[j < m, M[t, t+1] = 1]; If[i < n, M[t, t+m] = 1-2*Mod[j, 2]]]]]; Sqrt[Det[Array[M, {n*m-s, n*m-s}]]]]]; Table[Table[A[m, d-m], {m, 0, d}], {d, 0, 15}] // Flatten (* Jean-François Alcover, Dec 26 2013, translated from Maple *)

A096147 Prime denominators of the rational convergents to sqrt(3).

Original entry on oeis.org

3, 11, 41, 571, 2131, 110771, 1542841, 15558008491, 808717138331, 1663476485027525263506023431291963826940251, 33648911495192637123958375850447995878147331088460770783226682531
Offset: 1

Views

Author

Cino Hilliard, Jul 24 2004

Keywords

Comments

Next term is too large to include.
The next term has 79 digits. - Harvey P. Dale, Jul 06 2019
This is the prime subsequence of A002530. - Ray Chandler, Aug 01 2004
Primes p such that 3*p^2 - 2 is a square. - Vincenzo Librandi, May 21 2013

Crossrefs

Programs

  • Mathematica
    Select[Denominator[Convergents[Sqrt[3],300]],PrimeQ] (* Harvey P. Dale, Jul 06 2019 *)

Extensions

Offset corrected by Amiram Eldar, Jul 11 2024

A102341 Areas of 'close-to-equilateral' integer triangles.

Original entry on oeis.org

12, 120, 1848, 25080, 351780, 4890480, 68149872, 949077360, 13219419708, 184120982760, 2564481115560, 35718589344360, 497495864091732, 6929223155685600, 96511629630137568, 1344233586759971040, 18722758603319903340
Offset: 1

Views

Author

Johannes Koelman (Joc_Kay(AT)hotmail.com), Feb 20 2005

Keywords

Comments

A close-to-equilateral integer triangle is defined to be a triangle with integer sides and integer area such that the largest and smallest sides differ in length by unity. The first five close-to-equilateral integer triangles have sides (5, 5, 6), (17, 17, 16), (65, 65, 66), (241, 241, 240) and (901, 901, 902).
After these first five triangles, there are two more (namely (3361,3361,3360,4890480) and (12545,12545,12546,68149872)). - Nícolas V. Calsavara, Jul 13 2023
Then, the next four terms are {three sides a<=b<=c and area}: {46816, 46817, 46817, 949077360}, {174725, 174725, 174726, 13219419708}, {652080, 652081, 652081, 184120982760}, {2433601, 2433601, 2433602, 2564481115560}. Also, if we allow degenerate triangles (area 0), the first case would be {1,1,2,0}. We have 12 cases and a weak conjecture is that the total number of the 'close-to-equilateral' triangles is finite. - Zak Seidov, Feb 23 2005
This is an infinite series; two sides are equal in length to the hypotenuse of almost 30-60 triangles and the third side alternates between that length +/- 1. - Dan Sanders (dan(AT)ified.ca), Oct 22 2005
Heron's formula: a triangle with side lengths (x,y,z) has area A = sqrt(s*(s-x)*(s-y)*(s-z)) where s = (x+y+z)/2. For this sequence we assume integer side-lengths x = y = z +/- 1. Then for A to also be an integer, x+y+z must be even, so we can assume z = 2k for some positive integer k. Now s = (x+y+z)/2 = 3k +/- 1 and A = sqrt((3*k +/- 1)*k*k*(k +/- 1)) = k*sqrt(3*k^2 +/- 4*k + 1). To determine when this is an integer, set 3*k^2 +/- 4*k + 1 = d^2. If we multiply both sides by 3, it is easier to complete the square: (3*k +/- 2)^2 - 1 = 3*d^2. Now we are looking for solutions to the Pell equation c^2 - 3*d^2 = 1 with c = 3*k +/- 2, for which there are infinitely many solutions: use the upper principal convergents of the continued fraction expansion of sqrt(3) (A001075/A001353). - Danny Rorabaugh, Oct 16 2015

Examples

			a(2) = 120 because 120 is the area of a triangle with side lengths of 16, 17 and 17.
		

Crossrefs

For the continued fraction expansion of sqrt(3), cf. A002530, A002531, A040001.

Formula

(2/3) [ A007655(n+2) - (-1)^n*A001353(n+1) ] (conjectured). - Ralf Stephan, May 17 2007
Empirical g.f.: 12*x / ((x^2-14*x+1)*(x^2+4*x+1)). - Colin Barker, Apr 10 2013
a(n) = A001353(n+1)*A195499(n) = A001353(n+1)*A120892(n+1) - Danny Rorabaugh, Oct 16 2015

Extensions

More terms from Zak Seidov, Feb 23 2005
More terms from Dan Sanders (dan(AT)ified.ca), Oct 22 2005

A109437 a(-1) = a(0) = 0, a(1) = 1; a(n) = 5a(n-1) - 5a(n-2) + a(n-3) + 2*(-1)^(n+1), alternatively a(n) = 3a(n-1) + 3a(n-2) - a(n-3).

