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 38 results. Next

A339131 Odd composite integers m such that A056854(m-J(m,45)) == 2 (mod m) and gcd(m,45)=1, where J(m,45) is the Jacobi symbol.

Original entry on oeis.org

49, 121, 169, 289, 323, 329, 361, 377, 451, 529, 841, 961, 1081, 1127, 1369, 1681, 1819, 1849, 1891, 2033, 2209, 2303, 2809, 3481, 3653, 3721, 3751, 3827, 4181, 4489, 4901, 4961, 5041, 5329, 5491, 5671, 5777, 6137, 6241, 6601, 6721, 6889, 7381, 7921, 8149, 8557, 9409
Offset: 1

Views

Author

Ovidiu Bagdasar, Nov 24 2020

Keywords

Comments

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy the identity
V(p-J(p,D)) == 2 (mod p) when p is prime, b=1 and D=a^2-4.
This sequence contains the odd composite integers with V(m-J(m,D)) == 2 (mod m).
For a=7 and b=1, we have D=45 and V(m) recovers A056854(m).

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021)
  • D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted)

Crossrefs

Cf. A056854.
Cf. A339125 (a=1, b=-1), A339126 (a=3, b=-1), A339127 (a=5, b=-1), A339128 (a=7, b=-1), A339129 (a=3, b=1), A339130 (a=5, b=1).

Programs

  • Mathematica
    Select[Range[3, 10000, 2], CoprimeQ[#, 45] && CompositeQ[#] && Divisible[LucasL[4*(# - JacobiSymbol[#, 45])] - 2, #] &] (* Amiram Eldar, Nov 26 2020 *)

A339523 Odd composite integers m such that A056854(2*m-J(m,45)) == 7 (mod m) and gcd(m,45)=1, where J(m,45) is the Jacobi symbol.

Original entry on oeis.org

91, 203, 323, 329, 377, 451, 1001, 1081, 1183, 1547, 1729, 1771, 1819, 1891, 1967, 2033, 2093, 2639, 2821, 3197, 3311, 3653, 3731, 3827, 4181, 4669, 5551, 5671, 5777, 5887, 6601, 6721, 7471, 7931, 7973, 8149, 8557, 9541, 9737, 10877, 11309, 11663, 11977, 13201
Offset: 1

Views

Author

Ovidiu Bagdasar, Dec 07 2020

Keywords

Comments

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy V(k*p-J(p,D)) == V(k-1) (mod p) whenever p is prime, k is a positive integer, b=1 and D=a^2-4.
The composite integers m with the property V(k*m-J(m,D)) == V(k-1) (mod m) are called generalized Pell-Lucas pseudoprimes of level k+ and parameter a.
Here b=1, a=7, D=45 and k=2, while V(m) recovers A056854(m).

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021).
  • D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted).

Crossrefs

Cf. A056854, A071904, A339131 (a=7, b=1, k=1).
Cf. A339521 (a=3, b=1), A339522 (a=5, b=1).

Programs

  • Maple
    filter:= proc(m)
    uses LinearAlgebra:-Modular;
    local p,M;
      if igcd(m,45) <> 1 then return false fi;
      if isprime(m) then return false fi;
      p:= 2*m - numtheory:-jacobi(m,45);
      M:= Mod(m,[[0,1],[-1,7]],integer[8]);
      (MatrixPower(m,M,p) . <2,7>)[1] - 7 mod m = 0
    end proc:
    select(filter, [seq(i,i=9..10000,2)]); # Robert Israel, Dec 15 2020
  • Mathematica
    Select[Range[3, 15000, 2], CoprimeQ[#, 45] && CompositeQ[#] && Divisible[LucasL[4*(2*# - JacobiSymbol[#, 45])] - 7, #] &]

A339730 Odd composite integers m such that A056854(3*m-J(m,45)) == 47 (mod m) and gcd(m,45)=1, where J(m,45) is the Jacobi symbol.

Original entry on oeis.org

49, 161, 287, 323, 329, 341, 377, 451, 671, 737, 901, 1007, 1079, 1081, 1127, 1271, 1363, 1541, 1819, 1853, 1891, 1927, 2033, 2071, 2303, 2407, 2431, 2461, 2501, 2567, 2743, 3653, 3827, 4181, 4843, 5029, 5243, 5473, 5611, 5671, 5777, 6119, 6593, 6601, 6721, 6923
Offset: 1

Views

Author

Ovidiu Bagdasar, Dec 14 2020

Keywords

Comments

The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy V(k*p-J(p,D)) == V(k-1) (mod p) whenever p is prime, k is a positive integer, b=1 and D=a^2-4.
The composite integers m with the property V(k*m-J(m,D)) == V(k-1) (mod m) are called generalized Pell-Lucas pseudoprimes of level k+ and parameter a.
Here b=1, a=7, D=45 and k=3, while V(m) recovers A056854(m), with V(2)=47.

