A001254 -id:A001254 - cp's OEIS frontend

A219233 Alternating row sums of Riordan triangle A110162.

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)

A075150 a(n) = L(n)*C(n), L(n)=Lucas numbers (A000032), C(n)=reflected Lucas numbers (see comment to A061084).

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

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 05 2002

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-2,2,1).

Cf. A000032, A001254, A061084, A219233.

A075150:= func< n | (-1)^n*Lucas(n)^2 >; // G. C. Greubel, Jun 14 2025

CoefficientList[Series[(4 + 7*x - x^2)/(1 + 2*x - 2*x^2 - x^3), {x, 0, 30}], x]
LinearRecurrence[{-2,2,1},{4,-1,9},50] (* Harvey P. Dale, Nov 08 2011 *)

a(n) = round((2+(1/2*(-3-sqrt(5)))^n+(1/2*(-3+sqrt(5)))^n)) \\ Colin Barker, Oct 01 2016

Vec((4+7*x-x^2)/(1+2*x-2*x^2-x^3) + O(x^30)) \\ Colin Barker, Oct 01 2016

def A075150(n): return (-1)**n*lucas_number2(n,1,-1)**2 # G. C. Greubel, Jun 14 2025

a(n) = (-1)^n*A000032(2*n) + 2.
a(n) = -2*a(n-1) + 2*a(n-2) + a(n-3) with a(0)=4, a(1)=-1, a(2)=9.
G.f.: (4 + 7*x - x^2)/(1 + 2*x - 2*x^2 - x^3).
a(n) = (-1)^n*A001254(n). - R. J. Mathar, Jan 11 2012
a(n) = 2 + (1/2*(-3-sqrt(5)))^n + (1/2*(-3+sqrt(5)))^n. - Colin Barker, Oct 01 2016
From G. C. Greubel, Jun 14 2025: (Start)
a(n) = A000032(n)*A000032(-n) = (-1)^n*A000032(n)^2.
a(n) = A219233(n) + 2 + [n=0].
E.g.f.: 2*exp(-3*x/2)*cosh(sqrt(5)*x/2) + 2*exp(x). (End)

A105394 Decimal expansion of sum of reciprocals of squares of Lucas numbers.

Original entry on oeis.org

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

Offset: 1

Jonathan Vos Post, Apr 04 2005

This constant is transcendental (Duverney et al., 1997). - Amiram Eldar, Oct 30 2020

			1.207291996985747074417204...

Jonathan M. Borwein and Peter B. Borwein, Pi and the AGM, Wiley, 1987, p. 97.

Richard André-Jeannin, Irrationalité de la somme des inverses de certaines suites récurrentes, C. R. Acad. Sci. Paris Ser. I Math., Vol. 308, No. 19 (1989), pp. 539-541.
Paul S. Bruckman,, Problem H-347, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 20, No. (1982), p. 372; It All Adds Up, Solution to Problem H-347 by the proposer, ibid., Vol. 22, No. 1 (1984), pp. 94-96.
Daniel Duverney, Keiji Nishioka, Kumiko Nishioka and Iekata Shiokawa, Transcendence of Rogers-Ramanujan continued fraction and reciprocal sums of Fibonacci numbers, Proceedings of the Japan Academy, Series A, Mathematical Sciences, Vol. 73, No. 7 (1997), pp. 140-142.
Eric Weisstein's World of Mathematics, Lucas Number.
Eric Weisstein's World of Mathematics, Fibonacci Number.
Eric Weisstein's World of Mathematics, Reciprocal Fibonacci Constant.
Index entries for transcendental numbers

Cf. A000032, A001254 (squares of Lucas numbers).

Cf. A079586, A093540, A105393, A153415.

f[n_] := f[n] = RealDigits[ Sum[ 1/LucasL[k]^2, {k, 1, n}], 10, 100] // First; f[n=100]; While[f[n] != f[n-100], n = n+100]; f[n] (* Jean-François Alcover, Feb 13 2013 *)

Equals Sum_{n >= 1} 1/L(n)^2.
Equals (1/8)*( theta_3(beta)^4 - 1 ), 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. - Peter Bala, Nov 13 2019
Equals c*(2*c+1), where c = A153415 (follows from the identity Sum_{n=-oo..oo} 1/L(n^2) = (Sum_{n=-oo..oo} 1/L(2*n))^2, see Bruckman, 1982). - Amiram Eldar, Jan 27 2022

A005970 Partial sums of squares of Lucas numbers.

