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.

A014551 Jacobsthal-Lucas numbers.

Original entry on oeis.org

2, 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025, 2047, 4097, 8191, 16385, 32767, 65537, 131071, 262145, 524287, 1048577, 2097151, 4194305, 8388607, 16777217, 33554431, 67108865, 134217727, 268435457, 536870911, 1073741825, 2147483647, 4294967297, 8589934591
Offset: 0

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Author

Keywords

Comments

Also gives the number of points of period n in the subshift of finite type corresponding to the square matrix A=[1,2;1,0] (this is then given by trace(A^n)). - Thomas Ward, Mar 07 2001
Sequence is identical to its signed inverse binomial transform (autosequence of the second kind). - Paul Curtz, Jul 11 2008
a(n) can be expressed in terms of values of the Fibonacci polynomials F_n(x), computed at x=1/sqrt(2). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
Pisano period lengths: 1, 1, 2, 2, 4, 2, 6, 2, 6, 4, 10, 2, 12, 6, 4, 2, 8, 6, 18, 4, ... - R. J. Mathar, Aug 10 2012
Let F(x) = Product_{n >= 0} (1 - x^(3*n+1))/(1 - x^(3*n+2)). This sequence is the simple continued fraction expansion of the real number 1 + F(-1/2) = 2.83717 78068 73232 99799 ... = 2 + 1/(1 + 1/(5 + 1/(7 + 1/(17 + ...)))). See A111317. - Peter Bala, Dec 26 2012
With different signs, 2, -1, 5, -7, 17, -31, 65, -127, 257, -511, 1025, -2047, ... is the Lucas V(-1,-2) sequence. - R. J. Mathar, Jan 08 2013
The identity 2 = 2/2 + 2^2/(2*1) - 2^3/(2*1*5) - 2^4/(2*1*5*7) + 2^5/(2*1*5*7*17) + 2^6/(2*1*5*7*17*31) - - + + can be viewed as a generalized Engel-type expansion of the number 2 to the base 2. Compare with A062510. - Peter Bala, Nov 13 2013
For n >= 2, a(n) is the number of ways to tile a 2 X n strip, where the first two columns have an extra cell at the top, with 1 X 2 dominoes and 2 X 2 squares. Shown here is one of the a(7)=127 ways for the n=7 case:
._.
|_|_________.
| | | |_| |
||__|_|_|_|. - Greg Dresden, Sep 26 2021
Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965) and the French mathematician Édouard Lucas (1842-1891). - Amiram Eldar, Oct 02 2023

References

  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. pp. 180, 255.
  • Lind and Marcus, An Introduction to Symbolic Dynamics and Coding, Cambridge University Press, 1995. (General material on subshifts of finite type)
  • Kritkhajohn Onphaeng and Prapanpong Pongsriiam. Jacobsthal and Jacobsthal-Lucas Numbers and Sums Introduced by Jacobsthal and Tverberg. Journal of Integer Sequences, Vol. 20 (2017), Article 17.3.6.
  • Abdelmoumène Zekiri, Farid Bencherif, Rachid Boumahdi, Generalization of an Identity of Apostol, J. Int. Seq., Vol. 21 (2018), Article 18.5.1.

Crossrefs

Cf. A001045 (companion "autosequence"), A019322, A066845, A111317.
Cf. A135440 (first differences), A166920 (partial sums).
Cf. A006995.

