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 1-10 of 17 results. Next

A239544 (Round(c^prime(n)) - 1)/prime(n), where c is the tetranacci constant (A086088).

Original entry on oeis.org

14, 124, 390, 4118, 13690, 156122, 6351030, 22074820, 948652694, 11818395344, 41868809842, 528803858638, 24052859078262, 1108257471317098, 3982717894786008, 185987895674303758, 2422894681885464596, 8755616404517667662, 414985190213435939298
Offset: 4

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Author

Keywords

Comments

For n>=4, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. In particular, all terms are even.

Crossrefs

A058265 Decimal expansion of the tribonacci constant t, the real root of x^3 - x^2 - x - 1.

Original entry on oeis.org

1, 8, 3, 9, 2, 8, 6, 7, 5, 5, 2, 1, 4, 1, 6, 1, 1, 3, 2, 5, 5, 1, 8, 5, 2, 5, 6, 4, 6, 5, 3, 2, 8, 6, 6, 0, 0, 4, 2, 4, 1, 7, 8, 7, 4, 6, 0, 9, 7, 5, 9, 2, 2, 4, 6, 7, 7, 8, 7, 5, 8, 6, 3, 9, 4, 0, 4, 2, 0, 3, 2, 2, 2, 0, 8, 1, 9, 6, 6, 4, 2, 5, 7, 3, 8, 4, 3, 5, 4, 1, 9, 4, 2, 8, 3, 0, 7, 0, 1, 4
Offset: 1

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Author

Robert G. Wilson v, Dec 07 2000

Keywords

Comments

"The tribonacci constant, the only real solution to the equation x^3 - x^2 - x - 1 = 0, which is related to tribonacci sequences (in which U_n = U_n-1 + U_n-2 + U_n-3) as the Golden Ratio is related to the Fibonacci sequence and its generalizations. This ratio also appears when a snub cube is inscribed in an octahedron or a cube, by analogy once again with the appearance of the Golden Ratio when an icosahedron is inscribed in an octahedron. [John Sharp, 1997]"
The tribonacci constant corresponds to the Golden Section in a tripartite division 1 = u_1 + u_2 + u_3 of a unit line segment; i.e., if 1/u_1 = u_1/u_2 = u_2/u_3 = c, c is the tribonacci constant. - Seppo Mustonen, Apr 19 2005
The other two polynomial roots are the complex-conjugated pair -0.4196433776070805662759262... +- i* 0.60629072920719936925934... - R. J. Mathar, Oct 25 2008
For n >= 3, round(q^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Concerning orthogonal projections, the tribonacci constant is the ratio of the diagonal of a square to the width of a rhombus projected by rotating a square along its diagonal in 3D until the angle of rotation equals the apparent apex angle at approximately 57.065 degrees (also the corresponding angle in the formula generating A256099). See illustration in the links. - Peter M. Chema, Jan 02 2017
From Wolfdieter Lang, Aug 10 2018: (Start)
Real eigenvalue t of the tribonacci Q-matrix <<1, 1, 1>,<1, 0, 0>,<0, 1, 0>>.
Limit_{n -> oo} T(n+1)/T(n) = t (from the T recurrence), where T = {A000073(n+2)}_{n >= 0}. (End)
The nonnegative powers of t are t^n = T(n)*t^2 + (T(n-1) + T(n-2))*t + T(n-1)*1, for n >= 0, with T(n) = A000073(n), with T(-1) = 1 and T(-2) = -1, This follows from the recurrences derived from t^3 = t^2 + t + 1. See the examples below. For the negative powers see A319200. - Wolfdieter Lang, Oct 23 2018
Note that we have: t + t^(-3) = 2, and the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 16 2022
The roots of this cubic are found from those of y^3 - (4/3)*y - 38/27, after adding 1/3. - Wolfdieter Lang, Aug 24 2022
The algebraic number t - 1 has minimal polynomial x^3 + 2*x^2 - 2 over Q. The roots coincide with those of y^3 - (4/3)*y - 38/27, after subtracting 2/3. - Wolfdieter Lang, Sep 20 2022
The value of the ratio R/r of the radius R of a uniform ball to the radius r of a spherical hole in it with a common point of contact, such that the center of gravity of the object lies on the surface of the spherical hole (Schmidt, 2002). - Amiram Eldar, May 20 2023

