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.

Previous Showing 31-35 of 35 results.

A173731 a(n) = a(n-1) * (11*a(n-1) - a(n-2)) / (a(n-1) + 4*a(n-2)), a(0) = a(1) = 0, a(2) = 1.

Original entry on oeis.org

0, 0, 1, 11, 88, 638, 4466, 30856, 212135, 1455685, 9981840, 68428140, 469043796, 3214953456, 22035826813, 151036348463, 1035219958696, 7095506886986, 48633337477670, 333337879614520, 2284731883069955, 15659785467455305
Offset: 0

Views

Author

Michael Somos, Feb 23 2010

Keywords

Examples

			x^2 + 11*x^3 + 88*x^4 + 638*x^5 + 4466*x^6 + 30856*x^7 + 212135*x^8 + ...
		

Crossrefs

Programs

  • Magma
    [(4+Fibonacci(4*n+1)/3+Fibonacci(4*n + 3)/3-5* Fibonacci(2*n+1)) / 20: n in [0..25]]; // Vincenzo Librandi, Nov 30 2016
  • Mathematica
    Table[(4 + Fibonacci[4*n + 1]/3 + Fibonacci[4*n + 3]/3 - 5*Fibonacci[2*n + 1])/20, {n, 0, 25}] (* or *) LinearRecurrence[{11, -33, 33, -11, 1}, {0, 0, 1, 11, 88}, 25] (* G. C. Greubel, Nov 29 2016 *)
  • PARI
    {a(n) = (4 + fibonacci(4*n + 1)/3 + fibonacci(4*n + 3)/3 - 5 * fibonacci(2*n + 1)) / 20}
    

Formula

a(n) = (4 + A049685(n) - 5 * A122367(n)) / 20 = a(1 - n).
G.f.: x^2 / ((1 - x) * (1 - 3*x + x^2) * (1 - 7*x + x^2)) = ( 4 / (1 - x) - 5 * (1 - x) / (1 - 3*x + x^2) + (1 - x) / (1 - 7*x + x^2) ) / 20.
From G. C. Greubel, Nov 29 2016: (Start)
a(n) = 11*a(n-1) - 33*a(n-2) + 33*a(n-3) - 11*a(n-4) + a(n-5).
a(n) = (12 + Fibonacci(4*n + 1) + Fibonacci(4*n + 3) - 15*Fibonacci[2*n + 1] ) / 60. (End)

A333718 a(n) = L(8*n+4)/7, where L=A000032 (the Lucas sequence).

Original entry on oeis.org

1, 46, 2161, 101521, 4769326, 224056801, 10525900321, 494493258286, 23230657239121, 1091346396980401, 51270050000839726, 2408601003642486721, 113152977121196036161, 5315781323692571212846, 249728569236429650967601, 11731926972788501024264401, 551150839151823118489459246
Offset: 0

Views

Author

Greg Dresden and Tracy Z. Wu, Sep 03 2020

Keywords

Comments

a(n) is the denominator of the continued fraction [3*sqrt(5), 3*sqrt(5),..., 3*sqrt(5)] with 2n+1 terms.
a(n) = (2/7)*T(2*n+1, 7/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022

Examples

			The continued fraction [3*sqrt(5), 3*sqrt(5), 3*sqrt(5)] with 2*1 + 1 terms equals 141*sqrt(5)/46, and 46 is our a(1) term.
		

Crossrefs

Cf. A000032, A049685, first differences of A049668.

Programs

  • Mathematica
    Table[LucasL[8 n + 4]/7, {n, 0, 20}]

Formula

a(n) = 47*a(n-1) - a(n-2) for n>2.
G.f.: (1-x)/(1-47*x+x^2). - R. J. Mathar, Sep 03 2020

A143790 Integer Quotients of Lucas Numbers; a rectangular array by downward antidiagonals.

Original entry on oeis.org

1, 3, 1, 4, 6, 1, 7, 41, 19, 1, 11, 281, 341, 46, 1, 18, 1926, 6119, 2161, 124, 1, 29, 13201, 109801, 101521, 15251, 321, 1, 47, 90481, 1970299, 4769326, 1875749, 103361, 844, 1, 76, 620166, 35355581, 224056801, 230701876, 33281921, 711491, 2206, 1, 123
Offset: 1

Views

Author

Clark Kimberling, Sep 01 2008

Keywords

Comments

Every integer-valued quotient of Lucas numbers (excluding L(0)=2) is in this array.
(Row 1) = A000032 except for a(0)
(Row 2) = A049685
(Row 3) = A049629
(Column 2) = A110391 except for initial terms

Examples

			Q(3,2)=L(9)/L(3)=76/4=19.
		

Crossrefs

Cf. A000032.

Formula

Row 1: L(k)/L(1), where L(k)=A000032(k) = k-th Lucas number, for k>=1;
Row n: L(2nk-n)/L(n) for n>=2, k>=1.