Programs

Formula

a(n+1) = 2 * a(n) - (-1)^n * 3.
From Len Smiley, Dec 07 2001: (Start)
a(n) = 2^n + (-1)^n.
G.f.: (2-x)/(1-x-2*x^2). (End)
E.g.f.: exp(x) + exp(-2*x) produces a signed version. - Paul Barry, Apr 27 2003
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-1, 2*k)*3^(2*k)/2^(n-2). - Paul Barry, Feb 21 2003
0, 1, 5, 7 ... is 2^n - 2*0^n + (-1)^n, the 2nd inverse binomial transform of (2^n-1)^2 (A060867). - Paul Barry, Sep 05 2003
a(n) = 2*T(n, i/(2*sqrt(2))) * (-i*sqrt(2))^n with i^2=-1. - Paul Barry, Nov 17 2003
a(n) = A078008(n) + A001045(n+1). - Paul Barry, Feb 12 2004
a(n) = 2*A001045(n+1) - A001045(n). - Paul Barry, Mar 22 2004
a(0)=2, a(1)=1, a(n) = a(n-1) + 2*a(n-2) for n > 1. - Philippe Deléham, Nov 07 2006
a(2*n+1) = Product_{d|(2*n+1)} cyclotomic(d,2). a(2^k*(2*n+1)) = Product_{d|(2*n+1)} cyclotomic(2*d,2^(2^k)). - Miklos Kristof, Mar 12 2007
a(n) = 2^{(n-1)/2}F_{n-1}(1/sqrt(2)) + 2^{(n+2)/2}F_{n-2}(1/sqrt(2)). - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Dec 15 2008
E.g.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k*(k+1)/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
G.f.: U(0) where U(k) = 1 + (-1)^k/(2^k - 4^k*x*2/(2*x*2^k + (-1)^k/U(k+1))) ; (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 02 2012
a(n) = sqrt(9*(A001045)^2 + (-1)^n*2^(n+2)). - Vladimir Shevelev, Mar 13 2013
G.f.: 2 + G(0)*x*(1+4*x)/(2-x), where G(k) = 1 + 1/(1 - x*(9*k-1)/( x*(9*k+8) - 2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 13 2013
a(n) = [x^n] ( (1 + x + sqrt(1 + 2*x + 9*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
For n >= 1: a(n) = A006995(2^((n+2)/2)) when n is even, a(n) = A006995(3*2^((n-1)/2) - 1) when n is odd. - Bob Selcoe, Sep 04 2017
a(n) = J(n) + 4*J(n-1), a(0)=2, where J is A001045. - Yuchun Ji, Apr 23 2019
For n >= 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020
For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 06 2020
From Kai Wang, May 30 2020: (Start)
(2 - a(n+1)/a(n))/9 = Sum_{m>=n} (-2)^m/(a(m)*a(m+1)).
a(n) = 2*A001045(n+1) - A001045(n).
a(n)^2 = a(2*n) + 2*(-2)^n.
a(n)^2 = 9*A001045(n)^2 + 4*(-2)^n.
a(2*n) = 9*A001045(n)^2 + 2*(-2)^n.
2*A001045(m+n) = A001045(m)*a(n) + a(m)*A001045(n).
2*(-2)^n*A001045(m-n) = A001045(m)*a(n) - a(m)*A001045(n).
A001045(m+n) + (-2)^n*A001045(m-n) = A001045(m)*a(n).
A001045(m+n) - (-2)^n*A001045(m-n) = a(m)*A001045(n).
2*a(m+n) = 9*A001045(m)*A001045(n) + a(m)*a(n).
2*(-2)^n*a(m-n) = a(m)*a(n) - 9*A001045(m)*A001045(n).
a(m+n) - (-2)^n*a(m-n) = 9*A001045(m)*A001045(n).
a(m+n) + (-2)^n*a(m-n) = a(m)*a(n).
a(m+n)*a(m-n) - a(m)*a(m) = 9*(-2)^(m-n)*A001045(n)^2.
a(m+1)*a(n) - a(m)*a(n+1) = 9*(-2)^n*A001045(m-n). (End)
a(n) = F(n+1) + F(n-1) + Sum_{k=0..(n-2)} a(k)*F(n-1-k) for F(n) the Fibonacci numbers and for n > 1. - Greg Dresden, Jun 03 2020

A140253 a(2*n) = 2*(2*4^(n-1)-1) and a(2*n-1) = 2*4^(n-1)-1.

Original entry on oeis.org

-1, 1, 2, 7, 14, 31, 62, 127, 254, 511, 1022, 2047, 4094, 8191, 16382, 32767, 65534, 131071, 262142, 524287, 1048574, 2097151, 4194302, 8388607, 16777214, 33554431, 67108862, 134217727, 268435454, 536870911
Offset: 0

Views

Author

Paul Curtz, Jun 23 2008

Keywords

Comments

The inverse binomial transform is 1, 1, 4, -2, 10, -14, 34, -62 which leads to (-1)^(n+1)*A135440(n).
For n > 0: A266161(a(n)) = n and A266161(m) < n for m < a(n). - Reinhard Zumkeller, Dec 22 2015
Also, the decimal representation of the diagonal from the corner to the origin of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 673", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. - Robert Price, Jul 23 2017

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 170.

