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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A002532 a(n) = 2*a(n-1) + 5*a(n-2), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 9, 28, 101, 342, 1189, 4088, 14121, 48682, 167969, 579348, 1998541, 6893822, 23780349, 82029808, 282961361, 976071762, 3366950329, 11614259468, 40063270581, 138197838502, 476712029909, 1644413252328, 5672386654201, 19566839570042, 67495612411089
Offset: 0

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Author

Keywords

Comments

The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 6 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(6). - Cino Hilliard, Sep 25 2005
For n>=2, number of ordered partitions of n-1 into parts of sizes 1 and 2 where there are two types of 1 (singletons) and five types of 2 (twins). For example, the number of possible configurations of families of n-1 male (M) and female (F) offspring considering only single births and twins, where the birth order of M/F/pair-of-twins is considered and there are five types of twins; namely, both F (identical twins), both F (fraternal twins), both M (identical), both M (fraternal), or one F and one M - where birth order within a pair of twins itself is disregarded. In particular, for a(3)=9, two children could be either: (1) F, then M; (2) M, then F; (3) F,F; (4) M,M; (5) F,F identical twins; (6) F,F fraternal twins; (7) M,M identical twins; (8) M,M fraternal twins; or (9) M,F twins (emphasizing that birth order is irrelevant here when children are the same gender, when two children are within the same pair of twins and when pairs of twins have both the same gender(s) and identical-vs-fraternal characteristics). - Rick L. Shepherd, Sep 19 2004
Pisano period lengths: 1, 2, 3, 4, 4, 6, 24, 8, 3, 4, 120, 12, 56, 24, 12, 16, 288, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012

Examples

			G.f. = x + 2*x^2 + 9*x^3 + 28*x^4 + 101*x^5 + 342*x^6 + 1189*x^7 + ...
		

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
  • 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).
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Crossrefs

Cf. A015581 (similar application, but no distinguishing identical vs. fraternal twins).
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    [Floor(((1+Sqrt(6))^n-(1-Sqrt(6))^n)/(2*Sqrt(6))): n in [0..30]]; // Vincenzo Librandi, Aug 15 2011
    
  • Magma
    [n le 2 select n-1 else 2*Self(n-1) + 5*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
  • Maple
    A002532:=-z/(-1+2*z+5*z**2); # Conjectured by Simon Plouffe in his 1992 dissertation
    # second program
    seq(simplify(2^(n-1) * hypergeom([1 - (1/2)*n, 1/2 - (1/2)*n], [1 - n], -5)), n = 2..25); # Peter Bala, Jul 06 2025
  • Mathematica
    Expand[Table[((1 + Sqrt[6])^n - (1 - Sqrt[6])^n)/(2Sqrt[6]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
    a[n_]:=(MatrixPower[{{1,2},{1,-3}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]],{n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
    LinearRecurrence[{2,5},{0,1},30] (* Harvey P. Dale, Nov 03 2011 *)
  • PARI
    Vec(1/(1-2*x-5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Apr 17 2012
    
  • Sage
    from sage.combinat.sloane_functions import recur_gen2; it = recur_gen2(0,1,2,5); [next(it) for i in range(30)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [lucas_number1(n,2,-5) for n in range(0, 26)] # Zerinvary Lajos, Apr 22 2009
    