Examples

			1.8392867552141611325518525646532866004241787460975922467787586394042032220\
    81966425738435419428307014141979826859240974164178450746507436943831545\
    820499513796249655539644613666121540277972678118941041...
From _Wolfdieter Lang_, Oct 23 2018: (Start)
The coefficients of t^2, t, 1 for t^n begin, for n >= 0:
    n     t^2    t    1
    -------------------
    0      0     0    1
    1      0     1    0
    2      1     0    0
    1      1     1    1
    4      2     2    1
    5      4     3    2
    6      7     6    4
    7     13    11    7
    8     24    20   13
    9     44    37   24
   10     81    68   44
...  (End)
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.2.2.
  • Martin Gardner, The Second Scientific American Book of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, p. 101, Simon & Schuster, NY, 1961.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, Penguin Books, London, England, 1997, page 23.

Crossrefs

Cf. A000073, A019712 (continued fraction), A133400, A254231, A158919 (spectrum = floor(n*t)), A357101 (x^3-2*x^2-2).
Cf. A192918 (reciprocal), A276800 (square), A276801 (cube), A319200.
k-nacci constants: A001622 (Fibonacci), this sequence (tribonacci), A086088 (tetranacci), A103814 (pentanacci), A118427 (hexanacci), A118428 (heptanacci).

Programs

  • Maple
    Digits:=200; fsolve(x^3=x^2+x+1); # N. J. A. Sloane, Mar 16 2019
  • Mathematica
    RealDigits[ N[ 1/3 + 1/3*(19 - 3*Sqrt[33])^(1/3) + 1/3*(19 + 3*Sqrt[33])^(1/3), 100]] [[1]]
    RealDigits[Root[x^3-x^2-x-1,1],10,120][[1]] (* Harvey P. Dale, Mar 23 2019 *)
  • Maxima
    set_display(none)$ fpprec:100$ bfloat(rhs(solve(t^3-t^2-t-1,t)[3])); /* Dimitri Papadopoulos, Nov 09 2023 */
  • PARI
    default(realprecision, 20080); x=solve(x=1, 2, x^3 - x^2 - x - 1); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b058265.txt", n, " ", d));  \\ Harry J. Smith, May 30 2009
    
  • PARI
    q=(1+sqrtn(19+3*sqrt(33),3)+sqrtn(19-3*sqrt(33),3))/3 \\ Use \p# to set 'realprecision'. - M. F. Hasler, Mar 23 2014
    

Formula

t = (1/3)*(1+(19+3*sqrt(33))^(1/3)+(19-3*sqrt(33))^(1/3)). - Zak Seidov, Jun 08 2005
t = 1 - Sum_{k>=1} A057597(k+2)/(T_k*T_(k+1)), where T_n = A000073(n+1). - Vladimir Shevelev, Mar 02 2013
1/t + 1/t^2 + 1/t^3 = 1/A058265 + 1/A276800 + 1/A276801 = 1. - N. J. A. Sloane, Oct 28 2016
t = (4/3)*cosh((1/3)*arccosh(19/8)) + 1/3. - Wolfdieter Lang, Aug 24 2022
t = 2 - Sum_{k>=0} binomial(4*k + 2, k)/((k + 1)*2^(4*k + 3)). - Antonio Graciá Llorente, Oct 28 2024

A000078 Tetranacci numbers: a(n) = a(n-1) + a(n-2) + a(n-3) + a(n-4) for n >= 4 with a(0) = a(1) = a(2) = 0 and a(3) = 1.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 4, 8, 15, 29, 56, 108, 208, 401, 773, 1490, 2872, 5536, 10671, 20569, 39648, 76424, 147312, 283953, 547337, 1055026, 2033628, 3919944, 7555935, 14564533, 28074040, 54114452, 104308960, 201061985, 387559437, 747044834, 1439975216, 2775641472
Offset: 0