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021).
  • D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted).

Crossrefs

Cf. A056854, A071904, A339131 (a=7, b=1, k=1), A339523 (a=7, b=1, k=2).
Cf. A339728 (a=3, b=1), A339729 (a=5, b=1).

Programs

  • Mathematica
    Select[Range[3, 7000, 2], CoprimeQ[#, 45] && CompositeQ[#] && Divisible[LucasL[4*(3*# - JacobiSymbol[#, 45])] - 47, #] &]

A337781 Odd composite integers m such that U(m)^2 == 1 (mod m) and V(m) == 7 (mod m), where U(m)=A004187(m) and V(m)=A056854(m) are the m-th generalized Lucas and Pell-Lucas numbers of parameters a=7 and b=1, respectively.

Original entry on oeis.org

323, 329, 377, 451, 1081, 1189, 1819, 1891, 2033, 2737, 2849, 3059, 3289, 3653, 3689, 3827, 4181, 4879, 5671, 5777, 6479, 6601, 6721, 8149, 8533, 8557, 8569, 8651, 8701, 10199, 10877, 11309, 11339, 11521, 11663, 12341, 13201, 13489, 13861, 13981, 14701, 15251, 15301
Offset: 1

Views

Author

Ovidiu Bagdasar, Sep 20 2020

Keywords

Comments

For a, b integers, the following sequences are defined:
generalized Lucas sequences by U(n+2) = a*U(n+1)-b*U(n) and U(0)=0, U(1)=1,
generalized Pell-Lucas sequences by V(n+2) = a*V(n+1)-b*V(n) and V(0)=2, V(1)=a.
These satisfy the identities U(p)^2 == 1 and V(p) == a (mod p) for p prime and b=1,-1.
These numbers may be called weak generalized Lucas-Bruckner pseudoprimes of parameters a and b. The current sequence is defined for a=7 and b=1.

Crossrefs

Cf. A337630 (a=7, b=-1), A337778 (a=4, b=1), A337779 (a=5, b=1), A337780 (a=6, b=1).

Programs

  • Mathematica
    Select[Range[3, 20000, 2], CompositeQ[#] && Divisible[2*ChebyshevT[#, 7/2] - 7, #] && Divisible[ChebyshevU[#-1, 7/2]*ChebyshevU[#-1, 7/2] - 1, #] &]

A337782 Even composite integers m such that U(m)^2 == 1 (mod m) and V(m) == 7 (mod m), where U(m)=A004187(m) and V(m)=A056854(m) are the m-th generalized Lucas and Pell-Lucas numbers of parameters a=7 and b=1, respectively.

Original entry on oeis.org

4, 8, 44, 104, 136, 152, 232, 286, 442, 836, 1364, 1378, 2204, 2584, 2626, 2684, 2834, 3016, 3926, 4636, 5662, 7208, 7384, 7676, 7964, 8294, 9164, 9316, 11476, 12524, 14824, 15224, 17324, 20026, 20474, 21736, 21944, 22814, 23804, 24616, 26596, 27028, 27404, 31124
Offset: 1

Views

Author

Ovidiu Bagdasar, Sep 20 2020

Keywords

Comments

For a, b integers, the following sequences are defined:
generalized Lucas sequences by U(n+2)=a*U(n+1)-b*U(n) and U(0)=0, U(1)=1,
generalized Pell-Lucas sequences by V(n+2)=a*V(n+1)-b*V(n) and V(0)=2, V(1)=a.
These satisfy the identities U(p)^2 == 1 and V(p)==a (mod p) for p prime and b=1,-1.
These numbers may be called weak generalized Lucas-Bruckner pseudoprimes of parameters a and b. The current sequence is defined for a=7 and b=1.

Crossrefs

Cf. A337630 (a=7, b=-1), A337777 (a=3, b=1), A337781 (a=7, b=1).

Programs

  • Mathematica
    Select[Range[2, 20000, 2], CompositeQ[#] && Divisible[2*ChebyshevT[#, 7/2] - 7, #] && Divisible[ChebyshevU[#-1, 7/2]*ChebyshevU[#-1, 7/2] - 1, #] &]

Extensions

More terms from Amiram Eldar, Sep 21 2020

A338082 Odd composite integers m such that A056854(m) == 7 (mod m).