Crossrefs

Programs

  • Haskell
    import Data.List (transpose)
    a140253 n = a140253_list !! n
    a140253_list = -1 : concat
                        (transpose [a083420_list, map (* 2) a083420_list])
    -- Reinhard Zumkeller, Dec 22 2015
  • Maple
    A140253:=proc(n): if type(n, odd) then 2*4^(((n+1)/2)-1)-1 else 2*(2*4^((n/2)-1)-1) fi: end: seq(A140253(n),n=0..29); # Johannes W. Meijer, Jun 24 2011
  • Mathematica
    Table[(2^(n+1) - 3 + (-1)^(n+1))/2, {n, 0, 30}] (* Jean-François Alcover, Jun 05 2017 *)

Formula

a(2*n) = 2*A083420(n-1) and a(2*n+1) = A083420(n)
a(n+1) - a(n) = A014551(n); Jacobsthal-Lucas numbers.
a(2*n) + a(2*n+1) = 9*A002450(n)
a(n+1) - 2*a(n) = A010674(n+1); repeat 3, 0.
a(n) + A000034(n+1) = A000079(n); powers of 2.
a(n)= a(n-1) + 2*a(n-2) + 3. - Gary Detlefs, Jun 22 2010
a(n+1) = A000069(2^n); odious numbers. - Johannes W. Meijer, Jun 24 2011
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n>2, a(0) = -1, a(1) = 1, a(2) = 2. - Philippe Deléham, Feb 25 2012
G.f.: (x^2+3*x-1)/((1-2*x)*(1-x)*(1+x)). - Philippe Deléham, Feb 25 2012

Extensions

Edited, corrected and information added by Johannes W. Meijer, Jun 24 2011

A283070 Sierpinski tetrahedron or tetrix numbers: a(n) = 2*4^n + 2.

Original entry on oeis.org

4, 10, 34, 130, 514, 2050, 8194, 32770, 131074, 524290, 2097154, 8388610, 33554434, 134217730, 536870914, 2147483650, 8589934594, 34359738370, 137438953474, 549755813890, 2199023255554, 8796093022210, 35184372088834, 140737488355330, 562949953421314
Offset: 0

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Author

Peter M. Chema, Feb 28 2017

Keywords

Comments

Number of vertices required to make a Sierpinski tetrahedron or tetrix of side length 2^n. The sum of the vertices (balls) plus line segments (rods) of one tetrix equals the vertices of its larger, adjacent iteration. See formula.
Equivalently, the number of vertices in the (n+1)-Sierpinski tetrahedron graph. - Eric W. Weisstein, Aug 17 2017
Also the independence number of the (n+2)-Sierpinski tetrahedron graph. - Eric W. Weisstein, Aug 29 2021
Final digit alternates 4 and 0.

Crossrefs

Subsequence of A016957.
First bisection of A052548, A087288; second bisection of A049332, A133140, A135440.
Cf. A002023 (edge count).

Programs

Formula

G.f.: 2*(2 - 5*x)/((1 - x)*(1 - 4*x)).
a(n) = 5*a(n-1) - 4*a(n-2) for n > 1.
a(n+1) = a(n) + A002023(n).
a(n) = 2*A052539(n) = A188161(n) - 1 = A087289(n) + 1 = A056469(2*n+2) = A261723(4*n+1).
E.g.f.: 2*(exp(4*x) + exp(x)). - G. C. Greubel, Aug 17 2017

Extensions

Entry revised by Editors of OEIS, Mar 01 2017

A191370 a(n) = 2*(1+(-1)^n)/3 + 2*A010892(n-1).

Original entry on oeis.org

1, 2, 4, 2, 4, 8, 22, 44, 88, 170, 340, 680, 1366, 2732, 5464, 10922, 21844, 43688, 87382, 174764, 349528, 699050, 1398100, 2796200, 5592406, 11184812, 22369624, 44739242
Offset: 0

Views

Author

Paul Curtz, Jun 01 2011

Keywords

Comments

a(n) and successive differences define an infinite array:
1, 2, 4, 2, 4, 8, ...
1, 2, -2, 2, 4, 14, ...
1, -4, 4, 2, 10, 8, ...
-5, 8, -2, 8, -2, 14, ...
13, -10, 10, -10, 16, 2, ...
-23, 20, -20, 26, -14, 32, ...
...
Its main diagonal consists of the powers 2^n. The first upper diagonal is a signed sequence of 2's. The second upper diagonal contains essentially A135440.