Original entry on oeis.org

1, 10, 26, 75, 196, 520, 1361, 3570, 9346, 24475, 64076, 167760, 439201, 1149850, 3010346, 7881195, 20633236, 54018520, 141422321, 370248450, 969323026, 2537720635, 6643838876, 17393796000, 45537549121, 119218851370

Offset: 1

N. J. A. Sloane

Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972, p. 20.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A000032, A215602.

lucas := proc(n) option remember: if n=1 then RETURN(1) fi: if n=2 then RETURN(3) fi: lucas(n-1)+lucas(n-2) end: l[0] := 0: for i from 1 to 50 do l[i] := l[i-1]+lucas(i)^2; printf(`%d,`,l[i]) od: # James Sellers, May 29 2000

Accumulate[LucasL[Range[30]]^2] (* Harvey P. Dale, Dec 06 2019 *)

a(n) - a(n-1) = A001254(n).
G.f.: (1+7*x-4*x^2)/((1-x)*(1+x)*(1-3*x+x^2)). - Simon Plouffe in his 1992 dissertation
From Amiram Eldar, Jan 13 2022: (Start)
a(n) = Sum_{k=1..n} L(k)^2, where L(k) is the k-th Lucas number (A000032).
a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4), for n > 4.
a(n) = L(n)*L(n+1) - 2 = A215602(n) - 2. (End)

More terms from James Sellers, May 29 2000

Definition clarified by Harvey P. Dale, Dec 06 2019

A075151 a(n)=L(n)^2*C(n), L(n)=Lucas numbers (A000032), C(n)=reflected Lucas numbers (comment to A061084).

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 05 2002

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

Cf. A000032, A061084, A001254, A075155.

[((-1)^n*Lucas(n))^3: n in [0..30]]; // Vincenzo Librandi, Apr 22 2018

CoefficientList[Series[(8+23*x-24*x^2-x^3)/(1+3*x-6*x^2-3*x^3+x^4), {x, 0, 25}], x]
LinearRecurrence[{-3,6,3,-1},{8,-1,27,-64},30] (* Harvey P. Dale, Apr 06 2013 *)
Table[LucasL[-n]^3, {n, 0, 25}] (* Vincenzo Librandi, Apr 22 2018 *)

a(n) = 3*L(n)+(-1)^n*L(3n).
a(n) = -3a(n-1)+6a(n-2)+3a(n-3)-a(n-4), n>3.
G.f.: ( 8+23*x-24*x^2-x^3 ) / ( (x^2+x-1)*(x^2-4*x-1) ).
a(n) is asymptotic to (-phi)^(3n) where phi is the golden ratio (1+sqrt(5))/2. - Benoit Cloitre, Sep 07 2002
a(n) = ((-1)^n*L(n))^3 = L(-n)^3. - Ehren Metcalfe, Apr 21 2018

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)).

A201207 Half-convolution of sequence A000032 (Lucas) with itself.

Original entry on oeis.org

4, 2, 7, 11, 27, 41, 84, 137, 270, 435, 826, 1338, 2488, 4024, 7353, 11899, 21461, 34723, 61960, 100255, 177344, 286947, 503892, 815316, 1422892, 2302286, 3996619, 6466667, 11173935, 18079805, 31114236

Offset: 0

Wolfdieter Lang, Jan 03 2012

For the definition of the half-convolution of a sequence with itself see a comment on A201204. There the rule for the o.g.f. is given. Here the o.g.f. is (L(x)^2 + L2(x^2))/2, with the o.g.f. L(x)=(2-x)/(1-x-x^2) of A000032, and L2(x) = (4-7*x-x^2)/((1+x)*(1-3*x+x^2)) the o.g.f. of A001254. This leads to the o.g.f given in the formula section.

For the bisection of this sequence see A203570 and A203574.

Cf. A000032, A000045, A201204, A203570, A203574.

a(n) = Sum_{k=0..floor(n/2)} L(k)*L(n-k), n >= 0, with the Lucas numbers L(n)=A000032(n).
O.g.f.: (4-2*x-7*x^2+6*x^3-x^4+3*x^5)/((1-3*x^2+x^4)*(1+x^2)*(1-x-x^2)). See a comment above.
a(n) = (1/4)*(2*(2*n+5+(-1)^n)*F(n+1)-(2*n+3+(-1)^n)*F(n)) +(i^n+(-i)^n)/2, n >= 0, with the Fibonacci numbers F(n)=A000045(n) and the imaginary unit i=sqrt(-1). From the partial fraction decomposition of the o.g.f. and the Fibonacci recurrence.

A339669 Number of Fibonacci divisors of Lucas(n)^2 + 1.