Original entry on oeis.org

9, 15, 21, 35, 45, 63, 99, 105, 195, 231, 315, 323, 329, 369, 377, 423, 435, 451, 595, 665, 705, 805, 861, 903, 1081, 1189, 1443, 1551, 1819, 1833, 1869, 1891, 1935, 2033, 2211, 2345, 2465, 2737, 2849, 2871, 2961, 3059, 3289, 3653, 3689, 3745, 3827, 4005, 4059
Offset: 1

Views

Author

Ovidiu Bagdasar, Oct 08 2020

Keywords

Comments

If p is a prime, then A056854(p)==7 (mod p).
This sequence contains the odd composite integers for which the congruence holds.
The generalized Pell-Lucas sequences of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) when p is prime and b=-1,1.
For a=7 and b=1, V(m) recovers A056854(m).

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021)

Crossrefs

Cf. A005248, A335669 (a=3,b=-1), A335672 (a=3,b=1), A335673 (a=4,b=1), A335674 (a=5,b=1), A337233 (a=6,b=1).

Programs

  • Mathematica
    Select[Range[3, 20000, 2], CompositeQ[#] && Divisible[2*ChebyshevT[#, 7/2] - 7, #] &]
    Select[Range[9,5001,2],CompositeQ[#]&&Mod[LucasL[4#],#]==7&] (* Harvey P. Dale, Apr 28 2022 *)

A338312 Even composites m such that A056854(m)==7 (mod m).

Original entry on oeis.org

4, 8, 10, 20, 40, 44, 104, 136, 152, 170, 190, 232, 260, 286, 442, 580, 740, 836, 890, 1364, 1378, 1990, 2204, 2260, 2584, 2626, 2684, 2834, 3016, 3160, 3230, 3926, 4220, 4636, 5662, 6290, 7208, 7384, 7540, 7676, 7964, 8294, 8420, 9164, 9316, 9320, 10070, 11476
Offset: 1

Views

Author

Ovidiu Bagdasar, Oct 22 2020

Keywords

Comments

If p is a prime, then A056854(p)==7 (mod p).
This sequence contains the even composite integers for which the congruence holds.
The generalized Pell-Lucas sequence of integer parameters (a,b) defined by V(m+2)=a*V(m+1)-b*V(m) and V(0)=2, V(1)=a, satisfy the identity V(p)==a (mod p) whenever p is prime and b=-1,1.
For a=7 and b=1, V(m) recovers A056854(m).

References

  • D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (2020)
  • D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021)

Crossrefs

Cf. A338082 (sequence of odd terms), A337777 (a=3), A338311 (a=6).

Programs

  • Mathematica
    Select[Range[2, 20000, 2], CompositeQ[#] && Divisible[2*ChebyshevT[#, 7/2] - 7, #] &]

A005248 Bisection of Lucas numbers: a(n) = L(2*n) = A000032(2*n).

Original entry on oeis.org

2, 3, 7, 18, 47, 123, 322, 843, 2207, 5778, 15127, 39603, 103682, 271443, 710647, 1860498, 4870847, 12752043, 33385282, 87403803, 228826127, 599074578, 1568397607, 4106118243, 10749957122, 28143753123, 73681302247, 192900153618, 505019158607, 1322157322203
Offset: 0