Formula

From Mario Catalani (mario.catalani(AT)unito.it), Jun 14 2003: (Start)
a(2*n+1) = 5*a(n)^2 + a(n+1)^2.
6*a(2*n+1) = 5*A002533(n)^2 + A002533(n+1)^2. (End)
From Paul Barry, Sep 20 2003: (Start)
G.f.: x/(1-2*x-5*x^2).
E.g.f.: exp(x)*sinh(sqrt(6)*x)/sqrt(6).
a(n) = ((1+sqrt(6))^n - (1-sqrt(6))^n)/(2*sqrt(6)). (End)
a(n) = Sum_{k=0..n} binomial(n, 2*k+1)*6^k. - Paul Barry, Sep 29 2004
G.f.: G(0)*x/(2*(1-x)), where G(k)= 1 + 1/(1 - x*(6*k-1)/(x*(6*k+5) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
From Peter Bala, Jul 06 2025: (Start)
For n >= 0, a(n+1) = (2^n) * Sum_{k = 0..floor(n/2)} binomial(n-k, k) * (5/4)^k.
For n >= 2, a(n) = 2^(n-1) * hypergeom([1 - (1/2)*n, 1/2 - (1/2)*n], [1 - n], -5).
Sum_{n >= 1} (-5)^n/(a(n)*a(n+1)) = -(sqrt(6) - 1).
Sum_{n >= 1} 5^n/(a(n)*a(n+2)) = 5/4; Sum_{n >= 1} 5^n/(a(n)*a(n+4)) = 755/7056.
G.f. A(x) = x*exp(Sum_{n >= 1} a(2*n)/a(n)*x^n/n) = x + 2*x^2 + 9*x^3 + 28*x^4 + .... (End)

Extensions

More terms from Rick L. Shepherd, Sep 19 2004

A083099 a(n) = 2*a(n-1) + 6*a(n-2), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 10, 32, 124, 440, 1624, 5888, 21520, 78368, 285856, 1041920, 3798976, 13849472, 50492800, 184082432, 671121664, 2446737920, 8920205824, 32520839168, 118562913280, 432250861568, 1575879202816, 5745263575040
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003

Keywords

Comments

a(n+1) = a(n) + A083098(n+1). A083098(n+1)/a(n) converges to sqrt(7).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 7 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(7). - Cino Hilliard, Sep 25 2005
Pisano period lengths: 1, 1, 2, 1, 12, 2, 7, 1, 6, 12, 60, 2,168, 7, 12, 1,288, 6, 18, 12, ... - R. J. Mathar, Aug 10 2012
a(n) is divisible by 2^ceiling(n/2), see formula below. - Ralf Stephan, Dec 24 2013
Connect the center of a regular hexagon with side length 1 with its six vertices. a(n) is the number of paths of length n from the center to any of its vertices. Number of paths of length n from the center to itself is 6*a(n-1). - Jianing Song, Apr 20 2019

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.

Programs

  • Magma
    [n le 2 select n-1 else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018
    
  • Maple
    A083099 := proc(n)
        option remember;
        if n <= 1 then
            n;
        else
            2*procname(n-1)+6*procname(n-2) ;
        end if;
    end proc: # R. J. Mathar, Sep 23 2016
  • Mathematica
    CoefficientList[Series[x/(1-2x-6x^2), {x, 0, 25}], x] (* Adapted for offset 0 by Vincenzo Librandi, Feb 07 2014 *)
    Expand[Table[((1 + Sqrt[7])^n - (1 - Sqrt[7])^n)7/(14Sqrt[7]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
    LinearRecurrence[{2,6}, {0,1}, 25] (* Sture Sjöstedt, Dec 06 2011 *)
  • PARI
    a(n)=([0,1; 6,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, May 10 2016
    
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(x/(1-2*x-6*x^2))) \\ G. C. Greubel, Jan 24 2018
    
  • Sage
    [lucas_number1(n,2,-6) for n in range(0, 25)] # Zerinvary Lajos, Apr 22 2009
    
  • SageMath
    A083099=BinaryRecurrenceSequence(2,6,0,1)
    [A083099(n) for n in range(41)] # G. C. Greubel, Jun 01 2023

Formula

G.f.: x/(1 - 2*x - 6*x^2).
From Paul Barry, Sep 29 2004: (Start)
E.g.f.: (d/dx)(exp(x)*sinh(sqrt(7)*x)/sqrt(7));
a(n-1) = Sum_{k=0..n} binomial(n, 2k+1)*7^k. (End)
Simplified formula: a(n) = ((1 + sqrt(7))^n - (1 - sqrt(7))^n)/sqrt(28). - Al Hakanson (hawkuu(AT)gmail.com), Jan 05 2009
G.f.: G(0)*x/(2*(1-x)), where G(k) = 1 + 1/(1 - x*(7*k-1)/(x*(7*k+6) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(2n) = 2^n * A154245(n), a(2n+1) = 2^n * (5*A154245(n) - 9*A154245(n-1)). - Ralf Stephan, Dec 24 2013
a(n) = Sum_{k=1,3,5,...<=n} binomial(n,k)*7^((k-1)/2). - Vladimir Shevelev, Feb 06 2014
a(n) = i^(n-1)*6^((n-1)/2)*ChebyshevU(n-1, -i/sqrt(6)). - G. C. Greubel, Jun 01 2023

