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%I #26 Feb 22 2024 10:47:24
%S 2,5,17,45,122,320,842,2205,5777,15125,39602,103680,271442,710645,
%T 1860497,4870845,12752042,33385280,87403802,228826125,599074577,
%U 1568397605,4106118242,10749957120,28143753122,73681302245,192900153617,505019158605,1322157322202,3461452808000,9062201101802,23725150497405,62113250390417,162614600673845
%N Partial sums of A215602.
%C 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.
%C 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.
%H Paolo Xausa, <a href="/A215580/b215580.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-3,1).
%F a(2n) = L(4*n)-2, a(2*n+1) = L(4*n+2)-1, where L() are the Lucas numbers A000032.
%F G.f. ( -2+x-2*x^2 ) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - _R. J. Mathar_, Aug 21 2012
%F a(n) = A005248(n+1)-A000034(n). - _R. J. Mathar_, Aug 21 2012
%t LinearRecurrence[{3, 0, -3, 1}, {2, 5, 17, 45}, 35] (* _Paolo Xausa_, Feb 22 2024 *)
%Y Cf. A000032, A075269, A064831, A215602.
%Y Cf. A005248, A000034.
%K nonn,easy
%O 0,1
%A _J. M. Bergot_, Aug 16 2012
%E Edited by _N. J. A. Sloane_, Aug 17 2012
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