A215602 -id:A215602 - cp's OEIS frontend

A215580 Partial sums of A215602.

2, 5, 17, 45, 122, 320, 842, 2205, 5777, 15125, 39602, 103680, 271442, 710645, 1860497, 4870845, 12752042, 33385280, 87403802, 228826125, 599074577, 1568397605, 4106118242, 10749957120, 28143753122, 73681302245, 192900153617, 505019158605, 1322157322202, 3461452808000, 9062201101802, 23725150497405, 62113250390417, 162614600673845

Offset: 0

J. M. Bergot, Aug 16 2012

Dividing the terms of this sequence by Fibonacci or Lucas numbers yields symmetric sets of remainders of determinable lengths. For F(n) beginning at n=3: (a) F(2n) will have a set of remainders of length 2n in which the sum of the remainders is 3*(F(2n)-n). Example for F(2*6)=144: the set of remainders is {2,5,17,45,122,32,122,45,17,5,2,0} with 2*6=12 terms and a sum of 3*(144-6)=414. (b) For F(2n+1) there will be 2*(2n+1) terms having a sum equal to (2n+1)*(F(2n+1)-3). Example for F(2*4+1)=34: the remainders are {2,5,7,11,20,14,26,29,31,29,26,14,20,11,17,5,2,0} with 2*9 terms and a sum of 9*(34-1)=279.

Using Lucas numbers starting at n=2: (a) L(2n) has 4n remainders with sum (2n+1)*(L(2n)-6*n). Example for n=4 giving L(2*4)=47, has remainders {2,5,17,45,28,38,43,43,43,38,28,45,17,5,2,0} with a sum of (8+1)*(47)-6*4=399. (B) For L(2n+1) the length of the period is 2*(2n+1) and the sum of the remainders is 4*L(2n+1)-3*(2n+1). Example for n=3 for L(2*3+1)=29 has remainders {2,5,17,16,6,1,11,6,16,17,5,2,0} with length 2*7 and sum of terms 4*29-3*7=95.

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

Cf. A000032, A075269, A064831, A215602.

Cf. A005248, A000034.

LinearRecurrence[{3, 0, -3, 1}, {2, 5, 17, 45}, 35] (* Paolo Xausa, Feb 22 2024 *)

a(2n) = L(4*n)-2, a(2*n+1) = L(4*n+2)-1, where L() are the Lucas numbers A000032.
G.f. ( -2+x-2*x^2 ) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - R. J. Mathar, Aug 21 2012
a(n) = A005248(n+1)-A000034(n). - R. J. Mathar, Aug 21 2012

Edited by N. J. A. Sloane, Aug 17 2012

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

A075269 Product of Lucas numbers and inverted Lucas numbers: a(n)=A000032(n)*A075193(n).

Original entry on oeis.org

2, -3, 12, -28, 77, -198, 522, -1363, 3572, -9348, 24477, -64078, 167762, -439203, 1149852, -3010348, 7881197, -20633238, 54018522, -141422323, 370248452, -969323028, 2537720637, -6643838878, 17393796002, -45537549123, 119218851372, -312119004988, 817138163597

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Sep 11 2002

Index entries for linear recurrences with constant coefficients, signature (-2,2,1).

Cf. A000032, A002878, A075193.

CoefficientList[Series[(2 + x + 2x^2)/(1 + 2x - 2x^2 - x^3), {x, 0, 30}], x]
LinearRecurrence[{-2,2,1},{2,-3,12},30] (* Harvey P. Dale, Jun 30 2022 *)

a(n)=1+(-1)^n*(fibonacci(2*n)+fibonacci(2*n+2))

a(n) = 1 + (-1)^n*A002878(n).
From Michael Somos, Apr 07 2003: (Start)
G.f.: (2+x+2x^2)/((1+3x+x^2)(1-x)).
a(n) = -3a(n-1) - a(n-2)+5 = -2a(n-1) + 2a(n-2) + a(n-3) = a(-1-n). (End)
Sum_{n>=0} 1/a(n) = sqrt(5)/10. - Amiram Eldar, Jan 15 2022
a(n) = (-1)^n*A215602(n). - R. J. Mathar, Jul 09 2024
a(n) - a(n-1) = (-1)^n* A054888(n), n>0. - R. J. Mathar, Jul 09 2024

A348591 a(n) = L(n)*L(n+1) mod F(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.