Views

Author

Keywords

Comments

Drop initial 2; then iterates of A050411 do not diverge for these starting values. - David W. Wilson
All nonnegative integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = +4 together with b(n)=A001906(n), n>=0. - Wolfdieter Lang, Aug 31 2004
a(n+1) = B^(n)AB(1), n>=0, with compositions of Wythoff's complementary A(n):=A000201(n) and B(n)=A001950(n) sequences. See the W. Lang link under A135817 for the Wythoff representation of numbers (with A as 1 and B as 0 and the argument 1 omitted). E.g., 3=`10`, 7=`010`, 18=`0010`, 47=`00010`, ..., in Wythoff code. a(0) = 2 = B(1) in Wythoff code.
Output of Tesler's formula (as well as that of Lu and Wu) for the number of perfect matchings of an m X n Möbius band where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
Numbers having two 1's in their base-phi representation. - Robert G. Wilson v, Sep 13 2010
Pisano period lengths: 1, 3, 4, 3, 2, 12, 8, 6, 12, 6, 5, 12, 14, 24, 4, 12, 18, 12, 9, 6, ... - R. J. Mathar, Aug 10 2012
From Wolfdieter Lang, Feb 18 2013: (Start)
a(n) is also one half of the total number of round trips, each of length 2*n, on the graph P_4 (o-o-o-o) (the simple path with 4 points (vertices) and 3 lines (or edges)). See the array and triangle A198632 for the general case of the graph P_N (there N is n and the length is l=2*k).
O.g.f. for w(4,l) (with zeros for odd l): y*(d/dy)S(4,y)/S(4,y) with y=1/x and Chebyshev S-polynomials (coefficients A049310). See also A198632 for a rewritten form. One half of this o.g.f. for x -> sqrt(x) produces the g.f. (2-3x)/(1-3x+x^2) given below. (End)
Solutions (x, y) = (a(n), a(n+1)) satisfying x^2 + y^2 = 3xy - 5. - Michel Lagneau, Feb 01 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 7xy + y^2 + 45 = 0. - Colin Barker, Feb 16 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 18xy + y^2 + 320 = 0. - Colin Barker, Feb 16 2014
a(n) are the numbers such that a(n)^2-2 are Lucas numbers. - Michel Lagneau, Jul 22 2014
All sequences of this form, b(n+1) = 3*b(n) - b(n-1), regardless of initial values, which includes this sequence, yield this sequence as follows: a(n) = (b(j+n) + b(j-n))/b(j), for any j, except where b(j) = 0. Also note formula below relating this a(n) to all sequences of the form G(n+1) = G(n) + G(n-1). - Richard R. Forberg, Nov 18 2014
A non-simple continued fraction expansion for F(2n*(k+1))/F(2nk) k>=1 is a(n) + (-1)/(a(n) + (-1)/(a(n) + ... + (-1)/a(n))) where a(n) appears exactly k times (F(n) denotes the n-th Fibonacci number). E.g., F(16)/F(12) equals 7 + (-1)/(7 + (-1)/7). Furthermore, these a(n) are exactly the positive integers k such that the non-simple infinite continued fraction k + (-1)/(k + (-1)/(k + (-1)/(k + ...))) belongs to Q(sqrt(5)). Compare to Benoit Cloitre and Thomas Baruchel's comments at A002878. - Greg Dresden, Aug 13 2019
For n >= 1, a(n) is the number of cyclic up-down words of length 2*n over an alphabet of size 3. - Sela Fried, Apr 08 2025

Examples

			G.f. = 2 + 3*x + 7*x^2 + 18*x^3 + 47*x^4 + 123*x^5 + 322*x^6 + 843*x^7 + ... - _Michael Somos_, Aug 11 2009
		

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).
  • Richard P. Stanley, Enumerative combinatorics, Vol. 2. Volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Crossrefs

Cf. A000032, A002878 (odd-indexed Lucas numbers), A001906 (Chebyshev S(n-1, 3)), a(n) = sqrt(4+5*A001906(n)^2), A228842.
a(n) = A005592(n)+1 = A004146(n)+2 = A065034(n)-1.
First differences of A002878. Pairwise sums of A001519. First row of array A103997.
Cf. A153415, A201157. Also Lucas(k*n): A000032 (k = 1), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11), A089775 (k = 12).

Programs

  • Haskell
    a005248 n = a005248_list !! n
    a005248_list = zipWith (+) (tail a001519_list) a001519_list
    -- Reinhard Zumkeller, Jan 11 2012
  • Magma
    [Lucas(2*n) : n in [0..100]]; // Vincenzo Librandi, Apr 14 2011
    
  • Maple
    a:= n-> (<<2|3>>. <<3|1>, <-1|0>>^n)[1$2]: seq(a(n), n=0..30); # Alois P. Heinz, Jul 31 2008
    with(combinat): seq(5*fibonacci(n)^2+2*(-1)^n, n= 0..26);
  • Mathematica
    a[0] = 2; a[1] = 3; a[n_] := 3a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 27}] (* Robert G. Wilson v, Jan 30 2004 *)
    Fibonacci[1 + 2n] + 1/2 (-Fibonacci[2n] + LucasL[2n]) (* Tesler. Sarah-Marie Belcastro, Jul 04 2009 *)
    LinearRecurrence[{3, -1}, {2, 3}, 50] (* Sture Sjöstedt, Nov 27 2011 *)
    LucasL[Range[0,60,2]] (* Harvey P. Dale, Sep 30 2014 *)
  • PARI
    {a(n) = fibonacci(2*n + 1) + fibonacci(2*n - 1)}; /* Michael Somos, Jun 23 2002 */
    
  • PARI
    {a(n) = 2 * subst( poltchebi(n), x, 3/2)}; /* Michael Somos, Jun 28 2003 */
    
  • Sage
    [lucas_number2(n,3,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008
    

