A106729 -id:A106729 - cp's OEIS frontend

A122367 Dimension of 3-variable non-commutative harmonics (twisted derivative) of order n. The dimension of the space of non-commutative polynomials of degree n in 3 variables which are killed by all symmetric differential operators (where for a monomial w, d_{xi} ( xi w ) = w and d_{xi} ( xj w ) = 0 for i != j).

1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657, 75025, 196418, 514229, 1346269, 3524578, 9227465, 24157817, 63245986, 165580141, 433494437, 1134903170, 2971215073, 7778742049, 20365011074, 53316291173, 139583862445, 365435296162, 956722026041

Offset: 0

Mike Zabrocki, Aug 30 2006

Essentially identical to A001519.
From Matthew Lehman, Jun 14 2008: (Start)
Number of monotonic rhythms using n time intervals of equal duration (starting with n=0).
Representationally, let 0 be an interval which is "off" (rest),
1 an interval which is "on" (beep),
1 1 two consecutive "on" intervals (beep, beep),
1 0 1 (beep, rest, beep) and
1-1 two connected consecutive "on" intervals (beeeep).
For f(3)=13:
  0 0 0, 0 0 1, 0 1 0, 0 1 1, 0 1-1, 1 0 0, 1 0 1,
  1 1 0, 1-1 0, 1 1 1, 1 1-1, 1-1 1, 1-1-1.
(End)
Equivalent to the number of one-dimensional graphs of n nodes, subject to the condition that a node is either 'on' or 'off' and that any two neighboring 'on' nodes can be connected. - Matthew Lehman, Nov 22 2008
Sum_{n>=0} arctan(1/a(n)) = Pi/2. - Jaume Oliver Lafont, Feb 27 2009
This is the limit sequence for certain generalized Pell numbers. - Gregory L. Simay, Oct 21 2024

			a(1) = 2 because x1-x2, x1-x3 are both of degree 1 and are killed by the differential operator d_x1 + d_x2 + d_x3.
a(2) = 5 because x1*x2 - x3*x2, x1*x3 - x2*x3, x2*x1 - x3*x1, x1*x1 - x2*x1 - x2*x2 + x1*x2, x1*x1 - x3*x1 - x3*x3 + x3*x1 are all linearly independent and are killed by d_x1 + d_x2 + d_x3, d_x1 d_x1 + d_x2 d_x2 + d_x3 d_x3 and Sum_{j = 1..3} (d_xi d_xj, i).

Colin Barker, Table of n, a(n) for n = 0..1000
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876 (See Corollary 1 (ii)).
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010) # 10.8.2, Example 13.
N. Bergeron, C. Reutenauer, M. Rosas, and M. Zabrocki, Invariants and Coinvariants of the Symmetric Group in Noncommuting Variables, arXiv:math/0502082 [math.CO], 2005; Canad. J. Math. 60 (2008), no. 2, 266-296
C. Chevalley, Invariants of finite groups generated by reflections, Amer. J. Math. 77 (1955), 778-782.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013) 13.4.5.
Tanya Khovanova, Recursive Sequences
Ron Knott, Pi and the Fibonacci numbers. - _Jaume Oliver Lafont_, Feb 27 2009
Diego Marques and Alain Togbé, On the sum of powers of two consecutive Fibonacci numbers, Proc. Japan Acad. Ser. A Math. Sci., Volume 86, Number 10 (2010), 174-176.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory, Vol. 7, No. 5 (2011), pp. 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012), The John Selfridge Memorial Volume.
M. C. Wolf, Symmetric functions of noncommutative elements, Duke Math. J. 2 (1936), 626-637.
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A001519, A048575, A055105, A055107, A087903, A074664, A008277, A106729, A112340, A122368, A122369, A122370, A122371, A122372.

Cf. similar sequences listed in A238379.

