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 1-6 of 6 results.

A080879 a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=1, a(2)=6.

Original entry on oeis.org

1, 1, 6, 7, 44, 52, 328, 388, 2448, 2896, 18272, 21616, 136384, 161344, 1017984, 1204288, 7598336, 8988928, 56714752, 67094272, 423324672, 500798464, 3159738368, 3738010624, 23584608256, 27900891136, 176037912576, 208255086592, 1313964867584, 1554437128192
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> (<<0|1>, <-4|8>>^floor(n/2). <<1, 6+(n mod 2)>>)[1,1]:
    seq(a(n), n=0..30);  # Alois P. Heinz, Mar 18 2023
  • Mathematica
    LinearRecurrence[{0,8,0,-4},{1,1,6,7},30] (* Harvey P. Dale, Mar 10 2015 *)

Formula

a(2n) = A080876(2n+3)/2, a(2n+1) = A080876(2n+4)/4.
G.f.: (-x^3 - 2*x^2 + x + 1)/(4*x^4 - 8*x^2 + 1).
a(n) = ((9/16)*sqrt(3) - 7/16)*(1 + sqrt(3))^n + (-(9/16)*sqrt(3) - 7/16)*(1 - sqrt(3))^n + (-(19/48)*sqrt(3) + 15/16)*(-(1 + sqrt(3)))^n + ((19/48)*sqrt(3) + 15/16)*(-(1 - sqrt(3)))^n. - Richard Choulet, Dec 06 2008
a(n+4) = 8*a(n+2) - 4*a(n). - Richard Choulet, Dec 06 2008

A080881 a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=2, a(2)=10.

Original entry on oeis.org

1, 2, 10, 21, 106, 223, 1126, 2369, 11962, 25167, 127078, 267361, 1350010, 2840303, 14341798, 30173889, 152359738, 320551567, 1618589926, 3405371681, 17195050234, 36176882223, 182671192870, 384324217729, 1940602920634
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(-x^3-x^2+2x+1)/(4x^4-11x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[ {0,11,0,-4},{1,2,10,21},30] (* Harvey P. Dale, Jun 10 2024 *)

Formula

G.f.: (-x^3 - x^2 + 2*x + 1)/(4*x^4 - 11*x^2 + 1)
a(n + 4) = 11*a(n + 2) - 4*a(n) [From Richard Choulet, Dec 06 2008]
a(n) = (3/140*15^(1/2)*7^(1/2) + 1/4 + 3/56*7^(1/2) + 1/24*15^(1/2))*sqrt((11 + sqrt(105))/2)^n + ( - 3/140*15^(1/2)*7^(1/2) + 1/4 - 3/56*7^(1/2) + 1/24*15^(1/2))*sqrt((11 - sqrt(105))/2)^n + (3/140*15^(1/2)*7^(1/2) - 1/24*15^(1/2) - 3/56*7^(1/2) + 1/4)*( - sqrt((11 + sqrt(105))/2))^n + (1/4 + 3/56*7^(1/2) - 1/24*15^(1/2) - 3/140*15^(1/2)*7^(1/2))*( - sqrt((11 - sqrt(105))/2))^n [From Richard Choulet, Dec 07 2008]

A080882 a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=3, a(2)=7.

Original entry on oeis.org

1, 3, 7, 22, 52, 164, 388, 1224, 2896, 9136, 21616, 68192, 161344, 508992, 1204288, 3799168, 8988928, 28357376, 67094272, 211662336, 500798464, 1579869184, 3738010624, 11792304128, 27900891136, 88018956288, 208255086592
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Crossrefs

Programs

  • Maple
    a:= n-> (Matrix([[22,7,3,1]]). Matrix(4, (i,j)-> if (i=j-1) then 1 elif j=1 then [0,8,0,-4][i] else 0 fi)^(n))[1,4]: seq(a(n), n=0..26); # Alois P. Heinz, Aug 23 2008
  • Mathematica
    a[0]=1; a[1]=3; a[2]=7; a[3]=22; a[n_] := a[n] = 8*a[n-2] - 4*a[n-4]; Table[a[n], {n, 0, 26}] (* Jean-François Alcover, Jun 15 2015, after Richard Choulet *)
    LinearRecurrence[{0,8,0,-4},{1,3,7,22},30] (* Harvey P. Dale, Mar 23 2025 *)

Formula

a(2n)=A080879(2n+1)=A080876(2n+4)/4, a(2n+1)=A080879(2n+2)/2=A080876(2n+5)/4.
G.f.: (-2*x^3 - x^2 + 3*x + 1)/(4*x^4 - 8*x^2 + 1).
a(n + 4) = 8*a(n + 2) - 4*a(n). - Richard Choulet, Dec 06 2008
a(n) = (7/24*3^(1/2) + 1/2)*((1 + sqrt(3)))^n + ( - 7/24*3^(1/2) + 1/2)*((1 - sqrt(3)))^n + ( - 1/24*3^(1/2))*( - (1 + sqrt(3)))^n + (1/24*3^(1/2))*( - ((1 - sqrt(3))))^n. - Richard Choulet, Dec 06 2008

A080877 a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=1, a(2)=2.