A015519 a(n) = 2*a(n-1) + 7*a(n-2), with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 2, 11, 36, 149, 550, 2143, 8136, 31273, 119498, 457907, 1752300, 6709949, 25685998, 98341639, 376485264, 1441362001, 5518120850, 21125775707, 80878397364, 309637224677, 1185423230902, 4538307034543, 17374576685400
Offset: 0

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Author

Keywords

Comments

The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the denominators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 8 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(8). - Cino Hilliard, Sep 25 2005
Pisano period lengths: 1, 2, 8, 4, 24, 8, 3, 8, 24, 24, 15, 8, 168, 6, 24, 16, 16, 24, 120, 24, ... . - R. J. Mathar, Aug 10 2012

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

The following sequences (and others) belong to the same family: A000129, A001333, A002532, A002533, A002605, A015518, A015519, A026150, A046717, A063727, A083098, A083099, A083100, A084057.

Programs

  • Magma
    [ n eq 1 select 0 else n eq 2 select 1 else 2*Self(n-1)+7*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Aug 23 2011
    
  • Mathematica
    LinearRecurrence[{2,7},{0,1},30] (* Harvey P. Dale, Oct 09 2017 *)
  • PARI
    a(n)=([0,1; 7,2]^n*[0;1])[1,1] \\ Charles R Greathouse IV, May 10 2016
  • Sage
    [lucas_number1(n,2,-7) for n in range(0, 25)] # Zerinvary Lajos, Apr 22 2009
    

Formula

From Mario Catalani (mario.catalani(AT)unito.it), Apr 23 2003: (Start)
a(n) = a(n-1) + A083100(n-2), n>1.
A083100(n)/a(n+1) converges to sqrt(8). (End)
From Paul Barry, Jul 17 2003: (Start)
G.f.: x/ ( 1-2*x-7*x^2 ).
a(n) = ((1+2*sqrt(2))^n-(1-2*sqrt(2))^n)*sqrt(2)/8. (End)
E.g.f.: exp(x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Nov 20 2003
Second binomial transform is A000129(2n)/2 (A001109). - Paul Barry, Apr 21 2004
a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k)*(7/2)^k*2^(n-k-1). - Paul Barry, Jul 17 2004
a(n) = Sum_{k=0..n} binomial(n, 2*k+1)*8^k. - Paul Barry, Sep 29 2004
G.f.: G(0)*x/(2*(1-x)), where G(k)= 1 + 1/(1 - x*(8*k-1)/(x*(8*k+7) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013

A002533 a(n) = 2*a(n-1) + 5*a(n-2), with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 7, 19, 73, 241, 847, 2899, 10033, 34561, 119287, 411379, 1419193, 4895281, 16886527, 58249459, 200931553, 693110401, 2390878567, 8247309139, 28449011113, 98134567921, 338514191407, 1167701222419, 4027973401873, 13894452915841, 47928772841047, 165329810261299
Offset: 0

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Author

Keywords

Comments

The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 6 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(6). - Cino Hilliard, Sep 25 2005
a(n), n>0 = term (1,1) in the n-th power of the 2 X 2 matrix [1,3; 2,1]. - Gary W. Adamson, Aug 06 2010
a(n) is the number of compositions of n when there are 1 type of 1 and 6 types of other natural numbers. - Milan Janjic, Aug 13 2010
Pisano period lengths: 1, 1, 1, 4, 4, 1, 24, 4, 3, 4, 120, 4, 56, 24, 4, 8, 288, 3, 18, 4, ... - R. J. Mathar, Aug 10 2012
a(k*m) is divisible by a(m) if k is odd. - Robert Israel, May 03 2024