[Fibonacci(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jul 04 2015

a:=n->if n=0 then 1; elif n=1 then 2 else 3*a(n-1)-a(n-2); fi;
A122367List := proc(m) local A, P, n; A := [1,2]; P := [2];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(A), P[-1]]);
A := [op(A), P[-1]] od; A end: A122367List(30); # Peter Luschny, Mar 24 2022

Table[Fibonacci[2 n + 1], {n, 0, 30}] (* Vincenzo Librandi, Jul 04 2015 *)

Vec((1-x)/(1-3*x+x^2) + O(x^50)) \\ Michel Marcus, Jul 04 2015

G.f.: (1-q)/(1-3*q+q^2). More generally, (Sum_{d=0..n} (n!/(n-d)!*q^d)/Product_{r=1..d} (1 - r*q)) / (Sum_{d=0..n} q^d/Product_{r=1..d} (1 - r*q)) where n=3.
a(n) = 3*a(n-1) - a(n-2) with a(0) = 1, a(1) = 2.
a(n) = Fibonacci(2n+1) = A000045(2n+1). - Philippe Deléham, Feb 11 2009
a(n) = (2^(-1-n)*((3-sqrt(5))^n*(-1+sqrt(5)) + (1+sqrt(5))*(3+sqrt(5))^n)) / sqrt(5). - Colin Barker, Oct 14 2015
a(n) = Sum_{k=0..n} Sum_{i=0..n} binomial(k+i-1, k-i). - Wesley Ivan Hurt, Sep 21 2017
a(n) = A048575(n-1) for n >= 1. - Georg Fischer, Nov 02 2018
a(n) = Fibonacci(n)^2 + Fibonacci(n+1)^2. - Michel Marcus, Mar 18 2019
a(n) = Product_{k=1..n} (1 + 4*cos(2*k*Pi/(2*n+1))^2). - Seiichi Manyama, Apr 30 2021
From J. M. Bergot, May 27 2022: (Start)
a(n) = F(n)*L(n+1) + (-1)^n where L(n)=A000032(n) and F(n)=A000045(n).
a(n) = (L(n)^2 + L(n)*L(n+2))/5 - (-1)^n.
a(n) = 2*(area of a triangle with vertices at (L(n-1), L(n)), (F(n+1), F(n)), (L(n+1), L(n+2))) - 5*(-1)^n for n > 1. (End)
G.f.: (1-x)/(1-3x+x^2) = 1/(1-2x-x^2-x^3-x^4-...) - Gregory L. Simay, Oct 21 2024
E.g.f.: exp(3*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Nov 07 2024
From Peter Bala, May 04 2025: (Start)
a(n) = sqrt(2/5) * sqrt( 1 - T(2*n+1, - 3/2) ), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
a(2*n+1/2) = sqrt(5)*a(n)^2 - 2/sqrt(5).
a(3*n+1) = 5*a(n)^3 - 3*a(n); hence a(n) divides a(3*n+1).
a(4*n+3/2) = 5^(3/2)*a(n)^4 - 4*sqrt(5)*a(n)^2 + 2/sqrt(5).
a(5*n+2) = (5^2)*a(n)^5 - 5*5*a(n)^3 + 5*a(n); hence a(n) divides a(5*n+2).
See A034807 for the unsigned coefficients [1, 2; 1, 3; 1, 4, 2; 1, 5, 5; ...].
In general, for k >= 0, a(k*n + (k-1)/2) = a(-1/2) * T(k, a(n)/a(-1/2)), where a(n) = (2^(-1-n)*((3 - sqrt(5))^n *(-1 + sqrt(5)) + (1 + sqrt(5))*(3 + sqrt(5))^n)) / sqrt(5), as given above, and a(-1/2) = 2/sqrt(5).
The aerated sequence [b(n)]n>=1 = [1, 0, 2, 0, 5, 0, 13, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -5, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy.
Sum_{n >= 1} 1/(a(n) - 1/a(n)) = 1 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A001906(n) - 1/A001906(n+1).) (End)

A001254 Squares of Lucas numbers.

Original entry on oeis.org

4, 1, 9, 16, 49, 121, 324, 841, 2209, 5776, 15129, 39601, 103684, 271441, 710649, 1860496, 4870849, 12752041, 33385284, 87403801, 228826129, 599074576, 1568397609, 4106118241, 10749957124, 28143753121, 73681302249, 192900153616, 505019158609, 1322157322201, 3461452808004, 9062201101801, 23725150497409

Offset: 0

N. J. A. Sloane

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 36, 60.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 97.
Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001. [Note that Identity 34.7 on page 404 is wrong. - Alonso del Arte, Sep 07 2010]

G. C. Greubel, Table of n, a(n) for n = 0..2375 (terms 0..200 from T. D. Noe)
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Tanya Khovanova, Recursive Sequences
T. Mansour, A note on sum of k-th power of Horadam's sequence, arXiv:math/0302015 [math.CO], 2003.
P. Stanica, Generating functions, weighted and non-weighted sums for powers of second-order recurrence sequences, arXiv:math/0010149 [math.CO], 2000.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000204.
Cf. A007598, A079291.
With alternating signs, cf. A075150.
Bisection of A001638 and A006499. First differences of A005970.
Second row of array A103324.