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Author

Keywords

Comments

a(n) is the number of compositions of n-3 with no part greater than 4. Example: a(7) = 8 because we have 1+1+1+1 = 2+1+1 = 1+2+1 = 3+1 = 1+1+2 = 2+2 = 1+3 = 4. - Emeric Deutsch, Mar 10 2004
In other words, a(n) is the number of ways of putting stamps in one row on an envelope using stamps of denominations 1, 2, 3 and 4 cents so as to total n-3 cents [Pólya-Szegő]. - N. J. A. Sloane, Jul 28 2012
a(n+4) is the number of 0-1 sequences of length n that avoid 1111. - David Callan, Jul 19 2004
a(n) is the number of matchings in the graph obtained by a zig-zag triangulation of a convex (n-3)-gon. Example: a(8) = 15 because in the triangulation of the convex pentagon ABCDEA with diagonals AD and AC we have 15 matchings: the empty set, seven singletons and {AB,CD}, {AB,DE}, {BC,AD}, {BC,DE}, {BC,EA}, {CD,EA} and {DE,AC}. - Emeric Deutsch, Dec 25 2004
Number of permutations satisfying -k <= p(i)-i <= r, i=1..n-3, with k = 1, r = 3. - Vladimir Baltic, Jan 17 2005
For n >= 0, a(n+4) is the number of palindromic compositions of 2*n+1 into an odd number of parts that are not multiples of 4. In addition, a(n+4) is also the number of Sommerville symmetric cyclic compositions (= bilaterally symmetric cyclic compositions) of 2*n+1 into an odd number of parts that are not multiples of 4. - Petros Hadjicostas, Mar 10 2018
a(n) is the number of ways to tile a hexagonal double-strip (two rows of adjacent hexagons) containing (n-4) cells with hexagons and double-hexagons (two adjacent hexagons). - Ziqian Jin, Jul 28 2019
The term "tetranacci number" was coined by Mark Feinberg (1963; see A000073). - Amiram Eldar, Apr 16 2021
a(n) is the number of ways to tile a skew double-strip of n-3 cells using squares and all possible "dominos", as seen in Ziqian Jin's article, below. Here is the skew double-strip corresponding to n=15, with 12 cells:
_ ___ _ ___ _ ___
| | | | | | |
|__|___|_|___| |___|
| | | | | | |
|_|___|_|___|_|___|,
and here are the three possible "domino" tiles:
_ _
| | | |
| | | | | |
|_|, |_|, |_____|.
As an example, here is one of the a(15) = 1490 ways to tile the skew double-strip of 12 cells:
_ ___ _____ _______
| | | | | | |
|__|_ |_|_ | _| _|
| | | | | |
|_____|___|_|___|_|. - Greg Dresden, Jun 05 2024

Examples

			From _Petros Hadjicostas_, Mar 10 2018: (Start)
For n = 3, we get a(3+4) = a(7) = 8 palindromic compositions of 2*n+1 = 7 into an odd number of parts that are not a multiple of 4. They are the following: 7 = 1+5+1 = 3+1+3 = 2+3+2 = 1+2+1+2+1 = 2+1+1+1+2 = 1+1+3+1+1 = 1+1+1+1+1+1+1. If we put these compositions on a circle, they become bilaterally symmetric cyclic compositions of 2*n+1 = 7.
For n = 4, we get a(4+4) = a(8) = 15 palindromic compositions of 2*n + 1 = 9 into an odd number of parts that are not a multiple of 4. They are the following: 9 = 3+3+3 = 2+5+2 = 1+7+1 = 1+1+5+1+1 = 2+1+3+1+2 = 1+2+3+2+1 = 1+3+1+3+1 = 3+1+1+1+3 = 2+2+1+2+2 = 2+1+1+1+1+1+2 = 1+2+1+1+1+2+1 = 1+1+2+1+2+1+1 = 1+1+1+3+1+1+1 = 1+1+1+1+1+1+1+1+1.
As _David Callan_ points out in the comments above, for n >= 1, a(n+4) is also the number of 0-1 sequences of length n that avoid 1111. For example, for n = 5, a(5+4) = a(9) = 29 is the number of binary strings of length n that avoid 1111. Out of the 2^5 = 32 binary strings of length n = 5, the following do not avoid 1111: 11111, 01111, and 11110. (End)
		

References

  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • G. Pólya and G. Szegő, Problems and Theorems in Analysis, Springer-Verlag, NY, 2 vols., 1972, Vol. 1, p. 1, Problems 3 and 4.
  • J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, Princeton, NJ, 1978.
  • 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).

Crossrefs

Row 4 of arrays A048887 and A092921 (k-generalized Fibonacci numbers).
First differences are in A001631.
Cf. A008287 (quadrinomial coefficients) and A073817 (tetranacci with different initial conditions).

