



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
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OFFSET

0,1


COMMENTS

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*(1446)=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*(341)=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*293*7=95.


LINKS



FORMULA

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+x2*x^2 ) / ( (x1)*(1+x)*(x^23*x+1) ).  R. J. Mathar, Aug 21 2012


MATHEMATICA

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


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