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.
  • 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).
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Crossrefs

The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    [(1/2)*Floor((1+Sqrt(6))^n+(1-Sqrt(6))^n): n in [0..30]]; // Vincenzo Librandi, Aug 15 2011
    
  • Magma
    [n le 2 select 1 else 2*Self(n-1) + 5*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
  • Maple
    A002533:=(-1+z)/(-1+2*z+5*z**2); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    f[n_] := Simplify[((1 + Sqrt[6])^n + (1 - Sqrt[6])^n)/2]; Array[f, 28, 0] (* Or *)
    LinearRecurrence[{2, 5}, {1, 1}, 28] (* Or *)
    Table[ MatrixPower[{{1, 2}, {3, 1}}, n][[1, 1]], {n, 0, 25}]
    (* Robert G. Wilson v, Sep 18 2013 *)
  • PARI
    a(n)=([0,1; 5,2]^n*[1;1])[1,1] \\ Charles R Greathouse IV, May 10 2016
    
  • PARI
    x='x+O('x^30); Vec((1-x)/(1-2*x-5*x^2)) \\ G. C. Greubel, Jan 08 2018
    
  • Sage
    [lucas_number2(n,2,-5)/2 for n in range(0, 21)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n)/A002532(n), n>0, converges to sqrt(6). - Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003
From Mario Catalani (mario.catalani(AT)unito.it), May 03 2003: (Start)
G.f.: (1-x)/(1-2*x-5*x^2).
a(n) = (1/2)*((1+sqrt(6))^n + (1-sqrt(6))^n).
a(n)/A083694(n) converges to sqrt(3/2).
a(n)/A083695(n) converges to sqrt(2/3).
a(n) = a(n-1) + 3*A083694(n-1).
a(n) = a(n-1) + 2*A083695(n-1), n>0. (End)
Binomial transform of expansion of cosh(sqrt(6)*x) (A000400, with interpolated zeros). E.g.f.: exp(x)*cosh(sqrt(6)*x) - Paul Barry, May 09 2003
From Mario Catalani (mario.catalani(AT)unito.it), Jun 14 2003: (Start)
a(2*n+1) = 2*a(n)*a(n+1) - (-5)^n.
a(n)^2 - 6*A002532(n)^2 = (-5)^n. (End)
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k) * 6^k. - Paul Barry, Jul 25 2004
a(n) = Sum_{k=0..n} A098158(n,k)*6^(n-k). - Philippe Deléham, Dec 26 2007
If p(1)=1, and p(I)=6, for i>1, and if A is the Hessenberg matrix of order n defined by: A(i,j) = p(j-i+1) for i<=j, A(i,j)=-1 for i=j+1, and A(i,j)=0 otherwise. Then, for n>=1, a(n) = det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(6*k-1)/(x*(6*k+5) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013

A103435 a(n) = 2^n * Fibonacci(n).

Original entry on oeis.org

0, 2, 4, 16, 48, 160, 512, 1664, 5376, 17408, 56320, 182272, 589824, 1908736, 6176768, 19988480, 64684032, 209321984, 677380096, 2192048128, 7093616640, 22955425792, 74285318144, 240392339456, 777925951488, 2517421260800
Offset: 0

Views

Author

Ralf Stephan, Feb 08 2005

Keywords

Comments

Cardinality of set of bracelets of size at most n that are tiled with two types of colored squares and four types of colored dominoes.
a(n) is also the diagonal element of the matrix A(i,j) whose first row (i=1) and first column (j=1) are the Fibonacci numbers: A(1,k)=A(k,1)=fib(k) and whose generic element is the sum of element in adjacent (preceding) row and column minus the absolute value of their difference. So a(n) = A(n,n) = A(i-1,j)+A(i,j-1)-abs(A(i-1,j)-A(i,j-1)). - Carmine Suriano, May 13 2010
a(n) is the coefficient of x in the reduction by x^2->x+1 of the polynomial p(n,x) given for d=sqrt(x+1) by p(n,x)=((x+d)^n-(x-d)^n)/(2d), for n>=1. The constant terms under this reduction are the absolute values of terms of A086344. See A192232 for a discussion of reduction. - Clark Kimberling, Jun 29 2011
The exponential convolution of A000032 and A000045. - Vladimir Reshetnikov, Oct 06 2016

Examples

			a(5)=160=A(5,5)=A(4,5)+A(5,4)-abs[A(4,5)+A(5,4)]=80+80-0. - _Carmine Suriano_, May 13 2010
G.f. = 2*x + 4*x^2 + 16*x^3 + 48*x^4 + 160*x^5 + 512*x^6 + 1664*x^7 + ...
		