Programs

  • GAP
    a:=[0,0,0,1];; for n in [5..40] do a[n]:=a[n-1]+a[n-2]+a[n-3]+a[n-4]; od; a; # Muniru A Asiru, Mar 11 2018
  • Haskell
    import Data.List (tails, transpose)
    a000078 n = a000078_list !! n
    a000078_list = 0 : 0 : 0 : f [0,0,0,1] where
       f xs = y : f (y:xs) where
         y = sum $ head $ transpose $ take 4 $ tails xs
    -- Reinhard Zumkeller, Jul 06 2014, Apr 28 2011
    
  • Magma
    [n le 4 select Floor(n/4) else Self(n-1)+Self(n-2)+Self(n-3)+Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jan 29 2016
    
  • Maple
    a:= n-> (<<1|1|0|0>, <1|0|1|0>, <1|0|0|1>, <1|0|0|0>>^n)[1, 4]: seq(a(n), n=0..50); # Alois P. Heinz, Jun 12 2008
  • Mathematica
    CoefficientList[Series[x^3/(1-x-x^2-x^3-x^4), {x, 0, 50}], x]
    LinearRecurrence[{1,1,1,1}, {0,0,0,1}, 50]  (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)
    (* From Eric W. Weisstein, Nov 09 2017 *)
    Table[RootSum[-1 -# -#^2 -#^3 +#^4 &, 10#^n +157#^(n+1) -103 #^(n+2) +16#^(n+3) &]/563, {n, 0, 40}]
    Table[RootSum[#^4 -#^3 -#^2 -# -1 &, #^(n-2)/(-#^3 +6# -1) &], {n, 0, 40}] (* End *)
  • Maxima
    a(n):=sum(sum(if mod(5*k-i,4)>0 then 0 else binomial(k,(5*k-i)/4)*(-1)^((i-k)/4)*binomial(n-i+k-1,k-1),i,k,n),k,1,n); /* Vladimir Kruchinin, Aug 18 2010 */
    
  • PARI
    {a(n) = if( n<0, 0, polcoeff( x^3 / (1 - x - x^2 - x^3 - x^4) + x * O(x^n), n))}
    
  • Python
    A000078 = [0,0,0,1]
    for n in range(4, 100):
        A000078.append(A000078[n-1]+A000078[n-2]+A000078[n-3]+A000078[n-4])
    # Chai Wah Wu, Aug 20 2014
    

Formula

a(n) = A001630(n) - a(n-1). - Henry Bottomley
G.f.: x^3/(1 - x - x^2 - x^3 - x^4). - Simon Plouffe in his 1992 dissertation
G.f.: x^3 / (1 - x / (1 - x / (1 + x^3 / (1 + x / (1 - x / (1 + x)))))). - Michael Somos, May 12 2012
G.f.: Sum_{n >= 0} x^(n+3) * (Product_{k = 1..n} (k + k*x + k*x^2 + x^3)/(1 + k*x + k*x^2 + k*x^3)). - Peter Bala, Jan 04 2015
a(n) = term (1,4) in the 4 X 4 matrix [1,1,0,0; 1,0,1,0; 1,0,0,1; 1,0,0,0]^n. - Alois P. Heinz, Jun 12 2008
Another form of the g.f.: f(z) = (z^3 - z^4)/(1 - 2*z + z^5), then a(n) = Sum_{i=0..floor((n-3)/5)} (-1)^i*binomial(n-3-4*i, i)*2^(n - 3 - 5*i) - Sum_{i=0..floor((n-4)/5)} (-1)^i*binomial(n-4-4*i, i)*2^(n - 4 - 5*i) with natural convention Sum_{i=m..n} alpha(i) = 0 for m > n. - Richard Choulet, Feb 22 2010
a(n+3) = Sum_{k=1..n} Sum_{i=k..n} [(5*k-i mod 4) = 0] * binomial(k, (5*k-i)/4) *(-1)^((i-k)/4) * binomial(n-i+k-1,k-1), n > 0. - Vladimir Kruchinin, Aug 18 2010 [Edited by Petros Hadjicostas, Jul 26 2020, so that the formula agrees with the offset of the sequence]
Sum_{k=0..3*n} a(k+b) * A008287(n,k) = a(4*n+b), b >= 0 ("quadrinomial transform"). - N. J. A. Sloane, Nov 10 2010
G.f.: x^3*(1 + x*(G(0)-1)/(x+1)) where G(k) = 1 + (1+x+x^2+x^3)/(1-x/(x+1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
Starting (1, 2, 4, 8, ...) = the INVERT transform of (1, 1, 1, 1, 0, 0, 0, ...). - Gary W. Adamson, May 13 2013
a(n) ~ c*r^n, where c = 0.079077767399388561146007... and r = 1.92756197548292530426195... = A086088 (One of the roots of the g.f. denominator polynomial is 1/r.). - Fung Lam, Apr 29 2014
a(n) = 2*a(n-1) - a(n-5), n >= 5. - Bob Selcoe, Jul 06 2014
From Ziqian Jin, Jul 28 2019: (Start)
a(2*n+5) = a(n+4)^2 + a(n+3)^2 + a(n+2)^2 + 2*a(n+3)*(a(n+2) + a(n+1)).
a(n) - 1 = a(n-2) + 2*a(n-3) + 3*(a(n-4) + a(n-5) + ... + a(2) + a(1)), n >= 4. (End)
a(n) = (Sum_{i=0..n-1} a(i)*A073817(n-i))/(n-3) for n > 3. - Greg Dresden and Advika Srivastava, Sep 28 2019
a(n) = Sum_{r root of x^4-x^3-x^2-x-1} r^n/(4*r^3-3*r^2-2*r-1). - Fabian Pereyra, Dec 06 2024

