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 1-5 of 5 results.

A080871 a(n)*a(n+3) - a(n+1)*a(n+2) = 3, given a(0)=a(1)=1, a(2)=4.

Original entry on oeis.org

1, 1, 4, 7, 31, 55, 244, 433, 1921, 3409, 15124, 26839, 119071, 211303, 937444, 1663585, 7380481, 13097377, 58106404, 103115431, 457470751, 811826071, 3601659604, 6391493137, 28355806081, 50320119025, 223244789044
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Links

Crossrefs

Cf. A001519, A079496, A080872, A080873, A080874, A080875.
Bisections are A001091 and A070997.

Programs

  • Mathematica
    RecurrenceTable[{a[0]==a[1]==1,a[2]==4,a[n]==(3+a[n+1]a[n+2])/a[n+3]},a,{n,30}] (* Harvey P. Dale, Jun 08 2017 *)

Formula

a(n) = (3 + a(n-1)*a(n-2))/a(n-3) for n>2.
G.f.: (-x^3 - 4*x^2 + x + 1)/(x^4 - 8*x^2 + 1)
a(n+4) = 8*a(n+2)-a(n). [Richard Choulet, Dec 04 2008]
a(n) = (0.25 + sqrt(10)/20)*(sqrt(4 + sqrt(15)))^n + (0.25 + sqrt(10)/20)*(sqrt(4 - sqrt(15)))^n + ( - 1/20*10^(1/2) + 1/4)*( - sqrt(4 + sqrt(15)))^n + ( - 1/20*10^(1/2) + 1/4)*( - (sqrt(4 - sqrt(15))))^n. [Richard Choulet, Dec 06 2008]

A080873 a(n)*a(n+3) - a(n+1)*a(n+2) = 5, given a(0)=a(1)=1, a(2)=2.

Original entry on oeis.org

1, 1, 2, 7, 19, 69, 188, 683, 1861, 6761, 18422, 66927, 182359, 662509, 1805168, 6558163, 17869321, 64919121, 176888042, 642633047, 1751011099, 6361411349, 17333222948, 62971480443, 171581218381, 623353393081, 1698478960862
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Links

Crossrefs

Cf. A080871, A080872, A080874, A080875.

Programs

  • Mathematica
    CoefficientList[Series[(-3x^3-8x^2+x+1)/(x^4-10x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[{0,10,0,-1},{1,1,2,7},30] (* Harvey P. Dale, Feb 27 2023 *)

Formula

For n > 1: a(2*n-1) = 3*a(2*n-2) + 2*a(2*n-3) - a(2*n-4); a(2n) = 3*a(2*n-1) - a(2*n-2).
G.f.: (1 + x - 8*x^2 - 3*x^3) / (1 - 10*x^2 + x^4). - N. J. A. Sloane, Jul 19 2005
a(n+4) = 10*a(n+2) - a(n). [Richard Choulet, Dec 04 2008]
a(n) = (1/24*(3 + 3*3^(1/2)*2^(1/2) + 6*2^(1/2) + 2*3^(1/2))/(3^(1/2)*2^(1/2) + 2))*(sqrt(3) + sqrt(2))^n + (1/16*3^(1/2)*2^(1/2) + 1/8*2^(1/2) + 1/4 + 1/6*3^(1/2))*(sqrt(3) - sqrt(2))^n + (1/24*(3*3^(1/2)*2^(1/2) - 6*2^(1/2) + 3 - 2*3^(1/2))/(3^(1/2)*2^(1/2) + 2))*( - sqrt(3) - sqrt(2))^n + (1/16*3^(1/2)*2^(1/2) - 1/8*2^(1/2) + 1/4 - 1/6*3^(1/2))*( - sqrt(3) + sqrt(2))^n. [Richard Choulet, Dec 04 2008]

A080874 a(n)*a(n+3) - a(n+1)*a(n+2) = 5, given a(0)=a(1)=1, a(2)=3.

Original entry on oeis.org

1, 1, 3, 8, 29, 79, 287, 782, 2841, 7741, 28123, 76628, 278389, 758539, 2755767, 7508762, 27279281, 74329081, 270037043, 735782048, 2673091149, 7283491399, 26460874447, 72099131942, 261935653321, 713707828021, 2592895658763
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Links

Crossrefs

Cf. A080871, A080872, A080873, A080875.