Formula

a(n) = Fibonacci(2*n-1) + Fibonacci(2*n+1).
G.f.: (2-3*x)/(1-3*x+x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n, 3) - S(n-2, 3) = 2*T(n, 3/2) with S(n-1, 3) = A001906(n) and S(-2, x) = -1. U(n, x)=S(n, 2*x) and T(n, x) are Chebyshev's U- and T-polynomials.
a(n) = a(k)*a(n - k) - a(n - 2k) for all k, i.e., a(n) = 2*a(n) - a(n) = 3*a(n - 1) - a(n - 2) = 7*a(n - 2) - a(n - 4) = 18*a(n - 3) - a(n - 6) = 47*a(n - 4) - a(n - 8) etc., a(2n) = a(n)^2 - 2. - Henry Bottomley, May 08 2001
a(n) = A060924(n-1, 0) = 3*A001906(n) - 2*A001906(n-1), n >= 1. - Wolfdieter Lang, Apr 26 2001
a(n) ~ phi^(2*n) where phi=(1+sqrt(5))/2. - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(0)=2, a(1)=3, a(n) = 3*a(n-1) - a(n-2) = a(-n). - Michael Somos, Jun 28 2003
a(n) = phi^(2*n) + phi^(-2*n) where phi=(sqrt(5)+1)/2, the golden ratio. E.g., a(4)=47 because phi^(8) + phi^(-8) = 47. - Dennis P. Walsh, Jul 24 2003
With interpolated zeros, trace(A^n)/4, where A is the adjacency matrix of path graph P_4. Binomial transform is then A049680. - Paul Barry, Apr 24 2004
a(n) = (floor((3+sqrt(5))^n) + 1)/2^n. - Lekraj Beedassy, Oct 22 2004
a(n) = ((3-sqrt(5))^n + (3+sqrt(5))^n)/2^n (Note: substituting the number 1 for 3 in the last equation gives A000204, substituting 5 for 3 gives A020876). - Creighton Dement, Apr 19 2005
a(n) = (1/(n+1/2))*Sum_{k=0..n} B(2k)*L(2n+1-2k)*binomial(2n+1, 2k) where B(2k) is the (2k)-th Bernoulli number. - Benoit Cloitre, Nov 02 2005
a(n) = term (1,1) in the 1 X 2 matrix [2,3] . [3,1; -1,0]^n. - Alois P. Heinz, Jul 31 2008
a(n) = 2*cosh(2*n*psi), where psi=log((1+sqrt(5))/2). - Al Hakanson, Mar 21 2009
From Sarah-Marie Belcastro, Jul 04 2009: (Start)
a(n) - (a(n) - F(2n))/2 - F(2n+1) = 0. (Tesler)
Product_{r=1..n} (1 + 4*(sin((4r-1)*Pi/(4n)))^2). (Lu/Wu) (End)
a(n) = Fibonacci(2n+6) mod Fibonacci(2n+2), n > 1. - Gary Detlefs, Nov 22 2010
a(n) = 5*Fibonacci(n)^2 + 2*(-1)^n. - Gary Detlefs, Nov 22 2010
a(n) = A033888(n)/A001906(n), n > 0. - Gary Detlefs, Dec 26 2010
a(n) = 2^(2*n) * Sum_{k=1..2} (cos(k*Pi/5))^(2*n). - L. Edson Jeffery, Jan 21 2012
From Peter Bala, Jan 04 2013: (Start)
Let F(x) = Product_{n>=0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(3 - sqrt(5)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.31829 56058 81914 31334 ... = 2 + 1/(3 + 1/(7 + 1/(18 + ...))).
Also F(-alpha) = 0.64985 97768 07374 32950 has the continued fraction representation 1 - 1/(3 - 1/(7 - 1/(18 - ...))) and the simple continued fraction expansion 1/(1 + 1/((3-2) + 1/(1 + 1/((7-2) + 1/(1 + 1/((18-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((3^2-4) + 1/(1 + 1/((7^2-4) + 1/(1 + 1/((18^2-4) + 1/(1 + ...))))))).
Added Oct 13 2019: 1/2 + 1/2*F(alpha)/F(-alpha) = 1.5142923542... has the simple continued fraction expansion 1 + 1/((3 - 2) + 1/(1 + 1/((18 - 2) + 1/(1 + 1/(123 - 2) + 1/(1 + ...))))). (End)
G.f.: (W(0)+6)/(5*x), where W(k) = 5*x*k + x - 6 + 6*x*(5*k-9)/W(k+1) (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
Sum_{n >= 1} 1/( a(n) - 5/a(n) ) = 1. Compare with A001906, A002878 and A023039. - Peter Bala, Nov 29 2013
0 = a(n) * a(n+2) - a(n+1)^2 - 5 for all n in Z. - Michael Somos, Aug 24 2014
a(n) = (G(j+2n) + G(j-2n))/G(j), for n >= 0 and any j, positive or negative, except where G(j) = 0, and for any sequence of the form G(n+1) = G(n) + G(n-1) with any initial values for G(0), G(1), including non-integer values. G(n) includes Lucas, Fibonacci. Compare with A081067 for odd number offsets from j. - Richard R. Forberg, Nov 16 2014
a(n) = [x^n] ( (1 + 3*x + sqrt(1 + 6*x + 5*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
From J. M. Bergot, Oct 28 2015: (Start)
For n>0, a(n) = F(n-1) * L(n) + F(2*n+1) - (-1)^n with F(k) = A000045(k).
For n>1, a(n) = F(n+1) * L(n) + F(2*n-1) - (-1)^n.
For n>2, a(n) = 5*F(2*n-3) + 2*L(n-3) * L(n) + 8*(-1)^n. (End)
For n>1, a(n) = L(n-2)*L(n+2) -7*(-1)^n. - J. M. Bergot, Feb 10 2016
a(n) = 6*F(n-1)*L(n-1) - F(2*n-6) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, Apr 21 2017
a(n) = F(2*n) + 2*F(n-1)*L(n) with F(n)=A000045(n) and L(n)=A000032(n). - J. M. Bergot, May 01 2017
E.g.f.: exp(4*x/(1+sqrt(5))^2) + exp((1/4)*(1+sqrt(5))^2*x). - Stefano Spezia, Aug 13 2019
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(2*n+2) - F(2*n-2) = A001906(n+1) - A001906(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^2 = [1, 1; 1, 2].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 1/a(n) ) = 1/5.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 3/(a(n) + 2/(a(n))) ) = 1/6.
Sum_{n >= 1} (-1)^(n+1)/( a(n) + 9/(a(n) + 4/(a(n) + 1/(a(n)))) ) = 1/9.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 3*x^2 + 8*x^3 + 21*x^4 + ... is the o.g.f. for A001906. (End)
a(n) = n + 2 + Sum_{k=1..n-1} k*a(n-k). - Yu Xiao, May 30 2020
Sum_{n>=1} 1/a(n) = A153415. - Amiram Eldar, Nov 11 2020
Sum_{n>=0} 1/(a(n) + 3) = (2*sqrt(5) + 1)/10 (André-Jeannin, 1991). - Amiram Eldar, Jan 23 2022
a(n) = 2*cosh(2*n*arccsch(2)) = 2*cosh(2*n*asinh(1/2)). - Peter Luschny, May 25 2022
a(n) = (5/2)*(Sum_{k=-n..n} binomial(2*n, n+5*k)) - (1/2)*4^n. - Greg Dresden, Jan 05 2023
a(n) = Sum_{k>=0} Lucas(2*n*k)/(Lucas(2*n)^(k+1)). - Diego Rattaggi, Jan 12 2025