Extensions

Definition augmented (with 4 initial terms) by Daniel Forgues, Dec 02 2009
Deleted certain dangerous or potentially dangerous links. - N. J. A. Sloane, Jan 30 2021

A103814 Pentanacci constant: decimal expansion of limit of A001591(n+1)/A001591(n).

Original entry on oeis.org

1, 9, 6, 5, 9, 4, 8, 2, 3, 6, 6, 4, 5, 4, 8, 5, 3, 3, 7, 1, 8, 9, 9, 3, 7, 3, 7, 5, 9, 3, 4, 4, 0, 1, 3, 9, 6, 1, 5, 1, 3, 2, 7, 1, 7, 7, 4, 5, 6, 8, 6, 1, 3, 9, 3, 2, 3, 6, 9, 3, 4, 5, 0, 8, 4, 4, 2, 2, 5, 2, 7, 1, 2, 8, 7, 1, 8, 8, 6, 8, 8, 1, 7, 3, 4, 8, 1, 8, 6, 6, 5, 5, 5, 4, 6, 3, 0, 4, 7, 2, 0, 2, 1, 3, 0
Offset: 1

Views

Author

Jonathan Vos Post, Mar 29 2005

Keywords

Comments

The pentanacci constant P is the limit as n -> infinity of the ratio of Pentanacci(n+1)/Pentanacci(n) = A001591(n+1)/A001591(n), which is the principal root of x^5-x^4-x^3-x^2-x-1 = 0. Note that we have: P + P^-5 = 2.
The pentanacci constant corresponds to the Golden Section in a fivepartite division 1 = u_1 + u_2 + u_3 + u_4 + u_5 of a unit line segment, i.e., if 1/u_1 = u_1/u_2 = u_2/u_3 = u_3/u_4 + u_4/u_5 = c, c is the pentanacci constant. - Seppo Mustonen, Apr 19 2005
The other 4 roots of the polynomial 1+x+x^2+x^3+x^4-x^5 are the two complex-conjugated pairs -0.6783507129699967... +- i * 0.458536187273144499.. and 0.1953765946472540452... +- i * 0.848853640546245551858... - R. J. Mathar, Oct 25 2008
The continued fraction expansion is 1, 1, 28, 2, 1, 2, 1, 1, 1, 2, 4, 2, 1, 3, 1, 6, 1, 4, 1, 1, 5, 3, 2, 15, 69, 1, 1, 14, 1, 8, 1, 6,... - R. J. Mathar, Mar 09 2012
For n>=5, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Note that the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 07 2022

Examples

			1.965948236645485337189937375934401396151327177456861393236934508442...
		

References

  • Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, p. 101, Simon & Schuster, NY, 1961.

Crossrefs

Cf. A001591.
k-nacci constants: A001622 (Fibonacci), A058265 (tribonacci), A086088 (tetranacci), this sequence (pentanacci), A118427 (hexanacci), A118428 (heptanacci).

Programs

  • Mathematica
    RealDigits[Root[x^5-Total[x^Range[0,4]],1],10,120][[1]] (* Harvey P. Dale, Mar 22 2017 *)
  • PARI
    solve(x=1, 2, 1+x+x^2+x^3+x^4-x^5) \\ Michel Marcus, Mar 21 2014

A118427 Decimal expansion of hexanacci constant.

