A075150 -id:A075150 - cp's OEIS frontend

A001254 Squares of Lucas numbers.

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

A075155 Cubes of Lucas numbers.

Original entry on oeis.org

8, 1, 27, 64, 343, 1331, 5832, 24389, 103823, 438976, 1860867, 7880599, 33386248, 141420761, 599077107, 2537716544, 10749963743, 45537538411, 192900170952, 817138135549, 3461452853383, 14662949322176, 62113250509227, 263115950765039, 1114577054530568

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 06 2002

G. C. Greubel, Table of n, a(n) for n = 0..1000
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.
Index entries for linear recurrences with constant coefficients, signature (3,6,-3,-1).

Cf. A000032, A061084, A075150, A075151.

Third row of array A103324.

[ Lucas(n)^3 : n in [0..120]]; // Vincenzo Librandi, Apr 14 2011

CoefficientList[Series[(8 - 23*x - 24*x^2 + x^3)/((x^2 + 4*x - 1)*(x^2 - x - 1)), {x,0,50}], x] (* or *) Table[LucasL[n]^3, {n,0,30}] (* or *) LinearRecurrence[{3,6,-3,-1}, {8, 1, 27, 64}, 30] (* G. C. Greubel, Dec 21 2017 *)

a(n)=(fibonacci(n-1)+fibonacci(n+1))^3 \\ Charles R Greathouse IV, Feb 09 2016

from sympy import lucas
def a(n):  return lucas(n)**3
print([a(n) for n in range(25)]) # Michael S. Branicky, Aug 01 2021

a(n) = 3*(-1)^n*L(n) + L(3*n).
a(n) = (-1)^n*A075151(n).
a(n) = A000032(n)^3 = A000032(n) * A001254(n).
a(n) = L(n)*C(n)^2, L(n) = Lucas numbers (A000032), C(n) = reflected Lucas numbers (comment to A061084).
a(n) = 3*a(n-1) + 6*a(n-2) - 3*a(n-3) - a(n-4), n>=4.
G.f.: ( 8-23*x-24*x^2+x^3 )/( (x^2+4*x-1)*(x^2-x-1) ).
a(n) = 2*A001077(n) + 3*A061084(n+1). - R. J. Mathar, Nov 17 2011
a(n) = L(3*n) + (F(n+4) - F(n-4))*(-1)^n, n>3 and F(n)=A000045(n). - J. M. Bergot, Feb 09 2016
a(n) + Sum_{i=0..n+1} a(i) = 19/2 + (5/2)*L(3*n+2). - Greg Dresden, Feb 24 2025

Simpler definition from Ralf Stephan, Nov 01 2004

A219233 Alternating row sums of Riordan triangle A110162.

Original entry on oeis.org

1, -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

Offset: 0

Wolfdieter Lang, Nov 16 2012

If a(0) is put to 2 instead of 1 this becomes a(n) = (-1)^n*A005248(n), n >= 0. These are then the alternating row sums of triangle A127677.

Also abs(a(n)) is the number of rounded area of pentagon or pentagram in series arrangement. - Kival Ngaokrajang, Mar 27 2013

Colin Barker, Table of n, a(n) for n = 0..1000
Richard M. Low and Ardak Kapbasov, Non-Attacking Bishop and King Positions on Regular and Cylindrical Chessboards, Journal of Integer Sequences, Vol. 20 (2017), Article 17.6.1, Table 8.
Kival Ngaokrajang, Pentagram for n = 1..6
Eric Weisstein's World of Mathematics, Pentagram
Index entries for linear recurrences with constant coefficients, signature (-3,-1).

Cf. A099837 (row sums of A110162).

Cf. A000032, A000045, A001254, A005248, A075150, A127677.

A219233:= func< n | n eq 0 select 1 else (-1)^n*Lucas(2*n) >; // G. C. Greubel, Jun 13 2025

A219233[n_]:= (-1)^n*LucasL[2*n] - Boole[n==0]; (* G. C. Greubel, Jun 13 2025 *)

Vec((1-x^2)/(1+3*x+x^2) + O(x^40)) \\ Colin Barker, Oct 14 2015

def A219233(n): return (-1)**n*lucas_number2(2*n,1,-1) - int(n==0) # G. C. Greubel, Jun 13 2025

a(0) = 1 and a(n) = (-1)^n*(F(2*(n+1)) - F(2*(n-1))) = (-1)^n*L(2*n), n>=1, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
O.g.f.: (1-x^2)/(1+3*x+x^2).
G.f.: (W(0) -6)/(5*x) -1 , where W(k) = 5*x*k + x + 6 - 6*x*(5*k-9)/W(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
From Colin Barker, Oct 14 2015: (Start)
a(n) = -3*a(n-1) - a(n-2) for n>2.
a(n) = (1/2*(-3-sqrt(5)))^n + (1/2*(-3+sqrt(5)))^n for n>0. (End)
E.g.f.: 2*exp(-3*x/2)*cosh(sqrt(5)*x/2) - 1. - Stefano Spezia, Dec 26 2021
From G. C. Greubel, Jun 13 2025: (Start)
a(-n) = a(n).
a(n) = (-1)^n*A001254(n) - 2 - [n=0] = A075150(n) - 2 - [n=0]. (End)

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A001254 Squares of Lucas numbers.

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A075155 Cubes of Lucas numbers.

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A219233 Alternating row sums of Riordan triangle A110162.

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