Extensions

Additional comments from Michael Somos, Jun 23 2001

A004187 a(n) = 7*a(n-1) - a(n-2) with a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 7, 48, 329, 2255, 15456, 105937, 726103, 4976784, 34111385, 233802911, 1602508992, 10983760033, 75283811239, 516002918640, 3536736619241, 24241153416047, 166151337293088, 1138818207635569, 7805576116155895, 53500214605455696, 366695926122033977
Offset: 0

Views

Author

Keywords

Comments

Define the sequence T(a_0,a_1) by a_{n+2} is the greatest integer such that a_{n+2}/a_{n+1}= 0 . A004187 (with initial 0 omitted) is T(1,7).
This is a divisibility sequence.
For n>=2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 7's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
a(n) and b(n) := A056854(n) are the proper and improper nonnegative solutions of the Pell equation b(n)^2 - 5*(3*a(n))^2 = +4. see the cross-reference to A056854 below. - Wolfdieter Lang, Jun 26 2013
For n>=1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,2,3,4,5,6}. - Milan Janjic, Jan 25 2015
The digital root is A253298, which shares its digital root with A253368. - Peter M. Chema, Jul 04 2016
Lim_{n->oo} a(n+1)/a(n) = 2 + 3*phi = 1+ A090550 = 6.854101... - Wolfdieter Lang, Nov 16 2023

Examples

			a(2) = 7*a(1) - a(0) = 7*7 - 1 = 48. - _Michael B. Porter_, Jul 04 2016
		

Crossrefs

Cf. A000027, A001906, A001353, A004254, A001109, A049685, A033888. a(n)=sqrt((A056854(n)^2 - 4)/45).
Second column of array A028412.