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 236, p. 131.

Crossrefs

First differences of A014334.
Partial sums of A087131.

Programs

  • Magma
    [2^n *Fibonacci(n): n in [0..50]]; // Vincenzo Librandi, Apr 04 2011
    
  • Mathematica
    Expand[Table[((1 + Sqrt[5])^n - (1 - Sqrt[5])^n)5/(5 Sqrt[5]), {n, 0, 25}]] (* Zerinvary Lajos, Mar 22 2007 *)
    Table[2^n Fibonacci[n],{n,0,40}] (* or *) LinearRecurrence[{2,4},{0,2},40] (* Harvey P. Dale, Oct 14 2020 *)
  • PARI
    a(n)=fibonacci(n)<Charles R Greathouse IV, Feb 03 2014
    
  • PARI
    concat(0, Vec(2*x/(1-2*x-4*x^2) + O(x^99))) \\ Altug Alkan, May 11 2016

Formula

a(n) = A006483(n) + 1 = 2*A085449(n) = 2*A063727(n-1), n>0.
G.f.: 2*x / (1 - 2*x - 4*x^2).
a(n) = Sum_{i=0..n-1}( 2^i * Lucas(i) ).
a(n) = 2*a(n-1) + 4*a(n-2). - Carmine Suriano, May 13 2010
a(n) = a(-n) * -(-4)^n for all n in Z. - Michael Somos, Sep 20 2014
E.g.f.: 2*sinh(sqrt(5)*x)*exp(x)/sqrt(5). - Ilya Gutkovskiy, May 10 2016
Sum_{n>=1} 1/a(n) = (1/2) * A269991. - Amiram Eldar, Nov 17 2020
a(n) == 2*n (mod 10). - Amiram Eldar, Jan 15 2022
a(n) = Sum_{k=0..n} binomial(n,k) * Fibonacci(k) * Lucas(n-k) (Wall, 1987). - Amiram Eldar, Jan 27 2022

A083098 a(n) = 2*a(n-1) + 6*a(n-2).

Original entry on oeis.org

1, 1, 8, 22, 92, 316, 1184, 4264, 15632, 56848, 207488, 756064, 2757056, 10050496, 36643328, 133589632, 487039232, 1775616256, 6473467904, 23600633344, 86042074112, 313687948288, 1143628341248, 4169384372224, 15200538791936
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 22 2003

Keywords

Comments

a(n+1) = a(n) + 7*A083099(n-1); a(n+1)/A083099(n) converges to sqrt(7).
Binomial transform of expansion of cosh(sqrt(7)x) (A000420 with interpolated zeros: 1, 0, 7, 0, 49, 0, 343, 0, ...).
The same sequence may be obtained by the following process. Starting a priori with the fraction 1/1, the numerators of fractions built according to the rule: add top and bottom to get the new bottom, add top and 7 times the bottom to get the new top. The limit of the sequence of fractions is sqrt(7). - Cino Hilliard, Sep 25 2005
a(n) is the number of compositions of n when there are 1 type of 1 and 7 types of other natural numbers. - Milan Janjic, Aug 13 2010

References

  • John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, see p. 16.

