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 1-8 of 8 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

A062510 a(n) = 2^n + (-1)^(n+1).

Original entry on oeis.org

0, 3, 3, 9, 15, 33, 63, 129, 255, 513, 1023, 2049, 4095, 8193, 16383, 32769, 65535, 131073, 262143, 524289, 1048575, 2097153, 4194303, 8388609, 16777215, 33554433, 67108863, 134217729, 268435455, 536870913, 1073741823, 2147483649
Offset: 0

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Author

Jason Earls, Jun 24 2001

Keywords

Comments

The identity 2 = 2^2/3 + 2^3/(3*3) - 2^4/(3*3*9) - 2^5/(3*3*9*15) + + - - can be viewed as a generalized Engel-type expansion of the number 2 to the base 2. Compare with A014551. - Peter Bala, Nov 13 2013

References

  • D. M. Burton, Elementary Number Theory, Allyn and Bacon, Inc. Boston, MA, 1976, p. 29.

Crossrefs

Programs

  • Magma
    [2^n + (-1)^(n+1): n in [0..40]]; // Vincenzo Librandi, Aug 14 2011
  • Mathematica
    LinearRecurrence[{1,2},{0,3}, 30] (* or *) Table[2^n - (-1)^n, {n,0,30}] (* G. C. Greubel, Jan 15 2018 *)
  • PARI
    for(n=0,22,print(2^n+(-1)^(n+1)))
    

Formula

a(n) = 3*A001045(n). - Paul Curtz, Jan 17 2008
G.f.: 3*x / ( (1+x)*(1-2*x) )
G.f.: Q(0) where Q(k)= 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1) ))))); (continued fraction). - Sergei N. Gladkovskii, Apr 13 2013
E.g.f.: (exp(3*x) - 1)*exp(-x). - Ilya Gutkovskiy, Nov 20 2016

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jul 06 2001

A085058 a(n) = A001511(n+1) + 1.

Original entry on oeis.org

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

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Author

N. J. A. Sloane, Aug 11 2003

Keywords

Comments

Number of divisors of 2n+2 of the form 2^k. - Giovanni Teofilatto, Jul 25 2007
Number of steps for iteration of map x -> (3/2)*ceiling(x) to reach an integer when started at 2*n+1.
Also number of steps for iteration of map x -> (3/2)*floor(x) to reach an integer when started at 2*n+3. - Benoit Cloitre, Sep 27 2003
The first time that a(n) = e+1 is when n is of the form 2^e - 1. - Robert G. Wilson v, Sep 28 2003
Let 2^k(n) = largest power of 2 dividing tangent number A000182(n). Then a(n-1) = 2*n - k(n). - Yasutoshi Kohmoto, Dec 23 2006
a(n) is the number of integers generated by b(i+1) = (3+2n)*(b(i) + b(i-1))/2, following these two initial values, b(0) = b(1) = 1. Thereafter only non-integers are generated. - Richard R. Forberg, Nov 09 2014
a(n) is the 2-adic valuation of 4*n+4, which is equal to the number of trailing 1-bits of 4*n+3 in binary. - Ruud H.G. van Tol, Sep 11 2023

Crossrefs

Programs

  • Magma
    [Valuation(n+1, 2)+2: n in [0..100]]; // Vincenzo Librandi, Jan 16 2016
    
  • Maple
    f := x->(3/2)*ceil(x); g := proc(n) local t1,c; global f; t1 := f(n); c := 1; while not type(t1, 'integer') do c := c+1; t1 := f(t1); od; RETURN([c,t1]); end;
    a := n ->  A001511(n+1) + 1: A001511 := n -> padic[ordp](2*n, 2): seq(a(n), n=0..104); # Johannes W. Meijer, Dec 22 2012
  • Mathematica
    g = 3 Ceiling[ # ]/2 &; f[n_?OddQ] := Length @ NestWhileList[ g, g[n], !IntegerQ[ # ] & ]; Table[ f[n], {n, 1, 210, 2}]
  • PARI
    A085058(n)=if(n<0,0,c=2*n+7/2; x=0; while(frac(c)>0,c=3/2*floor(c); x++); x) \\ Benoit Cloitre, Sep 27 2003
    
  • PARI
    A085058(n)=if(n<0,0,c=(2*n+1)*3/2; x=1; while(frac(c)>0,c=3/2*ceil(c); x++); x) \\ Benoit Cloitre, Sep 27 2003
    
  • PARI
    a(n) = valuation(n+1,2)+2; \\ Michel Marcus, Jan 15 2016
    
  • Python
    def A085058(n): return (~(n+1) & n).bit_length()+2 # Chai Wah Wu, Apr 14 2023

Formula

a(n) = A007814(3^(n+1) - (-1)^(n+1)) = A007814(A105723(n+1)). - Reinhard Zumkeller, Apr 18 2005
a(n) = A001511(n+1) + 1 = A001511(2*n+2). - Ray Chandler, Jul 29 2007
a(n) = A007814(5^(n+1) - 1). - Ivan Neretin, Jan 15 2016
a(n) = A007814(4*(n+1)) = A007814(n+1) + 2. - Ruud H.G. van Tol, Sep 11 2023
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3. - Amiram Eldar, Sep 13 2024

Extensions

Edited by Franklin T. Adams-Watters, Dec 09 2013

A053524 a(n) = (6^n - (-2)^n)/8.

Original entry on oeis.org

0, 1, 4, 28, 160, 976, 5824, 35008, 209920, 1259776, 7558144, 45349888, 272097280, 1632587776, 9795518464, 58773127168, 352638730240, 2115832446976, 12694994550784, 76169967566848, 457019804876800, 2742118830309376, 16452712979759104
Offset: 0

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Author

Keywords

Comments

The ratio a(n+1)/a(n) converges to 6 as n approaches infinity. - Felix P. Muga II, Mar 10 2014

References

  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.1(b).
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, pp. 194-196.