Original entry on oeis.org

0, 1, 3, 12, 44, 165, 615, 2296, 8568, 31977, 119339, 445380, 1662180, 6203341, 23151183, 86401392, 322454384, 1203416145, 4491210195, 16761424636, 62554488348, 233456528757, 871271626679, 3251629977960, 12135248285160, 45289363162681, 169022204365563, 630799454299572
Offset: 0

Views

Author

Creighton Dement, Jun 28 2005

Keywords

Comments

See A105968 for a similar sequence. Observe the four periodic sequences (1,1,1,1,); (-1,-1,-1,-1); (1,-1,1,-1,); (-1,1,-1,1,); (a(n)) is the (Type 1A) jbasejfor-transform of the periodic sequence (1,1,1,1) with respect to the floretion given in the program code. A109438 is the (Type 1A) jbasejfor-transform of the periodic sequence (-1,-1,-1,-1) with respect to the floretion given in the program code. A001834 is the (Type 1A) jbasejfor-transform of the periodic sequence (1,-1,1,-1) with respect to the floretion given in the program code. A102871 is the (Type 1A) jbasejfor-transform of the periodic sequence (-1,1,-1,1) with respect to the floretion given in the program code.
Floretion Algebra Multiplication Program, FAMP Code: (-1)^(n+1)jbasejfor[ + .5'ii' + .5'kk' + .5'ij' + .5'ji' + .5'jk' + .5'kj'] 1vesfor = (1,1,1,1,)

Crossrefs

Programs

  • Maple
    with(numtheory):a := cfrac (tan(Pi/3),60): > b := cfrac (tan(Pi/6),60): > seq(nthnumer (b,i)*nthdenom (a,i), i=0..24 ); # Zerinvary Lajos, Feb 08 2007
  • Mathematica
    LinearRecurrence[{3,3,-1},{0,1,3},40] (* Harvey P. Dale, Apr 21 2018 *)
  • PARI
    {a(n) = local(s=1); if( n<0, n = -1 - n; s=-1); s * polcoeff( x / ((x + 1) * (x^2 -4*x + 1)) + x * O(x^n), n)} /* Michael Somos, Jul 27 2012 */

Formula

G.f.: x/((x+1)*(x^2-4*x+1)).
a(n) = A002530(n)*A002530(n+1). - Zerinvary Lajos, Feb 08 2007
a(-1 - n) = -a(n). a(2*n) = A011916(n). a(2*n + 1) = -A011916(-1 -n). - Michael Somos, Jul 27 2012
6*a(n) = A001353(n)+A001353(n+1)-(-1)^n. - R. J. Mathar, Sep 07 2016
a(n) = ((1 + sqrt(3))*(2 + sqrt(3))^n + (1 - sqrt(3))*(2 - sqrt(3))^n - 2*(-1)^n)/12. - Stefano Spezia, Sep 19 2023
a(n)+a(n+1) = A001353(n+1). - R. J. Mathar, Aug 31 2025

A041023 Denominators of continued fraction convergents to sqrt(15).

Original entry on oeis.org

1, 1, 7, 8, 55, 63, 433, 496, 3409, 3905, 26839, 30744, 211303, 242047, 1663585, 1905632, 13097377, 15003009, 103115431, 118118440, 811826071, 929944511, 6391493137, 7321437648, 50320119025
Offset: 0

Views

Author

Keywords

Comments

The following remarks assume an offset of 1. This is the sequence of Lehmer numbers U_n(sqrt(R),Q) for the parameters R = 6 and Q = -1; it is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all positive integers n and m. Consequently, this is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, May 28 2014

Crossrefs

Programs

  • Mathematica
    Denominator[NestList[(6/(6+#))&,0,60]] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2010 *)
    a0[n_] := (-((-5+Sqrt[15])*(4+Sqrt[15])^n)+(4-Sqrt[15])^n*(5+Sqrt[15]))/10 // Simplify
    a1[n_] := (-(4-Sqrt[15])^n+(4+Sqrt[15])^n)/(2*Sqrt[15]) // Simplify
    Flatten[MapIndexed[{a0[#], a1[#]} &,Range[20]]] (* Gerry Martens, Jul 11 2015 *)
    Convergents[Sqrt[15],30]//Denominator (* Harvey P. Dale, Aug 13 2016 *)

Formula

G.f.: (1+x-x^2)/(1-8*x^2+x^4). - Colin Barker, Jan 01 2012
From Peter Bala, May 28 2014: (Start)
The following remarks assume an offset of 1.
Let alpha = ( sqrt(6) + sqrt(10) )/2 and beta = ( sqrt(6) - sqrt(10) )/2 be the roots of the equation x^2 - sqrt(6)*x - 1 = 0. Then a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, while a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even.
a(n) = product {k = 1..floor((n-1)/2)} ( 6 + 4*cos^2(k*Pi/n) ).
Recurrence equations: a(0) = 0, a(1) = 1 and for n >= 1, a(2*n) = a(2*n - 1) + a(2*n - 2) and a(2*n + 1) = 6*a(2*n) + a(2*n - 1). (End)
From Gerry Martens, Jul 11 2015: (Start)
Interspersion of 2 sequences [a0(n),a1(n)] for n>0:
a0(n) = (-((-5+sqrt(15))*(4+sqrt(15))^n)+(4-sqrt(15))^n*(5+sqrt(15)))/10.
a1(n) = (-(4-sqrt(15))^n+(4+sqrt(15))^n)/(2*sqrt(15)). (End)
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