A075269 -id:A075269 - cp's OEIS frontend

A215602 a(n) = L(n)*L(n+1), where L = A000032 (Lucas numbers).

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, 2139295485798, 5600748293802, 14662949395603, 38388099893012, 100501350283428

Offset: 0

N. J. A. Sloane, Aug 17 2012

Colin Barker, Table of n, a(n) for n = 0..1000
Ömer Eğecioğlu, Elif Saygı, and Zülfükar Saygı, The Mostar and Wiener index of alternate Lucas cubes, Transactions on Combinatorics (2023) Vol. 12, No. 1, 37-46.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A215580. A075269 is a signed version.

Table[LucasL[n]*LucasL[n + 1], {n, 0, 33}] (* Amiram Eldar, Oct 06 2020 *)

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

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

G.f.: ( 2-x+2*x^2 ) / ( (1+x)*(x^2-3*x+1) ). - R. J. Mathar, Aug 21 2012
a(n) = A002878(n)+(-1)^n. - R. J. Mathar, Aug 21 2012
a(n) = F(n-1)*F(n) + F(n-1)*F(n+2) + F(n+1)*F(n) + F(n+1)*F(n+2), where F=A000045, F(-1)=1. - Bruno Berselli, Nov 03 2015
a(n) = F(2*n) + F(2*n+2) + (-1)^n with F(k)=A000045(k). - J. M. Bergot, Apr 15 2016
a(n) = ((-1)^n+(2^(-1-n)*((3-sqrt(5))^n*(-5+sqrt(5))+(3+sqrt(5))^n*(5+sqrt(5)))) / sqrt(5)). - Colin Barker, Oct 01 2016
Sum_{n>=0} (-1)^n/a(n) = sqrt(5)/10. - Amiram Eldar, Oct 06 2020

A215580 Partial sums of A215602.

Original entry on oeis.org

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

cp's OEIS Frontend

A215602 a(n) = L(n)*L(n+1), where L = A000032 (Lucas numbers).

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