[ Lucas(n)^2 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011

with(combinat):seq(5*fibonacci(n)^2+4*(-1)^n, n=0..26)

Table[LucasL[n]^2, {n, 0, 29}] (* Alonso del Arte, Apr 11 2011 *)
LinearRecurrence[{2, 2, -1}, {4, 1, 9}, 33] (* Jean-François Alcover, Jan 07 2019 *)

a(n)=5*fibonacci(n)^2 + 4*(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

from sympy import lucas
def a(n):  return lucas(n)**2
print([a(n) for n in range(33)]) # Michael S. Branicky, Apr 01 2021

a(n) = A000032(n)^2.
G.f.: ( 4-7*x-x^2 ) / ( (1+x)*(x^2-3*x+1) ). - Len Smiley, Nov 30 2001
From Ralf Stephan, Feb 08 2003: (Start)
a(n) = r^n + (1/r)^n + 2*(-1)^n, with r=(3+sqrt(5))/2.
a(n+3) = 2*a(n+2) + 2*a(n+1) - a(n). (End)
a(n) = L(2*n) + 2*(-1)^n = L(n-1)*L(n+1) + 5(-1)^n.
a(n) = 5*Fibonacci(n)^2 + 4*(-1)^n.
a(n) + a(n+1) = A106729(n). - R. J. Mathar, Nov 17 2011
E.g.f.: 2*exp(-x)*(exp(5*x/2)*cosh(sqrt(5)*x/2)+1). - Wolfdieter Lang, Jan 14 2012
a(n) = 1/4*( a(n-2) - a(n-1) - a(n+1) + a(n+2) ). The same recurrence holds for A007598. - Peter Bala, Aug 18 2015
For n>1, a(n)=(10*F(2*n-1) + 2*L(n-2)*L(n+1))/4 where F(n)=A000045(n), L(n)=A000204(n). - J. M. Bergot, Nov 25 2015
a(n) = (L(n-2)*L(n+2) + L(n-1)*L(n+1))/2 with L(k)=A000032(k). - J. M. Bergot, May 25 2017
From Peter Bala, Nov 13 2019: (Start)
Sum_{n >= 1} 1/a(n) = (1/8)*( theta_3(beta)^4 - 1 ) = A105394, where beta = (3 - sqrt(5))/2 and theta_3(q) = 1 + 2*Sum_{n >= 1} q^(n^2) is a theta function. See Borwein and Borwein, Exercise 7(f), p. 97.
Sum_{n >= 1} 1/(a(n) - 5) = (3 - sqrt(5))/6; Sum_{n >= 1} (-1)^n/(a(n) - 5) = (15 - sqrt(5))/30; Sum_{n >= 1} 1/(a(2*n) - 5) = (5 - sqrt(5))/10.
Sum_{n >= 1} 1/(a(n) - 25/a(n)) = 2/9.
Conjecture: Sum_{n >= 1} 1/(a(n) - 5*(-1)^n*F(2*k+1)^2) = 1/(2*a(2*k+1)) for k = 0,1,2,.... (End)
a(n) = 3*a(n-1) - a(n-2) + 10*(-1)^n. - Greg Dresden, May 18 2020

A081071 a(n) = Lucas(4n+2)-2 = Lucas(2n+1)^2.

Original entry on oeis.org

1, 16, 121, 841, 5776, 39601, 271441, 1860496, 12752041, 87403801, 599074576, 4106118241, 28143753121, 192900153616, 1322157322201, 9062201101801, 62113250390416, 425730551631121, 2918000611027441, 20000273725560976

Offset: 0

R. K. Guy, Mar 04 2003

Conjecture: a(n) = Fibonacci(4*n+3) + Sum_{k=2..2*n} Fibonacci(2*k). - Alex Ratushnyak, May 06 2012

The above conjecture is true for n >= 1. - Nguyen Tuan Anh, Aug 02 2025

Hugh C. Williams, Edouard Lucas and Primality Testing, John Wiley and Sons, 1998, p. 75.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Pridon Davlianidze, Problem B-1270, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 58, No. 2 (2020), p. 179; Four Telescopic Infinite Products, Solution to Problem B-1270 by Jason L. Smith, ibid., Vol. 59, No. 2 (2021), pp. 183-184.
Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A000032 (Lucas numbers), A000045, A001622, A002878 is Lucas(2n+1), A081069.