A269028 a(n) = 40*a(n - 1) - a(n - 2) for n>1, a(0) = 1, a(1) = 1.

Original entry on oeis.org

1, 1, 39, 1559, 62321, 2491281, 99588919, 3981065479, 159143030241, 6361740144161, 254310462736199, 10166056769303799, 406387960309415761, 16245352355607326641, 649407706263983649879, 25960062898203738668519, 1037753108221885563090881
Offset: 0

Views

Author

Ilya Gutkovskiy, Feb 18 2016

Keywords

Comments

In general, the ordinary generating function for the recurrence relation b(n) = k*b(n - 1) - b(n - 2) with n>1 and b(0)=1, b(1)=1, is (1 - (k - 1)*x)/(1 - k*x +x^2). This recurrence gives the closed form b(n) = (2^( -n - 1)*((k - 2)*(k - sqrt(k^2 - 4))^n + sqrt(k^2 - 4)*(k - sqrt(k^2 - 4))^n - (k - 2)*(sqrt(k^2 - 4) + k)^n + sqrt(k^2 - 4)*(sqrt(k^2 - 4) + k)^n))/sqrt(k^2 - 4).

Crossrefs

Programs

  • Magma
    [n le 2 select 1 else 40*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 19 2016
  • Mathematica
    Table[Cosh[n Log[20 + Sqrt[399]]] - Sqrt[19/21] Sinh[n Log[20 + Sqrt[399]]], {n, 0, 17}]
    Table[(2^(-n - 2) (38 (40 - 2 Sqrt[399])^n + 2 Sqrt[399] (40 - 2 Sqrt[399])^n - 38 (40 + 2 Sqrt[399])^n + 2 Sqrt[399] (40 + 2 Sqrt[399])^n))/Sqrt[399], {n, 0, 17}]
    LinearRecurrence[{40, -1}, {1, 1}, 17]

Formula

G.f.: (1 - 39*x)/(1 - 40*x + x^2).
a(n) = cosh(n*log(20 + sqrt(399))) - sqrt(19/21)*sinh(n*log(20 + sqrt(399))).
a(n) = (2^(-n - 2)*(38*(40 - 2*sqrt(399))^n + 2*sqrt(399)*(40 - 2*sqrt(399))^n - 38*(40 + 2*sqrt(399))^n + 2*sqrt(399)*(40 + 2*sqrt(399))^n))/sqrt(399).
Sum_{n>=0} 1/a(n) = 2.0262989201139499769986...

A278475 a(n) = floor(phi^7*a(n-1)) for n>0, a(0) = 1, where phi is the golden ratio (A001622).

Original entry on oeis.org

1, 29, 841, 24417, 708933, 20583473, 597629649, 17351843293, 503801085145, 14627583312497, 424703717147557, 12331035380591649, 358024729754305377, 10395048198255447581, 301814422479162285225, 8763013300093961719105, 254429200125204052139269, 7387209816931011473757905
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 23 2016

Keywords

Comments

In general, the ordinary generating function for the recurrence relation b(n) = floor(phi^k*b(n - 1)) with n>0 and b(0) = 1, is (1 - x)/(1 - (phi^k + (-phi)^(-k))*x + x^2) if k is even, and (1 - x - x^2)/((1 - x)*(1 - (phi^k + (-phi)^(-k))*x - x^2)) if k is odd.

Crossrefs

Cf. A001622.
Cf. similar sequences with recurrence relation b(n) = floor(phi^k*b(n-1)) for n>0, b(0) = 1: A000012 (k = 1), A001519 (k = 2), A024551 (k = 3), A049685 (k = 4), A214993 (k = 5), A007805 (k = 6), this sequence (k = 7).

Programs

  • Mathematica
    RecurrenceTable[{a[0] == 1, a[n] == Floor[GoldenRatio^7 a[n - 1]]}, a, {n, 17}]
    LinearRecurrence[{30, -28, -1}, {1, 29, 841}, 18]
  • PARI
    Vec( (1 - x - x^2)/((1 - x)*(1 - 29*x - x^2)) + O(x^50) ) \\ G. C. Greubel, Nov 24 2016

Formula

G.f.: (1 - x - x^2)/((1 - x)*(1 - 29*x - x^2)).
a(n) = 30*a(n-1) - 28*a(n-2) - a(n-3).
a(n) = ((-29 - 13*sqrt(5))^(-n)*(-7*(407 + 182*sqrt(5))*2^(n+3) + 13*(1885 + 843*sqrt(5))*(-29 - 13*sqrt(5))^n + 28*(25319 + 11323*sqrt(5))*(-843 - 377*sqrt(5))^n))/(377*(1885 + 843*sqrt(5))).
Previous Showing 31-35 of 35 results.