Crossrefs

Cf. A015518.

Programs

Formula

E.g.f.: (exp(6*x) - exp(-2*x))/8.
a(n) = 2^(n-3) * (3^n - (-1)^n) = 2^(n-3)*A105723(n).
a(n) = 4*a(n-1) + 12*a(n-2), with a(0)=0, a(1)=1.
G.f.: x / ((1+2*x)*(1-6*x)). - Colin Barker, Mar 11 2014

A102345 a(n) = 3^n + (-1)^n.

Original entry on oeis.org

2, 2, 10, 26, 82, 242, 730, 2186, 6562, 19682, 59050, 177146, 531442, 1594322, 4782970, 14348906, 43046722, 129140162, 387420490, 1162261466, 3486784402, 10460353202, 31381059610, 94143178826, 282429536482, 847288609442
Offset: 0

Views

Author

Graeme McRae, Feb 16 2005

Keywords

Comments

a(n) = A105723(n) + 2*(-1)^n; (a(n) + A105723(n))/2 = A000244(n). - Reinhard Zumkeller, Apr 18 2005

Crossrefs

Apart from leading term, same as A084182.

Programs

  • Mathematica
    Table[3^n+(-1)^n,{n,0,30}] (* or *) LinearRecurrence[{2,3},{2,2},30] (* Harvey P. Dale, Jun 19 2016 *)
  • Sage
    [lucas_number2(n,2,-3) for n in range(0, 26)] # Zerinvary Lajos, Apr 30 2009

Formula

a(n) = 2*a(n-1) + 3*a(n-2).
From Elmo R. Oliveira, Dec 18 2023: (Start)
G.f.: 2*(1-x)/((1+x)*(1-3*x)).
E.g.f.: exp(-x) + exp(3*x).
a(n) = 2*A046717(n). (End)

A164907 a(n) = (3*3^n-(-1)^n)/2.

Original entry on oeis.org

1, 5, 13, 41, 121, 365, 1093, 3281, 9841, 29525, 88573, 265721, 797161, 2391485, 7174453, 21523361, 64570081, 193710245, 581130733, 1743392201, 5230176601, 15690529805, 47071589413, 141214768241, 423644304721, 1270932914165
Offset: 0

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Author

Klaus Brockhaus, Aug 31 2009

Keywords

Comments

Interleaving of A096053 and A083884 without initial term 1.
Partial sums are (essentially) in A080926.
First differences are (essentially) in A105723.
a(n)+a(n+1) = A008776(n+1) = A099856(n+1) = A110593(n+2).
Binomial transform of A056450. Inverse binomial transform of A164908.

Crossrefs

Equals A046717 without initial term 1 and A080925 without initial term 0. Equals A084182 / 2 from second term onward.

Programs

Formula

a(n) = 2*a(n-1)+3*a(n-2) for n > 1; a(0) = 1, a(1) = 5.
G.f.: (1+3*x)/((1+x)*(1-3*x)).
a(n) = 3*a(n-1)+2*(-1)^n. - Carmine Suriano, Mar 21 2014

A274073 a(n) = 6^n-(-1)^n.

Original entry on oeis.org

0, 7, 35, 217, 1295, 7777, 46655, 279937, 1679615, 10077697, 60466175, 362797057, 2176782335, 13060694017, 78364164095, 470184984577, 2821109907455, 16926659444737, 101559956668415, 609359740010497, 3656158440062975, 21936950640377857, 131621703842267135
Offset: 0

Views

Author

Colin Barker, Jun 09 2016

Keywords

Crossrefs

Cf. A015540.
Sequences of the type k^n-(-1)^n: A062157 (k=0), A010673 (k=1), A062510 (k=2), A105723 (k=3), A247281 (k=4), A274072 (k=5), this sequence (k=6).

Programs

  • PARI
    concat(0, Vec(7*x/((1+x)*(1-6*x)) + O(x^30)))

Formula

O.g.f.: 7*x/((1+x)*(1-6*x)).
E.g.f.: exp(6*x) - exp(-x).
a(n) = 5*a(n-1) + 6*a(n-2) for n>1.
a(n) = 7*A015540(n).

A274072 a(n) = 5^n-(-1)^n.

Original entry on oeis.org

0, 6, 24, 126, 624, 3126, 15624, 78126, 390624, 1953126, 9765624, 48828126, 244140624, 1220703126, 6103515624, 30517578126, 152587890624, 762939453126, 3814697265624, 19073486328126, 95367431640624, 476837158203126, 2384185791015624, 11920928955078126
Offset: 0

Views

Author

Colin Barker, Jun 09 2016

Keywords

Crossrefs

Cf. A015531.
Sequences of the type k^n-(-1)^n: A062157 (k=0), A010673 (k=1), A062510 (k=2), A105723 (k=3), A247281 (k=4), this sequence (k=5), A274073 (k=6).

Programs

  • Mathematica
    LinearRecurrence[{4, 5}, {0, 6}, 30] (* Paolo Xausa, Oct 21 2024 *)
  • PARI
    concat(0, Vec(6*x/((1+x)*(1-5*x)) + O(x^30)))

Formula

O.g.f.: 6*x/((1+x)*(1-5*x)).
E.g.f.: exp(5*x) - exp(-x).
a(n) = 4*a(n-1) + 5*a(n-2) for n>1.
a(n) = 6*A015531(n).
Showing 1-8 of 8 results.