I:=[1, 16, 121]; [n le 3 select I[n] else 8*Self(n-1)-8*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Jun 26 2012

luc := proc(n) option remember: if n=0 then RETURN(2) fi: if n=1 then RETURN(1) fi: luc(n-1)+luc(n-2): end: for n from 0 to 40 do printf(`%d,`,luc(4*n+2)-2) od: # James Sellers, Mar 05 2003

CoefficientList[Series[-(1+8*x+x^2)/((x-1)*(x^2-7*x+1)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-8,1},{1,16,121},50] (* Vincenzo Librandi, Jun 26 2012 *)
LucasL[4*Range[0,20]+2]-2 (* Harvey P. Dale, Nov 25 2012 *)

x='x+O('x^30); Vec((1+8*x+x^2)/((1-x)*(x^2-7*x+1))) \\ G. C. Greubel, Dec 21 2017

a(n) = 8*a(n-1) - 8*a(n-2) + a(n-3).
G.f.: -(1+8*x+x^2)/((x-1)*(x^2-7*x+1)). - Colin Barker, Jun 26 2012
From Peter Bala, Nov 19 2019: (Start)
Sum_{n >= 1} 1/(a(n) + 5) = (3*sqrt(5) - 5)/30.
Sum_{n >= 1} 1/(a(n) - 5) = (15 - 4*sqrt(5) )/60.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 5) = 1/12.
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 25/a(n)) = (5 + 2*sqrt(5))/120. (End)
Sum_{n>=0} 1/a(n) = (1/sqrt(5)) * Sum_{n>=1} n/F(2*n), where F(n) is the n-th Fibonacci number (A000045). - Amiram Eldar, Oct 05 2020
Product_{n>=1} (1 - 5/a(n)) = phi^2/4, where phi is the golden ratio (A001622) (Davlianidze, 2020). - Amiram Eldar, Dec 04 2024
From Enrique Navarrete, Mar 24 2025: (Start)
20 + 5*a(n) = A106729(n)^2.
Limit_{n->oo} a(n+1)/a(n) = (7 + 3*sqrt(5))/2. (End)

More terms from James Sellers, Mar 05 2003

A106789 Sum of two consecutive squares of Lucas 3-step numbers (A001644).

Original entry on oeis.org

10, 10, 58, 170, 562, 1962, 6562, 22202, 75242, 254330, 860474, 2911226, 9848050, 33316090, 112707970, 381286954, 1289885834, 4363653034, 14762129274, 49939929610, 168945571442, 571538767370, 1933501811618, 6540989771354

Offset: 0

Jonathan Vos Post, May 16 2005

A106729 is sum of two consecutive squares of Lucas numbers (A001254), for which L(n)^2 + L(n+1)^2 = 5*{F(n)^2 + F(n+1)^2} = 5*A001519(n). Sum of two consecutive squares of Lucas 3-step numbers can be expressed in terms of tribonacci numbers, but not quite as neatly, as derived from the identity A001644(n) = T(n) + 2*T(n-1) + 3*T(n-2) = 3*T(n+1) - 2*T(n) - T(n-1) where the tribonacci numbers T(n) = A000073(n).

			a(0) = A001644(0)^2 + A001644(1)^2 = 3^2 + 1^2 = 9 + 1 = 10.
a(1) = A001644(1)^2 + A001644(2)^2 = 1^2 + 3^2 = 1 + 9 = 10.
a(2) = A001644(2)^2 + A001644(3)^2 = 3^2 + 7^2 = 9 + 49 = 58.
a(3) = A001644(3)^2 + A001644(4)^2 = 7^2 + 11^2 = 49 + 121 = 170 = 13^2 + 1.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000073, A001644.

Cf. A001254, A001519, A106729.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6) )); // G. C. Greubel, Apr 21 2019

CoefficientList[Series[2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2 -6*x^3+x^4+x^6), {x,0,40}], x] (* G. C. Greubel, Apr 21 2019 *)
Total/@Partition[LinearRecurrence[{1,1,1},{3,1,3},40]^2,2,1] (* Harvey P. Dale, Apr 03 2022 *)

my(x='x+O('x^40)); Vec(2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6)) \\ G. C. Greubel, Apr 21 2019

(2*(5-5*x+4*x^2-18*x^3-x^4-5*x^5)/(1-2*x-3*x^2-6*x^3+x^4+x^6)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 21 2019

a(n) = A001644(n)^2 + A001644(n+1)^2.

