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%I #15 Feb 23 2017 22:14:39
%S 0,0,0,0,1,1,1,1,2,2,2,2,2,2,3,3,3,3,4,4,4,4,4,4,5,5,5,5,6,6,6,6,6,6,
%T 7,7,7,7,8,8,8,8,8,8,9,9,9,9,10,10,10,10,10,10,11,11,11,11,12,12,12,
%U 12,12,12,13,13,13,13,14,14,14,14,14,14,15,15,15,15,16,16,16,16,16,16,17,17,17
%N a(n) = floor(n^2/10) - floor((n-1)^2/10).
%C Note that for n >=1 there is a pattern that keeps steadily alternating between 4 terms and 6 terms for the each two consecutive groups. The terms value remains the same within each 4-term or 6-term group, while during the switch from the 4-group to the 6-group and then back from the 6-group to the 4-group, etc., the term value is getting bumped by 1.
%C Assuming this obeys the recurrence a(n) = a(n-10) + 2, this has generating function G(x) = x^4*(1+x^4)/[(-1+x)^2*(x+1)*(x^4 + x^3 + x^2 + x + 1)*(x^4 - x^3 + x^2 - x + 1)] = (1 - 3x^2 - 3x^3)/[10(x^4 + x^3 + x^2 + x + 1)]+1/[10(x+1)] + 1/[5(-1+x)^2] +(-1 + 2x - 3x^2 - x^3)/[10(x^4 - x^3 + x^2 - x + 1)] + 3/[10(-1+x)]. The first term can be rewritten as a linear superposition of A104384(n), A104384(n+2), A103483(n+3); the second, ~1/(x+1), with the alternating A033999, the third component ~1/(x-1)^2 with a(n)=n+1, the next ~1/(x^4 - x^3 + x^2 - x + 1) = A014019 and the last is proportional to 1/(1-x) = A000012. So a(n) is a sum of these sequences. - _R. J. Mathar_, Jan 16 2008
%H G. C. Greubel, <a href="/A134119/b134119.txt">Table of n, a(n) for n = 0..1000</a>
%F Empirical g.f.: x^4*(x^4+1) / (x^11 - x^10 - x + 1). - _Colin Barker_, Aug 08 2013
%t Table[Floor[n^2/10] - Floor[(n - 1)^2/10], {n, 0, 50}] (* _G. C. Greubel_, Feb 22 2017 *)
%o (PARI) a(n)= floor(n^2/10) - floor((n-1)^2/10)
%K nonn
%O 0,9
%A _Alexander R. Povolotsky_, Jan 12 2008
%E More terms from _N. J. A. Sloane_, Jan 22 2008
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