Crossrefs

The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
  • Mathematica
    CoefficientList[Series[(1+6x)/(1-2x-6x^2), {x, 0, 25}], x]
    LinearRecurrence[{2, 6}, {1, 1}, 25] (* Sture Sjöstedt, Dec 06 2011 *)
    a[n_] := Simplify[((1 + Sqrt[7])^n + (1 - Sqrt[7])^n)/2]; Array[a, 25, 0] (* Robert G. Wilson v, Sep 18 2013 *)
  • PARI
    x='x+O('x^30); Vec((1-x)/(1-2*x-6*x^2)) \\ G. C. Greubel, Jan 08 2018
    
  • Sage
    [lucas_number2(n,2,-6)/2 for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009
    

Formula

G.f.: (1-x)/(1-2*x-6*x^2).
a(n) = (1+sqrt(7))^n/2 + (1-sqrt(7))^n/2.
E.g.f.: exp(x)*cosh(sqrt(7)x).
a(n) = Sum_{k=0..n} A098158(n,k)*7^(n-k). - Philippe Deléham, Dec 26 2007
If p[1]=1, and p[i]=7, (i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n) = det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(7*k-1)/(x*(7*k+6) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013

A083100 a(n) = 2*a(n-1) + 7*a(n-2).

Original entry on oeis.org

1, 9, 25, 113, 401, 1593, 5993, 23137, 88225, 338409, 1294393, 4957649, 18976049, 72655641, 278143625, 1064876737, 4076758849, 15607654857, 59752621657, 228758827313, 875786006225, 3352883803641, 12836269650857, 49142725927201
Offset: 0

Views

Author

Mario Catalani (mario.catalani(AT)unito.it), Apr 23 2003

Keywords

Comments

a(n) = a(n-1) + 8*A015519(n). a(n)/A015519(n+1) converges to sqrt(8).
a(n-1) is the number of compositions of n when there is 1 type of 1 and 8 types of other natural numbers. - Milan Janjic, Aug 13 2010

Crossrefs

Essentially a duplicate of A084058.
The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.

Programs

  • Magma
    I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1) + 7*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 08 2018
  • Mathematica
    CoefficientList[Series[(1 + 7 x)/(1 - 2 x - 7 x^2), {x, 0, 25}], x] (* Or *) a[n_] := Simplify[((1 + Sqrt[8])^n + (1 - Sqrt[8])^n)/2]; Array[a, 25, 0] (* Or *) LinearRecurrence[{2, 7}, {1, 1}, 28] (* Or *) Table[ MatrixPower[{{1, 2}, {4, 1}}, n][[1, 1]], {n, 0, 25}] (* Robert G. Wilson v, Sep 18 2013 *)
  • PARI
    a(n)=([0,1; 7,2]^n*[1;9])[1,1] \\ Charles R Greathouse IV, Apr 06 2016
    
  • PARI
    x='x+O('x^30); Vec((1+7*x)/(1-2*x-7*x^2)) \\ G. C. Greubel, Jan 08 2018
    

Formula

G.f.: (1+7*x)/(1-2*x-7*x^2).
a(n) = binomial transform of 1,8,8,64,64,512. - Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009
If p[1]=1, and p[i]=8,(i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=det A. - Milan Janjic, Apr 29 2010
G.f.: G(0)/(2*x) - 1/x, where G(k)= 1 + 1/(1 - x*(8*k-1)/(x*(8*k+7) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013

A087131 a(n) = 2^n*Lucas(n), where Lucas = A000032.

Original entry on oeis.org

2, 2, 12, 32, 112, 352, 1152, 3712, 12032, 38912, 125952, 407552, 1318912, 4268032, 13811712, 44695552, 144637952, 468058112, 1514668032, 4901568512, 15861809152, 51329892352, 166107021312, 537533612032, 1739495309312
Offset: 0

Views

Author

Paul Barry, Aug 16 2003

Keywords

Comments

Number of ways to tile an n-bracelet with two types of colored squares and four types of colored dominoes.
Inverse binomial transform of even Lucas numbers (A014448).
From L. Edson Jeffery, Apr 25 2011: (Start)
Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,4)=
(0 0 0 0 1)
(0 0 0 2 0)
(0 0 2 0 1)
(0 2 0 2 0)
(2 0 2 0 1).
Then a(n)=(Trace(A^n)-1)/2. Also a(n)=Trace((2*A_(5,1))^n), where A_(5,1)=[(0,1); (1,1)] is also a unit-primitive matrix. (End)
Also the number of connected dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, May 02 2017
Also the number of total dominating sets in the n-sun graph for n >= 3. - Eric W. Weisstein, Apr 27 2018

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, identity 237, p. 132.

