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 11-15 of 15 results.

A219164 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 5.

Original entry on oeis.org

5, 527, 77132286527, 35395236908668169265765137996816180039862527
Offset: 0

Views

Author

Peter Bala, Nov 13 2012

Keywords

Comments

Bisection of A003487.
The next term -- a(4) -- has 175 digits. - Harvey P. Dale, Jun 09 2017

Crossrefs

Cf. A001999, A003487, A003501, A219162, A219163, A219165.

Programs

  • Mathematica
    NestList[#^4-4#^2+2&,5,5] (* Harvey P. Dale, Jun 09 2017 *)

Formula

Let alpha = 1/2*(5 + sqrt(21)). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003487(2*n) = A003501(4^n).
Product_{n>=0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(7/3).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,5/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 5. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

A219165 Recurrence equation a(n+1) = a(n)^4 - 4*a(n)^2 + 2 with a(0) = 6.

Original entry on oeis.org

6, 1154, 1773462177794, 9892082352510403757550172975146702122837936996354
Offset: 0

Views

Author

Peter Bala, Nov 13 2012

Keywords

Comments

Bisection of A003423.

Crossrefs

Cf. A001999, A003423, A003499, A219162, A219163, A219164.

Formula

Let alpha = 3 + 2*sqrt(2). Then a(n) = (alpha)^(4^n) + (1/alpha)^(4^n).
a(n) = A003423(2*n) = A003499(4^n).
Product {n >= 0} ((1 + 2/a(n))/(1 - 2/a(n)^2)) = sqrt(2).
From Peter Bala, Dec 06 2022: (Start)
a(n) = 2*T(4^n,3), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Let b(n) = a(n) - 6. The sequence {b(n)} appears to be a strong divisibility sequence, that is, gcd(b(n),b(m)) = b(gcd(n,m)) for n, m >= 1. (End)

A002000 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 7.

Original entry on oeis.org

7, 322, 33385282, 37210469265847998489922, 51522323599677629496737990329528638956583548304378053615581043535682
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

References

  • 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).

Links

Crossrefs

Cf. A000032, A001999, A006267, A219161, A271223.

Programs

  • Magma
    [n eq 1 select 7 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]]; // Vincenzo Librandi, Feb 09 2017
  • Maple
    a := proc(n) option remember; if n = 0 then 7 else a(n-1)^3 - 3*a(n-1) end if; end;
seq(a(n), n = 0..4); # Peter Bala, Nov 15 2022
  • Mathematica
    RecurrenceTable[{a[0] == 7, a[n] == a[n - 1]^3 - 3 a[n - 1]}, a, {n, 0, 8}]
(* Vincenzo Librandi, Feb 09 2017 *)
NestList[#(#^2-3)&,7,4] (* Harvey P. Dale, Aug 11 2021 *)

Formula

From Peter Bala, Feb 01 2017: (Start)
a(n) = ((7 + sqrt(45))/2)^(3^n) + ((7 - sqrt(45))/2)^(3^n).
a(n) = 2*T(3^n,7/2), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
Product_{n >= 0} (1 + 2/(a(n) - 1)) = 3*sqrt(5)/5.
Cf. A001999 and A219161. (End)
From Peter Bala, Nov 15 2022: (Start)
a(n) = Lucas(4*(3^n)).
a(n+1) == a(n) (mod 3^(n+1)) (a particular case of the Gauss congruences for the Lucas numbers).
Conjecture: a(n+1) == a(n) (mod 3^(n+r+2)) for n >= r.
The least positive residue of a(n) mod(3^n) = 3^n - 2 = A058481(n). In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to -2.
Product_{k = 0..n} (a(k) - 1) = (1/3)*Lucas(6*(3^n)). (End)

A282180 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 8.

Original entry on oeis.org

8, 488, 116212808, 1569502402942700328379688, 3866214585126515728777536857817155683642224883875510905654220958052649608
Offset: 0

Views

Author

Vincenzo Librandi, Feb 10 2017

Keywords

Links

Crossrefs

Cf. similar sequences with initial value k: A001999 (k=3), A219160 (k=4), A219161 (k=5), A112845 (k=6), A002000 (k=7), this sequence (k=8), A282181 (k=9), A006242 (k=10), A006243 (k=198).

Programs

  • Magma
    [n eq 1 select 8 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]];
  • Mathematica
    RecurrenceTable[{a[0] == 8, a[n] == a[n-1]^3 - 3 a[n-1]}, a, {n, 8}]

Formula

a(n) = (4 + sqrt(15))^(3^n) + (4 - sqrt(15))^(3^n). - Bruno Berselli, Feb 10 2017
a(n) = -2*cos(3^n * arccos(-4)). - Daniel Suteu, Feb 10 2017

A111366 Numbers such that the sum of the digits of floor(phi^n) is also the sum of the digits of the n-th Fibonacci number (in base 10), where phi is the golden ratio.

Original entry on oeis.org

1, 6, 13, 61, 73, 92, 97, 198, 212, 217, 222, 270, 349, 380, 404, 438, 524, 630, 649, 836, 937, 1446, 1477, 1513, 1532, 1729, 2005, 2046, 2060, 2077, 2209, 2348, 2660, 2862, 2934, 3265, 3649, 3889, 4093, 4609, 4686, 4945, 5180, 5444, 5497, 5749, 5929, 6102
Offset: 1

Views

Author

Stefan Steinerberger, Nov 07 2005

Keywords

Comments

Questions: (1) Is this sequence infinite? (2) Are the gaps between the elements of this sequence bounded from above? (3) If this sequence is infinite, what is its asymptotic growth? (4) Consider the definition of this sequence for other values c instead of the golden ratio. What are the properties of this modified sequence?

Examples

			trunc(phi^6) = 17, the 6th Fibonacci number is 8; the sum of their digits is the same, thus 6 is in the sequence.

Crossrefs

Cf. A066212, A001999.

Programs

  • Mathematica
    $MaxExtraPrecision = 10^9; fQ[n_] := Plus @@ IntegerDigits@Floor@(GoldenRatio^n) == Plus @@ IntegerDigits@Fibonacci@n; Select[ Range[6108], fQ[ # ] &] (* Robert G. Wilson v *)

Extensions

Edited, corrected and extended by Robert G. Wilson v, Nov 16 2005
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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