Original entry on oeis.org

0, 1, 0, 3, 5, 3, 18, 3, 52, 3, 141, 3, 374, 3, 984, 3, 2581, 3, 6762, 3, 17708, 3, 46365, 3, 121390, 3, 317808, 3, 832037, 3, 2178306, 3, 5702884, 3, 14930349, 3, 39088166, 3, 102334152, 3, 267914293, 3, 701408730, 3, 1836311900, 3, 4807526973, 3, 12586269022, 3, 32951280096, 3, 86267571269, 3

Offset: 0

J. M. Bergot and Robert Israel, Jan 25 2022

			a(5) = L(5)*L(6) mod F(7) = 11*18 mod 13 = 3.

Robert Israel, Table of n, a(n) for n = 0..4761
Index entries for linear recurrences with constant coefficients, signature (-1,3,3,-1,-1).

Cf. A000032, A000045, A215602, A333599, A347861, A348592.

F:= combinat:-fibonacci:
L:= n -> F(n-1)+F(n+1):
map(n -> L(n)*L(n+1) mod F(n+2), [$0..30]);

a[n_] := Mod[LucasL[n] * LucasL[n + 1], Fibonacci[n + 2]]; Array[a, 50, 0] (* Amiram Eldar, Jan 26 2022 *)

from gmpy2 import fib, lucas2
def A348591(n): return (lambda x,y:int(x[0]*x[1] % y))(lucas2(n+1),fib(n+2)) # Chai Wah Wu, Jan 26 2022

a(n) = 3 if n >= 3 is odd.
a(n) = A000045(n+2)-3 if n >= 2 is even.
a(n) + a(n+1) - 3*a(n+2) - 3*a(n+3) + a(n+4) + a(n+5) = 0 for n >= 2.
G.f.: -x*(2*x^5-5*x^3-x-1)/((x+1)*(x^2+x-1)*(x^2-x-1)). - Alois P. Heinz, Jan 26 2022

A216243 Partial sums of the squares of Lucas numbers (A000032).

Original entry on oeis.org

4, 5, 14, 30, 79, 200, 524, 1365, 3574, 9350, 24479, 64080, 167764, 439205, 1149854, 3010350, 7881199, 20633240, 54018524, 141422325, 370248454, 969323030, 2537720639, 6643838880, 17393796004, 45537549125, 119218851374, 312119004990, 817138163599, 2139295485800

Offset: 0

R. J. Mathar, Mar 14 2013

Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1).

Cf. A001654.

A001254 := proc(n)
        A000032(n)^2 ;
end proc;
A := proc(n)
        add( A001254(i),i=0..n) ;
end proc:

Accumulate[LucasL[Range[0,30]]^2] (* or *) LinearRecurrence[{3,0,-3,1},{4,5,14,30},30] (* Harvey P. Dale, Oct 13 2019 *)

a(n) = Sum_{i=0..n} A001254(i) = A002878(n) +A176040(n) = A215602(n)+2.
G.f.: ( -4+7*x+x^2 ) / ( (x-1)*(1+x)*(x^2-3*x+1) ).
a(n) = -7*A064831(n) -A064831(n-1) +4*A064831(n+1).
a(n) = L(2*n+1) + 2 + (-1)^n, for L(n) the Lucas sequence A000032(n). - Greg Dresden, Jan 26 2021

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A215580 Partial sums of A215602.

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A005970 Partial sums of squares of Lucas numbers.

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A075269 Product of Lucas numbers and inverted Lucas numbers: a(n)=A000032(n)*A075193(n).

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A348591 a(n) = L(n)*L(n+1) mod F(n+2) where F=A000045 is the Fibonacci numbers and L = A000032 is the Lucas numbers.

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A216243 Partial sums of the squares of Lucas numbers (A000032).

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