Original entry on oeis.org

1, 9, 8, 3, 5, 8, 2, 8, 4, 3, 4, 2, 4, 3, 2, 6, 3, 3, 0, 3, 8, 5, 6, 2, 9, 2, 9, 3, 3, 9, 1, 4, 2, 5, 7, 5, 2, 7, 3, 0, 0, 8, 0, 8, 6, 5, 5, 6, 8, 8, 2, 1, 7, 5, 3, 2, 1, 6, 3, 5, 9, 0, 6, 5, 6, 5, 6, 7, 0, 2, 2, 7, 8, 0, 1, 4, 1, 7, 2, 4, 0, 2, 9, 8, 6, 5, 7, 5, 0, 7, 0, 2, 2, 6, 8, 9, 9, 7, 9, 7, 3, 2, 7, 7, 5
Offset: 1

Views

Author

Eric W. Weisstein, Apr 27 2006

Keywords

Comments

The continued fraction expansion starts 1, 1, 59, 1, 10, 2, 1, 6, 2, 1, 6, 1, 1, 7, 1, 71, 7, 1, 6, 8, ... - R. J. Mathar, Mar 09 2012
For n>=7, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Note that we have: c + c^(-6) = 2, and the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 06 2022

Examples

			1.9835828434243263303...
		

References

  • Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, Simon & Schuster, NY, 1961.

Crossrefs

Cf. A001592.
k-nacci constants: A001622 (Fibonacci), A058265 (tribonacci), A086088 (tetranacci), A103814 (pentanacci), this sequence (hexanacci), A118428 (heptanacci).

Programs

  • Mathematica
    RealDigits[ Root[ x^6 - x^5 - x^4 - x^3 - x^2 - x - 1, 2] , 10, 105] // First (* Jean-François Alcover, Feb 07 2013 *)

A118428 Decimal expansion of heptanacci constant.

Original entry on oeis.org

1, 9, 9, 1, 9, 6, 4, 1, 9, 6, 6, 0, 5, 0, 3, 5, 0, 2, 1, 0, 9, 7, 7, 4, 1, 7, 5, 4, 5, 8, 4, 3, 7, 4, 9, 6, 3, 4, 7, 9, 3, 1, 8, 9, 6, 0, 0, 5, 3, 1, 5, 7, 9, 9, 5, 2, 4, 4, 7, 8, 2, 1, 5, 3, 4, 0, 0, 9, 5, 1, 9, 8, 0, 3, 0, 9, 6, 2, 2, 1, 8, 3, 5, 6, 3, 1, 4, 1, 5, 7, 7, 0, 2, 2, 7, 1, 9, 0, 1, 7, 0, 9, 9, 1, 6
Offset: 1

Views

Author

Eric W. Weisstein, Apr 27 2006

Keywords

Comments

Other roots of the equation x^7 - x^6 - ... - x - 1 see in A239566. For n>=7, round(c^prime(n)) == 1 (mod 2*prime(n)). Proof in Shevelev link. - Vladimir Shevelev, Mar 21 2014
Note that we have: c + c^(-7) = 2, and the k-nacci constant approaches 2 when k approaches infinity (Martin Gardner). - Bernard Schott, May 07 2022

Examples

			1.9919641966050350210...
		

References

  • Martin Gardner, The Second Scientific American Book Of Mathematical Puzzles and Diversions, "Phi: The Golden Ratio", Chapter 8, p. 101, Simon & Schuster, NY, 1961.

Crossrefs

k-nacci constants: A001622 (Fibonacci), A058265 (tribonacci), A086088 (tetranacci), A103814 (pentanacci), A118427 (hexanacci), this sequence (heptanacci).

Programs

  • Mathematica
    RealDigits[x/.FindRoot[x^7+Total[-x^Range[0,6]]==0,{x,2}, WorkingPrecision-> 110]][[1]] (* Harvey P. Dale, Dec 13 2011 *)
  • PARI
    polrootsreal(x^7 - x^6 - x^5 - x^4 - x^3 - x^2 - x - 1)[1] \\ Charles R Greathouse IV, Feb 11 2025

A184823 a(n) = n + floor(n/t) + floor(n/t^2) + floor(n/t^3), where t is the tetranacci constant.