Crossrefs

First differences of A006483 and A103435.

Programs

  • Magma
    [2] cat [2^n*Lucas(n): n in [1..30]]; // G. C. Greubel, Dec 18 2017
  • Mathematica
    Table[Tr[MatrixPower[{{2, 2}, {2, 0}}, x]], {x, 1, 20}] (* Artur Jasinski, Jan 09 2007 *)
    Join[{2}, Table[2^n LucasL[n], {n, 20}]] (* Eric W. Weisstein, May 02 2017 *)
    Join[{2}, 2^# LucasL[#] & [Range[20]]] (* Eric W. Weisstein, May 02 2017 *)
    LinearRecurrence[{2, 4}, {2, 12}, {0, 20}] (* Eric W. Weisstein, Apr 27 2018 *)
    CoefficientList[Series[(2 (-1 + x))/(-1 + 2 x + 4 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 27 2018 *)
  • PARI
    for(n=0,30, print1(if(n==0, 2, 2^n*(fibonacci(n+1) + fibonacci(n-1))), ", ")) \\ G. C. Greubel, Dec 18 2017
    
  • PARI
    first(n) = Vec(2*(1-x)/(1-2*x-4*x^2) + O(x^n)) \\ Iain Fox, Dec 19 2017
    
  • Sage
    [lucas_number2(n,2,-4) for n in range(0, 25)] # Zerinvary Lajos, Apr 30 2009
    

Formula

a(n) = 2*A084057(n).
Recurrence: a(n) = 2a(n-1) + 4a(n-2), a(0)=2, a(1)=2.
G.f.: 2*(1-x)/(1-2*x-4*x^2).
a(n) = (1+sqrt(5))^n + (1-sqrt(5))^n.
For n>=2, a(n) = Trace of matrix [({2,2},{2,0})^n]. - Artur Jasinski, Jan 09 2007
a(n) = 2*[A063727(n)-A063727(n-1)]. - R. J. Mathar, Nov 16 2007
a(n) = (5*A052899(n)-1)/2. - L. Edson Jeffery, Apr 25 2011
a(n) = [x^n] ( 1 + x + sqrt(1 + 2*x + 5*x^2) )^n for n >= 1. - Peter Bala, Jun 23 2015
Sum_{n>=1} 1/a(n) = (1/2) * A269992. - Amiram Eldar, Nov 17 2020
From Amiram Eldar, Jan 15 2022: (Start)
a(n) == 2 (mod 10).
a(n) = 5 * A014334(n) + 2.
a(n) = 10 * A014335(n) + 2. (End)

Extensions

Edited by Ralf Stephan, Feb 08 2005

A091914 a(n) = 2*a(n-1) + 12*a(n-2).

Original entry on oeis.org

1, 2, 16, 56, 304, 1280, 6208, 27776, 130048, 593408, 2747392, 12615680, 58200064, 267788288, 1233977344, 5681414144, 26170556416, 120518082560, 555082842112, 2556382674944, 11773759455232, 54224111009792, 249733335482368
Offset: 0

Views

Author

Paul Barry, Feb 12 2004

Keywords

Comments

Binomial transform of 1, 1, 13, 13, 169, 169, ....
The inverse binomial transform of 2^n*c(n), where c(n) is the solution to c(n) = c(n-1) + k*c(n-2), a(0)=1, a(1)=1 is 1, 1, 4k+1, 4k+1, (4k+1)^2, ...

