
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.
