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.

A085376 Ratio-dependent insertion sequence I(0.36704) (see the link below).

Original entry on oeis.org

1, 3, 11, 30, 109, 297, 1079, 2940, 10681, 29103, 105731, 288090, 1046629, 2851797, 10360559, 28229880, 102558961, 279447003, 1015229051, 2766240150, 10049731549, 27382954497, 99482086439, 271063304820, 984771132841
Offset: 1

Views

Author

John W. Layman, Jun 26 2003

Keywords

Comments

This sequence is the ratio-determined insertion sequence (RDIS) "twin" of I(0.37802)=A080874 and "child" of I(0.33344)=A001835 and I(0.38208)=A001906 in the RDIS recurrence tree (see the link for an explanation of terms). See A082630, A082981, A085348 and A085349 for recent examples of RDIS sequences.
Conjecture: partial sums of A129445. - Sean A. Irvine, Jul 14 2022

Crossrefs

Formula

It is conjectured that a(n) = 10*a(n-2) - a(n-4).
Apparently a(n)*a(n+3) = -3 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004

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

Crossrefs

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]

A080872 a(n)*a(n+3) - a(n+1)*a(n+2) = 4, given a(0)=a(1)=1, a(2)=5.

Original entry on oeis.org

1, 1, 5, 9, 49, 89, 485, 881, 4801, 8721, 47525, 86329, 470449, 854569, 4656965, 8459361, 46099201, 83739041, 456335045, 828931049, 4517251249, 8205571449, 44716177445, 81226783441, 442644523201, 804062262961, 4381729054565, 7959395846169, 43374646022449, 78789896198729, 429364731169925
Offset: 0

Views

Author

Paul D. Hanna, Feb 22 2003

Keywords

Crossrefs

Bisections are A001079 and A072256.

Programs

  • Mathematica
    CoefficientList[Series[(-x^3-5 x^2+x+1)/(x^4-10 x^2+1),{x,0,30}],x] (* or *) LinearRecurrence[{0,10,0,-1},{1,1,5,9},30] (* Harvey P. Dale, May 06 2012 *)
  • PARI
    Vec( (-x^3 - 5*x^2 + x + 1)/(x^4 - 10*x^2 + 1) + O(x^66) ) \\ Joerg Arndt, Jan 29 2016

Formula

G.f.: (-x^3 - 5*x^2 + x + 1)/(x^4 - 10*x^2 + 1).
a(n) = (3+sqrt(3))/12*(sqrt(3)-sqrt(2))^n+(3-sqrt(3))/12*(-sqrt(3)+sqrt(2))^n+(3+sqrt(3))/12*(sqrt(3)+sqrt(2))^n+(3-sqrt(3))/12*(-sqrt(3)-sqrt(2))^n. [Richard Choulet, Dec 03 2008]
a(n+4) = 10*a(n+2)-a(n). [Richard Choulet, Dec 04 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

Crossrefs

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]

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

Crossrefs

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]
Showing 1-5 of 5 results.