Programs

  • Magma
    [Fibonacci(4*n)/3 : n in [0..30]]; // Vincenzo Librandi, Jun 07 2011
    
  • Magma
    /* By definition: */ [n le 2 select n-1 else 7*Self(n-1)-Self(n-2): n in [1..23]]; // Bruno Berselli, Dec 24 2012
  • Maple
    seq(combinat:-fibonacci(4*n)/3, n = 0 .. 30); # Robert Israel, Jan 26 2015
  • Mathematica
    LinearRecurrence[{7,-1},{0,1},30] (* Harvey P. Dale, Jul 13 2011 *)
    CoefficientList[Series[x/(1 - 7*x + x^2), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 23 2012 *)
  • Maxima
    a[0]:0$ a[1]:1$ a[n]:=7*a[n-1] - a[n-2]$ A004187(n):=a[n]$ makelist(A004187(n),n,0,30); /* Martin Ettl, Nov 11 2012 */
    
  • MuPAD
    numlib::fibonacci(4*n)/3 $ n = 0..25; // Zerinvary Lajos, May 09 2008
    
  • PARI
    a(n)=fibonacci(4*n)/3 \\ Charles R Greathouse IV, Mar 09 2012
    
  • PARI
    concat(0, Vec(x/(1-7*x+x^2) + O(x^99))) \\ Altug Alkan, Jul 03 2016
    
  • Sage
    [lucas_number1(n,7,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
    
  • Sage
    [fibonacci(4*n)/3 for n in range(0, 21)] # Zerinvary Lajos, May 15 2009
    

Formula

G.f.: x/(1-7*x+x^2).
a(n) = F(4*n)/3 = A033888(n)/3, where F=A000045 (the Fibonacci sequence).
a(n) = S(2*n-1, sqrt(9))/sqrt(9) = S(n-1, 7); S(n, x) := U(n, x/2), Chebyshev polynomials of the 2nd kind, A049310.
a(n) = Sum_{i = 0..n-1} C(2*n-1-i, i)*5^(n-i-1). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
[A049685(n-1), a(n)] = [1,5; 1,6]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = A167816(4*n). - Reinhard Zumkeller, Nov 13 2009
a(n) = (((7+sqrt(45))/2)^n-((7-sqrt(45))/2)^n)/sqrt(45). - Noureddine Chair, Aug 31 2011
a(n+1) = Sum_{k = 0..n} A101950(n,k)*6^k. - Philippe Deléham, Feb 10 2012
a(n) = (A081072(n)/3)-1. - Martin Ettl, Nov 11 2012
From Peter Bala, Dec 23 2012: (Start)
Product {n >= 1} (1 + 1/a(n)) = (1/5)*(5 + 3*sqrt(5)).
Product {n >= 2} (1 - 1/a(n)) = (1/14)*(5 + 3*sqrt(5)). (End)
From Peter Bala, Apr 02 2015: (Start)
Sum_{n >= 1} a(n)*x^(2*n) = -A(x)*A(-x), where A(x) = Sum_{n >= 1} Fibonacci(2*n)* x^n.
1 + 5*Sum_{n >= 1} a(n)*x^(2*n) = F(x)*F(-x) = G(x)*G(-x), where F(x) = 1 + A(x) and G(x) = 1 + 5*A(x).
1 + Sum_{n >= 1} a(n)*x^(2*n) = H(x)*H(-x) = I(x)*I(-x), where H(x) = 1 + Sum_{n >= 1} Fibonacci(2*n + 3)*x^n and I(x) = 1 + x + x*Sum_{n >= 1} Fibonacci(2*n - 1)*x^n. (End)
E.g.f.: 2*exp(7*x/2)*sinh(3*sqrt(5)*x/2)/(3*sqrt(5)). - Ilya Gutkovskiy, Jul 03 2016
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*9^k*binomial(n+k, 2*k+1). - Peter Bala, Jul 17 2023
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k*7^(n-2*k)*binomial(n-k, k). - Greg Dresden, Aug 03 2024
From Peter Bala, Jul 22 2025: (Start)
The following products telescope:
Product {n >= 2} (1 + (-1)^n/a(n)) = (3/14)*(3 + sqrt(5)).
Product {n >= 1} (1 - (-1)^n/a(n)) = (1/3)*(3 + sqrt(5)).
Product_{n >= 1} (a(2*n) + 1)/(a(2*n) - 1) = (3/5)*sqrt(5). (End)

Extensions

Entry improved by comments from Michael Somos and Wolfdieter Lang, Aug 02 2000

A019670 Decimal expansion of Pi/3.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Dec 11 1996

