A134119 - OEIS

A134119

a(n) = floor(n^2/10) - floor((n-1)^2/10).

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, 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, 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

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OFFSET

0,9

COMMENTS

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.

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

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).

FORMULA

Empirical g.f.: x^4*(x^4+1) / (x^11 - x^10 - x + 1). - Colin Barker, Aug 08 2013

The above conjectured g.f. is correct. - Sela Fried, Dec 08 2024

MATHEMATICA

Table[Floor[n^2/10] - Floor[(n - 1)^2/10], {n, 0, 50}] (* G. C. Greubel, Feb 22 2017 *)

PROG

(PARI) a(n)= floor(n^2/10) - floor((n-1)^2/10)

CROSSREFS

Cf. A000012, A014019, A033999, A056865.