Original entry on oeis.org

1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, 22, 23, 25, 28, 30, 31, 33, 35, 37, 38, 41, 43, 45, 46, 48, 51, 52, 55, 57, 59, 60, 62, 64, 66, 68, 70, 72, 74, 75, 78, 79, 82, 83, 86, 87, 89, 90, 93, 94, 97, 98, 101, 103, 104, 107, 108, 111, 112, 115, 116, 118, 119, 122, 124, 126, 128, 130, 131, 133, 135, 138, 139, 141, 143, 145, 146, 148, 151, 153, 155, 157, 159, 160, 162, 165, 167, 168, 170, 172, 174, 175, 178, 180, 182, 183, 186, 187, 189, 190, 194, 195, 197, 198, 201, 202, 204, 208, 209, 211, 212, 215, 216, 218, 220, 223, 224
Offset: 1

Views

Author

Paul D. Hanna, Jan 23 2011

Keywords

Comments

This is one of four sequences that partition the positive integers.
Given t is the tetranacci constant, then the following sequences are disjoint:
. A184823(n) = n + [n/t] + [n/t^2] + [n/t^3],
. A184824(n) = n + [n*t] + [n/t] + [n/t^2],
. A184825(n) = n + [n*t] + [n*t^2] + [n/t],
. A184826(n) = n + [n*t] + [n*t^2] + [n*t^3], where []=floor.
This is a special case of Clark Kimberling's results given in A184812.

Examples

			Let t be the tetranacci constant, then t = 1 + 1/t + 1/t^2 + 1/t^3 and:
t = 1.92756197548..., t^2 = 3.71549516932..., t^3 = 7.16184720848..., t^4 = 13.8049043532...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(t=real(polroots(1+x+x^2+x^3-x^4)[2])); n+floor(n/t)+floor(n/t^2)+floor(n/t^3)}

Formula

Limit a(n)/n = t = 1.9275619754829253042619058...
a(n) = n + floor(n*p/s) + floor(n*q/s) + floor(n*r/s), where p=t, q=t^2, r=t^3, s=t^4, and t is the tetranacci constant.

A184824 a(n) = n + floor(n*t) + floor(n/t) + floor(n/t^2), where t is the tetranacci constant.

Original entry on oeis.org

2, 6, 9, 14, 17, 21, 24, 29, 32, 36, 39, 44, 47, 50, 54, 58, 61, 65, 69, 73, 76, 80, 84, 88, 91, 95, 100, 102, 106, 110, 114, 117, 121, 125, 129, 132, 136, 140, 144, 147, 152, 154, 158, 161, 166, 169, 173, 176, 181, 184, 188, 191, 196, 200, 203, 207, 210, 214, 217, 222, 225, 229, 232, 237, 240, 244, 248, 252, 255, 258, 262, 266, 269, 273, 277, 281, 284, 288, 292, 296, 300, 304, 307, 310, 314, 318, 322, 325, 329, 333, 337, 340, 345, 348, 352, 355, 359, 362, 366, 369, 374, 377, 381, 384, 389, 392, 396, 401, 404, 408
Offset: 1

Views

Author

Paul D. Hanna, Jan 23 2011

Keywords

Comments

This is one of four sequences that partition the positive integers.
Given t is the tetranacci constant, then the following sequences are disjoint:
. A184823(n) = n + [n/t] + [n/t^2] + [n/t^3],
. A184824(n) = n + [n*t] + [n/t] + [n/t^2],
. A184825(n) = n + [n*t] + [n*t^2] + [n/t],
. A184826(n) = n + [n*t] + [n*t^2] + [n*t^3], where []=floor.
This is a special case of Clark Kimberling's results given in A184812.

Examples

			Let t be the tetranacci constant, then t^2 = 1 + t + 1/t + 1/t^2 and:
t = 1.92756197548..., t^2 = 3.71549516932..., t^3 = 7.16184720848..., t^4 = 13.8049043532...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(t=real(polroots(1+x+x^2+x^3-x^4)[2])); n+floor(n*t)+floor(n/t)+floor(n/t^2)}

Formula

Limit a(n)/n = t^2 = 3.7154951693276375317543272...
a(n) = n + floor(n*p/r) + floor(n*q/r) + floor(n*s/r), where p=t, q=t^2, r=t^3, s=t^4, and t is the tetranacci constant.

A184825 a(n) = n + floor(n*t) + floor(n*t^2) + floor(n/t), where t is the tetranacci constant.