Keywords

Comments

With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area of the 0-winding sectors enclosed by a closed Brownian planar path, of a given length t, according to Desbois, p. 1. - Jonathan Vos Post, Jan 23 2011
Polar angle (or apex angle) of the cone that subtends exactly one quarter of the full solid angle. See comments in A238238. - Stanislav Sykora, Jun 07 2014
60 degrees in radians. - M. F. Hasler, Jul 08 2016
Volume of a quarter sphere of radius 1. - Omar E. Pol, Aug 17 2019
Also smallest positive zero of Sum_{k>=1} cos(k*x)/k = -log(2*|sin(x/2)|). Proof of this identity: Sum_{k>=1} cos(k*x)/k = Re(Sum_{k>=1} exp(k*x*i)/k) = Re(-log(1-exp(x*i))) = -log(2*|sin(x/2)|), x != 2*m*Pi, where i = sqrt(-1). - Jianing Song, Nov 09 2019
The area of a circle circumscribing a unit-area regular dodecagon. - Amiram Eldar, Nov 05 2020

Examples

			Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
From _Peter Bala_, Nov 16 2016: (Start)
Case n = 1. Pi/3 = 18 * Sum_{k >= 0} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ).
Using the methods of Borwein et al. we can find the following asymptotic expansion for the tails of this series: for N divisible by 6 there holds Sum_{k >= N/6} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ) ~ 1/N^3 + 6/N^5 + 1671/N ^7 - 241604/N^9 + ..., where the sequence [1, 0, 6, 0, 1671, 0, -241604, 0, ...] is the sequence of coefficients in the expansion of ((1/18)*cosh(2*x)/cosh(3*x)) * sinh(3*x)^2 = x^2/2! + 6*x^4/4! + 1671*x^6/6! - 241604*x^8/8! + .... Cf. A024235, A278080 and A278195. (End)
		

References

  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.

Crossrefs

Integral_{x=0..oo} 1/(1+x^m) dx: A013661 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), this sequence (m=6), A352125 (m=8), A094888 (m=10).

Programs

Formula

A third of A000796, a sixth of A019692, the square root of A100044.
Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - Charles R Greathouse IV, Sep 08 2011
Product_{k >= 1}(1-(6k)^(-2))^(-1). - Fred Daniel Kline, May 30 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/3 = Sum {k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k = 2F1(1/2,1/2;3/2;1/4). Similar series expansions hold for Pi^2 (A002388), Pi^3 (A091925) and Pi/(2*sqrt(2)) (A093954.)
The integer sequences A(n) := 4^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k ) both satisfy the second-order recurrence equation u(n) = (20*n^2 + 4*n + 1)*u(n-1) - 8*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/3 = 1 + 1/(24 - 8*3^3/(89 - 8*2*5^3/(193 - 8*3*7^3/(337 - ... - 8*(n - 1)*(2*n - 1)^3/((20*n^2 + 4*n + 1) - ... ))))). Cf. A002388 and A093954. (End)
Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - Michel Marcus, Mar 29 2016
Equals Product_{n >= 1} A016910(n) / A136017(n). - Fred Daniel Kline, Jun 09 2016
Equals Integral_{x=-oo..oo} sech(x)/3 dx. - Ilya Gutkovskiy, Jun 09 2016
From Peter Bala, Nov 16 2016: (Start)
Euler's series transformation applied to the series representation Pi/3 = Sum_{k >= 0} (-1)^k/(6*k + 1) + (-1)^k/(6*k + 5) given above by Greathouse produces the faster converging series Pi/3 = (1/2) * Sum_{n >= 0} 3^n*n!*( 1/(Product_{k = 0..n} (6*k + 1)) + 1/(Product_{k = 0..n} (6*k + 5)) ).
The series given above by Greathouse is the case n = 0 of the more general result Pi/3 = 9^n*(2*n)! * Sum_{k >= 0} (-1)^(k+n)*( 1/(Product_{j = -n..n} (6*k + 1 + 6*j)) + 1/(Product_{j = -n..n} (6*k + 5 + 6*j)) ) for n = 0,1,2,.... Cf. A003881. See the example section for notes on the case n = 1.(End)
Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - Dimitris Valianatos, May 13 2017
Equals A019699/4 or A019693/2. - Omar E. Pol, Aug 17 2019
Equals Integral_{x >= 0} (sin(x)/x)^4 = 1/2 + Sum_{n >= 0} (sin(n)/n)^4, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - Bernard Schott, Mar 12 2022
Pi/3 = -Sum_{n >= 1} i/(n*P(n, 1/sqrt(-3))*P(n-1, 1/sqrt(-3))), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximation Pi/3 = 1.04719755(06...) correct to 8 decimal places. - Peter Bala, Mar 16 2024
Equals Integral_{x >= 0} (2*x^2 + 1)/((x^2 + 1)*(4*x^2 + 1)) dx. - Peter Bala, Feb 12 2025
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