Crossrefs

Programs

  • Maple
    A010892 := proc(n) op( 1+(n mod 6),[1,1,0,-1,-1,0]) ; end proc:
    A191370 := proc(n) 2^n/3+2*(-1)^n/3+2*A010892(n-1) ; end proc:
    seq(A191370(n),n=0..30) ; # R. J. Mathar, Jun 06 2011
  • Mathematica
    LinearRecurrence[{2,0,-1,2},{1,2,4,2},30] (* Harvey P. Dale, Sep 06 2022 *)

Formula

a(n+3) = 3*2^n - a(n), n >= 0.
a(n+1) - 2*a(n) = -6*A131531(n+1).
a(3*n) = A007613(n), a(1+3*n) = 2*A007613(n), a(2+3*n) = 4*A007613(n).
a(n+6) = a(n) + 21*2^n.
a(n) = ((2^n + 2*(-1)^n)*2^n - 2*i*sqrt(3)*((1+i*sqrt(3))^n - (1-i*sqrt(3))^n))/(3*2^n), where i=sqrt(-1); a(n+1) = 2*(A001045(n) + A010892(n)). - Bruno Berselli, Jun 06 2011
G.f.: ( -1+5*x^3 ) / ( (2*x-1)*(1+x)*(x^2-x+1) ). - R. J. Mathar, Jun 06 2011
a(n) = 2*a(n-1) - a(n-3) + 2*a(n-4). - Paul Curtz, Jun 07 2011
a(n) = A113405(n+3) - 5*A113405(n). - R. J. Mathar, Jun 24 2011

A340660 A000079 is the first row. For the second row, subtract A001045. For the third row, subtract A001045 from the second one, etc. The resulting array is read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 1, 4, 1, 0, 3, 8, 1, -1, 2, 5, 16, 1, -2, 1, 2, 11, 32, 1, -3, 0, -1, 6, 21, 64, 1, -4, -1, -4, 1, 10, 43, 128, 1, -5, -2, -7, -4, -1, 22, 85, 256, 1, -6, -3, -10, -9, -12, 1, 42, 171, 512, 1, -7, -4, -13, -14, -23, -20, -1, 86, 341, 1024
Offset: 0

Views

Author

Paul Curtz, Jan 15 2021

Keywords

Comments

Every row has the signature (1,2).
(Among consequences: a(n) read by antidiagonals is
1,
1, 2,
1, 1, 4,
1, 0, 3, 8,
1, -1, 2, 5, 16
1, -2, 1, 2, 11, 32,
1, -3, 0, -1, 6, 21, 64,
... .
The row sums and their first two difference table terms are
1, 3, 6, 12, 23, 45, 88, ... = A086445(n+1) - 1
2, 3, 6, 11, 22, 43, 86, ... = A005578(n+2)
1, 3, 5, 11, 21, 43, 85, ... = A001045(n+2).
The antidiagonal sums are
b(n) = 1, 1, 3, 2, 5, 3, 9, 4, 15, 5, 27, 6, 49, 7, ... .)

Examples

			Square array:
1,  2,  4,   8,  16,  32,  64,  128, ... = A000079(n)
1,  1,  3,   5,  11,  21,  43,   85, ... = A001045(n+1)
1,  0,  2,   2,   6,  10,  22,   42, ... = A078008(n)
1, -1,  1,  -1,   1,  -1,   1,   -1, ... = A033999(n)
1, -2,  0,  -4,  -4, -12, -20,  -44, ... = -A084247(n)
1, -3, -1,  -7,  -9, -23, -41,  -87, ... = (-1)^n*A140966(n+1)
1, -4, -2, -10, -14, -34, -62, -130, ... = -A135440(n)
1, -5, -3, -13, -19, -45, -83, -173, ... = -A155980(n+3) or -A171382(n+1)
...
		

Crossrefs

Programs

  • Maple
    A:= (n, k)-> (<<0|1>, <2|1>>^k. <<1, 2-n>>)[1$2]:
    seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Jan 21 2021
  • Mathematica
    A340660[m_, n_] := LinearRecurrence[{1, 2}, {1, m}, {n}]; Table[Reverse[Table[A340660[m, n + m - 2] // First, {m, 2, -n + 3, -1}]], {n, 1, 11}] // Flatten (* Robert P. P. McKone, Jan 28 2021 *)
  • PARI
    T(n, k) = 2^k - n*(2^k - (-1)^k)/3;
    matrix(10,10,n,k,T(n-1,k-1)) \\ Michel Marcus, Jan 19 2021

Formula

A(n,k) = 2^k - n*round(2^k/3).
Showing 1-5 of 5 results.