Original entry on oeis.org

5, 13, 20, 27, 34, 42, 49, 56, 63, 71, 77, 85, 92, 99, 105, 113, 120, 127, 134, 142, 149, 156, 163, 171, 177, 185, 193, 199, 206, 213, 221, 227, 235, 242, 250, 256, 264, 271, 278, 285, 293, 299, 306, 313, 321, 327, 335, 342, 350, 356, 364, 371, 378, 386, 393, 400, 406, 414, 421, 428, 435, 443, 450, 457, 464, 472, 478, 486, 493, 500, 506, 514, 521, 528, 535, 543, 550, 557, 564, 572, 579, 586, 593, 600, 607, 614, 622, 628, 636, 643, 651, 657, 665, 672, 679, 686, 693, 700, 707, 714, 722, 728, 736, 743, 751, 757
Offset: 1

Views

Author

Paul D. Hanna, Jan 23 2011

Keywords

Comments

This is one of four sequences that partition the positive integers.
Given t is the tetranacci constant, then the following sequences are disjoint:
. A184823(n) = n + [n/t] + [n/t^2] + [n/t^3],
. A184824(n) = n + [n*t] + [n/t] + [n/t^2],
. A184825(n) = n + [n*t] + [n*t^2] + [n/t],
. A184826(n) = n + [n*t] + [n*t^2] + [n*t^3], where []=floor.
This is a special case of Clark Kimberling's results given in A184812.

Examples

			Let t be the tetranacci constant, then t^3 = 1 + t + t^2 + 1/t and:
t = 1.92756197548..., t^2 = 3.71549516932..., t^3 = 7.16184720848..., t^4 = 13.8049043532...
		

Crossrefs

Programs

  • Mathematica
    With[{t=x/.Last[Solve[x^4==Total[x^Range[0,3]],x]]},Table[n+Floor[n t]+Floor[n t^2]+Floor[n/t],{n,120}]]  (* Harvey P. Dale, Feb 02 2011 *)
  • PARI
    {a(n)=local(t=real(polroots(1+x+x^2+x^3-x^4)[2])); n+floor(n*t)+floor(n*t^2)+floor(n/t)}

Formula

Lim_{n->infinity} a(n)/n = t^3 = 7.1618472084864470579236869...
a(n) = n + floor(n*p/q) + floor(n*r/q) + floor(n*s/q), where p=t, q=t^2, r=t^3, s=t^4, and t is the tetranacci constant.

A184826 a(n) = n + floor(n*t) + floor(n*t^2) + floor(n*t^3) where t is the tetranacci constant.

Original entry on oeis.org

12, 26, 40, 53, 67, 81, 96, 109, 123, 137, 150, 164, 179, 192, 205, 219, 233, 246, 261, 275, 289, 302, 316, 330, 344, 358, 372, 385, 398, 412, 427, 440, 454, 468, 482, 495, 509, 524, 537, 551, 565, 578, 591, 606, 620, 633, 647, 661, 675, 689, 703, 717, 730, 744, 758, 772, 785, 799, 813, 826, 840, 855, 869, 882, 896, 910, 923, 938, 952, 965, 978, 992, 1006, 1019, 1034, 1048, 1062, 1075, 1089, 1103, 1117, 1131, 1144, 1158, 1171, 1185, 1200, 1213, 1227, 1241, 1255, 1268, 1283, 1297, 1310, 1324, 1337, 1351
Offset: 1

Views

Author

Paul D. Hanna, Jan 23 2011

Keywords

Comments

This is one of four sequences that partition the positive integers.
Given t is the tetranacci constant, then the following sequences are disjoint:
. A184823(n) = n + [n/t] + [n/t^2] + [n/t^3],
. A184824(n) = n + [n*t] + [n/t] + [n/t^2],
. A184825(n) = n + [n*t] + [n*t^2] + [n/t],
. A184826(n) = n + [n*t] + [n*t^2] + [n*t^3], where []=floor.
This is a special case of Clark Kimberling's results given in A184812.

Examples

			Let t be the tetranacci constant, then t^4 = 1 + t + t^2 + t^3 and:
t = 1.92756197548..., t^2 = 3.71549516932..., t^3 = 7.16184720848..., t^4 = 13.8049043532...
		

Crossrefs

Programs

  • Mathematica
    Module[{t=x/.FindRoot[x^4-x^3-x^2-x-1==0,{x,2},WorkingPrecision->200], t2,t3},t2=t^2;t3=t^3;Table[n+Floor[t*n]+Floor[t2*n]+Floor[t3*n], {n,100}]] (* Harvey P. Dale, Oct 18 2012 *)
  • PARI
    {a(n)=local(t=real(polroots(1+x+x^2+x^3-x^4)[2])); n+floor(n*t)+floor(n*t^2)+floor(n*t^3)}

Formula

Limit a(n)/n = t^4 = 13.804904353297009893939920...
a(n) = n + floor(n*q/p) + floor(n*r/p) + floor(n*s/p), where p=t, q=t^2, r=t^3, s=t^4, and t is the tetranacci constant.
Showing 1-10 of 17 results. Next