G.f.: 2*(5 - 5*x + 4*x^2 - 18*x^3 - x^4 - 5*x^5)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)).

A122678 Invariant number of polygons of n-circum-C_5 H_5 systems.

Original entry on oeis.org

1, 6, 21, 61, 166, 441, 1161, 3046, 7981, 20901, 54726, 143281, 375121, 982086, 2571141, 6731341, 17622886, 46137321, 120789081, 316229926, 827900701, 2167472181, 5674515846, 14856075361, 38893710241, 101825055366, 266581455861, 697919312221, 1827176480806

Offset: 1

N. J. A. Sloane, Sep 23 2006

Note that the index in Eq. (5) of the paper must be 2k-1, not 2k as published, to reproduce the numbers. - R. J. Mathar, Nov 23 2014

Colin Barker, Table of n, a(n) for n = 1..1000
J. Brunvoll, S. J. Cyvin and B. N. Cyvin, Azulenoids, MATCH, No. 34, 1996, 91-108.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Vec(-x*(1+x)^2/((x-1)*(x^2-3*x+1)) + O(x^40)) \\ Colin Barker, Nov 03 2016

a(n) = 5*Fibonacci(2*k-1)-4 = A106729(k-1)-4.
G.f.: -x*(1+x)^2 / ( (x-1)*(x^2-3*x+1) ). - R. J. Mathar, Nov 23 2014
From Colin Barker, Nov 03 2016: (Start)
a(n) = 2^(-1-n)*(-2^(3+n)-(-5+sqrt(5))*(3+sqrt(5))^n+(3-sqrt(5))^n*(5+sqrt(5))).
a(n) = 4*a(n-1)-4*a(n-2)+a(n-3) for n>3. (End)

A190757 Lucas Aurifeuillian primitive part A of Lucas(10*n - 5).

Original entry on oeis.org

1, 1, 101, 71, 181, 39161, 24571, 12301, 1158551, 87382901, 21211, 373270451, 28143378001, 32414581, 1322154751061, 9062194370461, 1550853481, 2819407321151, 265272771839851, 2366632711, 137083914639998701, 85417012034751151, 3455782010101

Offset: 1

Arkadiusz Wesolowski, Dec 29 2012

nonn

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..250
John Brillhart, Peter L. Montgomery and Robert D. Silverman, Tables of Fibonacci and Lucas factorizations, Math. Comp. 50 (1988), pp. 251-260, S1.
C. K. Caldwell, "Top Twenty" page, Lucas Aurifeuillian primitive part

Cf. A061447, A190781.

lst = {1}; n = 23; Do[f = LucasL[k]; Do[f = f/GCD[f, lst[[d]]], {d, Most@Divisors[k]}]; AppendTo[lst, f], {k, 2, 10*n - 5}]; Table[GCD[lst[[5*k]], 5*Fibonacci[k]*(Fibonacci[k] - 1) + 1], {k, 1, 2*n - 1, 2}]

a(n) = GCD(A061447(10*n-5), A027941(n-1)*A106729(n-1) + 1).

A190781 Lucas Aurifeuillian primitive part B of Lucas(10*n - 5).

Original entry on oeis.org

11, 31, 151, 911, 541, 3641, 272611, 18451, 12760031, 7947701, 767131, 4106261531, 28144128251, 34379101, 120196353941, 823837075741, 51164521, 4215154433351, 2918000731816531, 73998076231, 12462174208709101, 939587136717207031, 3467131047901

Offset: 1

Arkadiusz Wesolowski, Dec 29 2012

nonn

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..250
John Brillhart, Peter L. Montgomery and Robert D. Silverman, Tables of Fibonacci and Lucas factorizations, Math. Comp. 50 (1988), pp. 251-260, S1.
C. K. Caldwell, "Top Twenty" page, Lucas Aurifeuillian primitive part

Cf. A061447, A190757.

lst = {1}; n = 23; Do[f = LucasL[k]; Do[f = f/GCD[f, lst[[d]]], {d, Most@Divisors[k]}]; AppendTo[lst, f], {k, 2, 10*n - 5}]; Table[GCD[lst[[5*k]], 5*Fibonacci[k]*(Fibonacci[k] + 1) + 1], {k, 1, 2*n - 1, 2}]

a(n) = GCD(A061447(10*n-5), A032908(n)*A106729(n-1) + 1).