Crossrefs

Programs

  • GAP
    a := [1,2];; for n in [3..30] do a[n] := 2*a[n-1] + 12*a[n-2]; od; a; # Muniru A Asiru, Jan 31 2018
  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!(1/(1-2*x-12*x^2))) // G. C. Greubel, Jan 30 2018
    
  • Maple
    a := proc(n) option remember: if n=0 then 1 elif n=1 then 2 elif n>=2 then 2*procname(n-1) + 12*procname(n-2) fi; end: # Muniru A Asiru, Jan 31 2018
  • Mathematica
    LinearRecurrence[{2,12},{1,2},30] (* or *) With[{s=Sqrt[13]},Table[ Simplify[ -(((13+s)((1-s)^n-(1+s)^n))/(26(1+s)))],{n,30}]] (* Harvey P. Dale, May 25 2013 *)
  • PARI
    my(x='x+O('x^30)); Vec(1/(1-2*x-12*x^2)) \\ G. C. Greubel, Jan 30 2018
    
  • Sage
    [lucas_number1(n,2,-12) for n in range(1, 30)] # Zerinvary Lajos, Apr 22 2009
    

Formula

a(n) = A000079(n)*A006130(n).
G.f.: 1/(1-2*x-12*x^2).
a(n) = ((1+sqrt(13))*(1+sqrt(13))^n - (1-sqrt(13))*(1-sqrt(13))^n) /(2*sqrt(13)).
a(n) = Sum_{k=0..floor(n/2)} C(n+1,2*k+1) * 13^k. - Paul Barry, Jan 15 2007

A085449 Horadam sequence (0,1,4,2).

Original entry on oeis.org

0, 1, 2, 8, 24, 80, 256, 832, 2688, 8704, 28160, 91136, 294912, 954368, 3088384, 9994240, 32342016, 104660992, 338690048, 1096024064, 3546808320, 11477712896, 37142659072, 120196169728, 388962975744, 1258710630400
Offset: 0

Views

Author

Ross La Haye, Aug 18 2003

Keywords

Comments

a(n) / a(n-1) converges to sqrt(5) + 1 as n approaches infinity. sqrt(5) + 1 can also be written as Phi^3 - 1, 2 * Phi, Phi^2 + Phi - 1 and (L(n) / F(n)) + 1, where L(n) is the n-th Lucas number and F(n) is the n-th Fibonacci number as n approaches infinity.
Binomial transform is A001076. - Paul Barry, Aug 25 2003

Examples

			a(4) = 24 because a(3) = 8, a(2) = 2, s = 2, r = 4 and (2 * 8) + (4 * 2) = 24.
G.f. = x + 2*x^2 + 8*x^3 + 24*x^4 + 80*x^5 + 256*x^6 + 832*x^7 + ... - _Michael Somos_, Mar 07 2021
		

Crossrefs

Essentially the same as A063727.

Programs

  • GAP
    a:=[0,1];; for n in [3..30] do a[n]:=2*a[n-1]+4*a[n-2]; od; a; # Muniru A Asiru, Oct 09 2018
  • Magma
    [2^(n-1)*Fibonacci(n): n in [0..50]]; // G. C. Greubel, Oct 08 2018
    
  • Maple
    a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=2*(a[n-1]+2*a[n-2]) od: seq(a[n], n=0..26); # Zerinvary Lajos, Mar 17 2008
  • Mathematica
    Table[2^(n-1)*Fibonacci[n], {n,0,50}] (* G. C. Greubel, Oct 08 2018 *)
  • PARI
    vector(50, n, n--; 2^(n-1)*fibonacci(n)) \\ G. C. Greubel, Oct 08 2018
    

Formula

a(n) = s*a(n-1) + r*a(n-2); for n > 1, where a(0) = 0, a(1) = 1, s = 2, r = 4.
From Paul Barry, Aug 25 2003: (Start)
G.f.: x/(1-2*x-4*x^2).
a(n) = sqrt(5)*((1+sqrt(5))^n - (1-sqrt(5))^n)/10.
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k+1)5^k . (End)
The signed version 0, 1, -2, ... has a(n)=sqrt(5)((sqrt(5)-1)^n-(-sqrt(5)-1)^n)/10. It is the second inverse binomial transform of A085449. - Paul Barry, Aug 25 2003
a(n) = 2^(n-1)*Fib(n). - Paul Barry, Mar 22 2004
Sum_{n>=1} 1/a(n) = A269991. - Amiram Eldar, Feb 01 2021
a(n) = -(-4)^n*a(-n) for all integer n. - Michael Somos, Mar 07 2021
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