Original entry on oeis.org

1, 1, 2, 3, 8, 14, 40, 72, 208, 376, 1088, 1968, 5696, 10304, 29824, 53952, 156160, 282496, 817664, 1479168, 4281344, 7745024, 22417408, 40553472, 117379072, 212340736, 614604800, 1111830528, 3218112512, 5821620224, 16850255872
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Crossrefs

Cf. A154626, A098648 (bisections). [From R. J. Mathar, Oct 26 2009]

Programs

  • Mathematica
    LinearRecurrence[{0,6,0,-4},{1,1,2,3},50] (* or *) CoefficientList[ Series[ (-3x^3-4x^2+x+1)/(4x^4-6x^2+1),{x,0,50}],x] (* Harvey P. Dale, May 02 2011 *)

Formula

G.f.: (-3*x^3 - 4*x^2 + x + 1)/(4*x^4 - 6*x^2 + 1)
a(n + 4) = 6*a(n + 2) - 4*a(n) [From Richard Choulet, Dec 06 2008]
a(n) = ( - 1/20*5^(1/2) + 1/16*5^(1/2)*2^(1/2) - 1/16*2^(1/2) + 1/4)*(sqrt(3 + sqrt(5)))^n + (1/20*5^(1/2) + 1/16*5^(1/2)*2^(1/2) + 1/16*2^(1/2) + 1/4)*(sqrt(3 - sqrt(5)))^n + ( - 1/20*5^(1/2) - 1/16*5^(1/2)*2^(1/2) + 1/16*2^(1/2) + 1/4)*( - (sqrt(3 + sqrt(5))))^n + (1/20*5^(1/2) - 1/16*5^(1/2)*2^(1/2) - 1/16*2^(1/2) + 1/4)*( - (sqrt(3 - sqrt(5))))^n [From Richard Choulet, Dec 07 2008]

A080878 a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=1, a(2)=3.

Original entry on oeis.org

1, 1, 3, 4, 14, 20, 72, 104, 376, 544, 1968, 2848, 10304, 14912, 53952, 78080, 282496, 408832, 1479168, 2140672, 7745024, 11208704, 40553472, 58689536, 212340736, 307302400, 1111830528, 1609056256, 5821620224, 8425127936, 30482399232
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Examples

			G.f. = 1 + x + 3*x^2 + 4*x^3 + 14*x^4 + 20*x^5 + 72*x^6 + 104*x^7 + 376*x^8 + ...
		

Crossrefs

Programs

  • Mathematica
    a[ n_] := If[ n < 0, 2^n, 1] SeriesCoefficient[ (1 + x - 3*x^2 - 2*x^3)/(1 - 6*x^2 + 4*x^4), {x, 0, Abs@n}]; (* Michael Somos, May 25 2014 *)
    a[ n_] := 2^Quotient[ n - 1, 2] If[ OddQ@n, Fibonacci@n, LucasL@n]; (* Michael Somos, May 25 2014 *)
    LinearRecurrence[{0,6,0,-4},{1,1,3,4},40] (* Harvey P. Dale, Dec 07 2014 *)
  • PARI
    {a(n) = if( n<0, 2^n, 1) * polcoeff( (1 + x - 3*x^2 - 2*x^3) / (1 - 6*x^2 + 4*x^4) + x * O(x^abs(n)), abs(n))}; /* Michael Somos, May 25 2014 */
    
  • PARI
    {a(n) = 2^((n - 1)\2) * if( n%2, fibonacci(n), fibonacci(n-1) + fibonacci(n+1))}; /* Michael Somos, May 25 2014 */

Formula

G.f.: (1 + x - 3*x^2 - 2*x^3) / (1 - 6*x^2 + 4*x^4). a(n) = 6*a(n-2) - 4*a(n-4). - Michael Somos, Mar 05 2003
a(2n) = A080877(2n+1), a(2n+1) = A080877(2n+2)/2.
a(n) = (1/20*10^(1/2) + 1/4)*(sqrt(3 + sqrt(5)))^n + (1/20*10^(1/2) + 1/4)*(sqrt(3 - sqrt(5)))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(3 + sqrt(5))))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(3 - sqrt(5))))^n. - Richard Choulet, Dec 07 2008
a(-n) = a(n) / 2^n. a(2*n) = A098648(n). a(2*n + 1) = A082761(n). - Michael Somos, May 25 2014
0 = a(n)*(+2*a(n+2)) + a(n+1)*(+2*a(n+1) - 7*a(n+2) + a(n+3)) + a(n+2)*(+a(n+2)) for all n in Z. - Michael Somos, May 25 2014

A080880 a(n)*a(n+3) - a(n+1)*a(n+2) = 2^n, given a(0)=1, a(1)=2, a(2)=2.

Original entry on oeis.org

1, 2, 2, 5, 6, 17, 22, 65, 86, 257, 342, 1025, 1366, 4097, 5462, 16385, 21846, 65537, 87382, 262145, 349526, 1048577, 1398102, 4194305, 5592406, 16777217, 22369622, 67108865, 89478486, 268435457, 357913942, 1073741825, 1431655766, 4294967297
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{0,5,0,-4},{1,2,2,5},40] (* Harvey P. Dale, Nov 30 2012 *)

Formula

a(2n)=(4^n+2)/3, a(2n+1)=4^n+1.
G.f.: (-5*x^3 - 3*x^2 + 2*x + 1)/(4*x^4 - 5*x^2 + 1)
a(n) = 5/6 + 5/12*2^n - 1/6*( - 1)^n - 1/12*( - 2)^n [From Richard Choulet, Dec 07 2008]
a(n + 4) = 5*a(n + 2) - 4*a(n) [From Richard Choulet, Dec 06 2008]

Extensions

More terms from Ralf Stephan, Jul 25 2003
Showing 1-6 of 6 results.