A106791 Sum of two consecutive squares of Lucas 4-step numbers (A073817).

Original entry on oeis.org

17, 10, 58, 274, 901, 3277, 12402, 46282, 171170, 635953, 2364489, 8785386, 32637202, 121265666, 450571589, 1674090725, 6220049810, 23110593298, 85867345570, 319039636721, 1185390110881, 4404311472106, 16364198176874

Offset: 0

Jonathan Vos Post, May 16 2005

A106729 is sum of two consecutive squares of Lucas numbers (A001254), for which L(n)^2 + L(n+1)^2 = 5*{F(n)^2 + F(n+1)^2} = 5*A001519(n). A106789 is sum of two consecutive squares of Lucas 3-step numbers (A001644). Sum of two consecutive squares of Lucas 4-step numbers can be expressed in terms of tetranacci numbers, but not quite as neatly.

			a(0) = A073817(0)^2 + A073817(1)^2 = 4^2 + 1^2 = 16 + 1 = 17.
a(1) = A073817(1)^2 + A073817(2)^2 = 1^2 + 3^2 = 1 + 9 = 10.
a(2) = A073817(2)^2 + A073817(3)^2 = 3^2 + 7^2 = 9 + 49 = 58.
a(3) = A073817(3)^2 + A073817(4)^2 = 7^2 + 15^2 = 49 + 225 = 274.
a(4) = A073817(4)^2 + A073817(5)^2 = 15^2 + 26^2 = 225 + 676 = 901 = 30^2 + 1.
a(5) = A073817(5)^2 + A073817(6)^2 = 26^2 + 51^2 = 676 + 2601 = 3277.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,4,6,12,-4,-6,0,-2,0,1).

Cf. A001254, A001519, A001644, A073817, A106729.

a:=[17,10,58,274,901,3277,12402, 46282,171170,635953];; for n in [11..40] do a[n]:=2*a[n-1]+4*a[n-2]+6*a[n-3]+12*a[n-4]-4*a[n-5] -6*a[n-6]-2*a[n-8]+a[n-10]; od; a; # G. C. Greubel, Apr 23 2019

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (17-24*x-30*x^2+16*x^3-143*x^4-21*x^5 +46*x^6-32*x^7+2*x^8+17*x^9)/(1-2*x-4*x^2 -6*x^3-12*x^4+4*x^5+6*x^6+2*x^8 -x^10) )); // G. C. Greubel, Apr 23 2019

LinearRecurrence[{2,4,6,12,-4,-6,0,-2,0,1}, {17,10,58,274,901,3277,12402, 46282,171170,635953}, 40] (* G. C. Greubel, Apr 23 2019 *)

my(x='x+O('x^40)); Vec((17-24*x-30*x^2+16*x^3-143*x^4-21*x^5 +46*x^6-32*x^7+2*x^8+17*x^9)/(1-2*x-4*x^2-6*x^3-12*x^4+4*x^5+6*x^6+2*x^8 -x^10)) \\ G. C. Greubel, Apr 23 2019

((17-24*x-30*x^2+16*x^3-143*x^4-21*x^5 +46*x^6-32*x^7+2*x^8+ 17*x^9)/(1-2*x-4*x^2-6*x^3-12*x^4+4*x^5+6*x^6+2*x^8 -x^10)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Apr 23 2019

a(n) = A073817(n)^2 + A073817(n+1)^2.
a(n) = 5*A073817(n)^2 + 4*A073817(n)*A073817(n-4) + A073817(n-4)^2.
G.f.: (17-24*x-30*x^2+16*x^3-143*x^4-21*x^5+46*x^6-32*x^7+2*x^8+17*x^9)/( (1- 3*x-3*x^2+x^3+x^4)*(1+x+2*x^2+2*x^3-2*x^4+x^5-x^6)). - Colin Barker, Dec 17 2012

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A001254 Squares of Lucas numbers.

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A081071 a(n) = Lucas(4*n+2)-2 = Lucas(2*n+1)^2.

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A106789 Sum of two consecutive squares of Lucas 3-step numbers (A001644).

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A122678 Invariant number of polygons of n-circum-C_5 H_5 systems.

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A190757 Lucas Aurifeuillian primitive part A of Lucas(10*n - 5).

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A190781 Lucas Aurifeuillian primitive part B of Lucas(10*n - 5).

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A081071 a(n) = Lucas(4n+2)-2 = Lucas(2n+1)^2.