Programs

  • Mathematica
    LinearRecurrence[{0,10,0,-1},{1,1,3,8},30] (* Harvey P. Dale, Sep 18 2016 *)

Formula

G.f.: (1+x-7x^2-2x^3)/(1-10x^2+x^4). a(n)=10a(n-2)-a(n-4). - Michael Somos, Mar 04 2003.
a(n) = ( - 1/24*3^(1/2)*2^(1/2) + 1/4 - 1/16*2^(1/2) + 1/8*3^(1/2))*(sqrt(3) + sqrt(2))^n + (1/24*3^(1/2)*2^(1/2) + 1/4 + 1/16*2^(1/2) + 1/8*3^(1/2))*(sqrt(3) - sqrt(2))^n + ( - 1/24*3^(1/2)*2^(1/2) - 1/8*3^(1/2) + 1/16*2^(1/2) + 1/4)*( - sqrt(3) - sqrt(2))^n + ( - 1/16*2^(1/2) + 1/4 + 1/24*3^(1/2)*2^(1/2) - 1/8*3^(1/2))*( - sqrt(3) + sqrt(2))^n [From Richard Choulet, Dec 04 2008]

A080875 a(n)*a(n+3) - a(n+1)*a(n+2) = 5, given a(0)=a(1)=1, a(2)=6.

Original entry on oeis.org

1, 1, 6, 11, 71, 131, 846, 1561, 10081, 18601, 120126, 221651, 1431431, 2641211, 17057046, 31472881, 203253121, 375033361, 2421980406, 4468927451, 28860511751, 53252096051, 343904160606, 634556225161, 4097989415521
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Links

Crossrefs

Cf. A080871, A080872, A080873, A080874.
Bisections are A023038 and A077417.

Programs

  • Mathematica
    LinearRecurrence[{0,12,0,-1},{1,1,6,11},30] (* Harvey P. Dale, Jul 14 2024 *)

Formula

G.f.: (-x^3 - 6*x^2 + x + 1)/(x^4 - 12*x^2 + 1).
a(n+4) = 12*a(n+2)-a(n). [Richard Choulet, Dec 04 2008]
a(n) = (1/4 + ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*(sqrt(6 + sqrt(35)))^n + (1/4 + ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*(sqrt(6 - sqrt(35)))^n + (1/4 - ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*( - sqrt(6 + sqrt(35)))^n + (1/4 - ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*( - (sqrt(6 - sqrt(35))))^n. [Richard Choulet, Dec 06 2008]

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

Original entry on oeis.org

0, 1, 6, 15, 64, 153, 638, 1519, 6320, 15041, 62566, 148895, 619344, 1473913, 6130878, 14590239, 60689440, 144428481, 600763526, 1429694575, 5946945824, 14152517273, 58868694718, 140095478159, 582740001360, 1386802264321, 5768531318886, 13727927165055
Offset: 1

Views

Author

Bruno Berselli, Dec 10 2013

Keywords

Comments

For n>1, partial sums of A080872 starting from A080872(1).

Examples

			153 is in the sequence because 3*153*154/2+1 = 188^2.

Links

Crossrefs

Sequence A129444 gives n+1.
Cf. A000217, A080872, A129445 (square roots of 3*A000217(a(n))+1), A132596 (numbers m such that 3*A000217(m) is a square).
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; this sequence for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6.

Programs

  • Mathematica
    LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 6, 15, 64}, 30]

Formula

G.f.: x^2*(1 + 5*x - x^2 - x^3) / ((1 - x)*(1 - 10*x^2 + x^4)).
a(n) = a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=0, a(2)=1, a(3)=6, a(4)=15, a(5)=64.
a(n) = -1/2 + ( (-3*(-1)^n + 2*sqrt(6))*(5 + 2*sqrt(6))^floor(n/2) - (3*(-1)^n + 2*sqrt(6))*(5 - 2*sqrt(6))^floor(n/2) )/12.
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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