Original entry on oeis.org

2, 2, 3, 1, 3, 2, 3, 2, 5, 1, 5, 2, 4, 2, 5, 1, 5, 2, 4, 2, 6, 1, 6, 2, 4, 2, 6, 1, 6, 2, 4, 2, 6, 1, 7, 2, 5, 2, 6, 1, 6, 2, 4, 2, 7, 1, 7, 2, 5, 2, 7, 1, 6, 2, 5, 2, 7, 1, 6, 2, 4, 2, 8, 1, 9, 2, 5, 2, 6, 1, 6, 2, 4, 2, 7, 1, 9, 2, 6, 2, 7, 1, 7, 2, 5, 2, 7, 1, 6

Offset: 0

Michel Lagneau, Dec 12 2020

nonn

Particular attention must be paid to the regularity properties of the number of divisors of Lucas(n)^2 + 1 observed for n < 156, when a(n) = 1 or 2. From this observation, we propose two conjectures verified for n < 156.
Conjecture 1: a(6*n+3) = 1.
Conjecture 2: a(6*n+1) = a(6*n+5) = 2.
The table in the links shows an array where terms are arranged in a table of 12 columns and 13 rows. We see the periods when a(n) = 1 and 2.

			a(8) = 5 because the divisors of Lucas(8)^2 + 1 = 47^2 + 1 = 2210 are {1, 2, 5, 10, 13, 17, 26, 34, 65, 85, 130, 170, 221, 442, 1105, 2210} with 5 Fibonacci divisors: 1, 2, 5, 13 and 34.

Antti Karttunen, Table of n, a(n) for n = 0..17800
Michel Lagneau, Table

Cf. A000032, A000045, A001254, A339461.

with(combinat,fibonacci):nn:=100:F:={}:
Lucas:=n->2*fibonacci(n-1)+fibonacci(n):
for k from 0 to nn do:
  F:=F union {fibonacci(k)}:
od:
   for m from 0 to 90 do:
    l:=Lucas(m)^2+1:d:=numtheory[divisors](l):n0:=nops(d):
    lst:= F intersect d: n1:=nops(lst):printf(`%d, `,n1):
   od:

Array[DivisorSum[LucasL[#]^2 + 1, 1 &, AnyTrue[Sqrt[5 #^2 + 4 {-1, 1}], IntegerQ] &] &, 89, 0] (* Michael De Vlieger, Dec 12 2020 *)

a(n) = { my(l2 = 5*fibonacci(n)^2 + 4*(-1)^n + 1, k = 1, m = 2, res = 1, g); while(m <= l2, if(l2 % m == 0, res++); g = m; m += k; k = g; ); res } \\ David A. Corneth, Dec 12 2020

A105941 Powers of Lucas numbers.

Original entry on oeis.org

1, 2, 3, 4, 7, 8, 9, 11, 16, 18, 27, 29, 32, 47, 49, 64, 76, 81, 121, 123, 128, 199, 243, 256, 322, 324, 343, 512, 521, 729, 841, 843, 1024, 1331, 1364, 2048, 2187, 2207, 2209, 2401, 3571, 4096, 5776, 5778, 5832, 6561, 8192, 9349, 14641, 15127, 15129, 16384

Offset: 1

Jonathan Vos Post, Apr 27 2005

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 56.
Thomas Koshy, "Fibonacci and Lucas Numbers and Applications", Wiley, New York, 2001.
V. E. Hoggatt, Jr., Fibonacci and Lucas Numbers. Houghton, Boston, MA, 1969.

T. D. Noe, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Lucas Number.

A000032 Lucas numbers. A001254 Squares of Lucas numbers. A075155 Cubes of Lucas numbers. A099923 Fourth powers of Lucas numbers. A103325 Fifth powers of Lucas numbers. A103324 Square array T(n, k) read by antidiagonals: powers of Lucas numbers. A105317 Powers of Fibonacci numbers.

Cf. A000032, A001254, A075155, A099923, A103325, A103324, A105317.

lim = 10^5; t = Table[f = LucasL[n]; If[f == 1, {1}, f^Range[Floor[Log[lim]/Log[f]]]], {n, 0, Floor[Log[GoldenRatio, lim]]}]; Union[Flatten[t]] (* T. D. Noe, Sep 27 2011 *)

{a(n)} = {A000204} U {A001254} U {A075155} U {A099923} U {A103325}... L(n)^2 = L(2n) + 2(-1)^n = L(n-1)*L(n+1) + 5(-1)^n. L(n)^3 = L(3n) + 3(-1)^n*L(n). L(n)^4 = L(4n) + 4(-1)^n*L(2n) + 6. L(n)^5 = L(5n) + 5(-1)^n*L(3n) + 10L(n).

Corrected by T. D. Noe, Sep 26 2011

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

cp's OEIS Frontend

A219233 Alternating row sums of Riordan triangle A110162.

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A075150 a(n) = L(n)*C(n), L(n)=Lucas numbers (A000032), C(n)=reflected Lucas numbers (see comment to A061084).

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A105394 Decimal expansion of sum of reciprocals of squares of Lucas numbers.

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A005970 Partial sums of squares of Lucas numbers.

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A075151 a(n)=L(n)^2*C(n), L(n)=Lucas numbers (A000032), C(n)=reflected Lucas numbers (comment to A061084).

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

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A201207 Half-convolution of sequence A000032 (Lucas) with itself.

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