

A134119


a(n) = floor(n^2/10)  floor((n1)^2/10).


1



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 4term or 6term group, while during the switch from the 4group to the 6group and then back from the 6group to the 4group, etc., the term value is getting bumped by 1.
Assuming this obeys the recurrence a(n) = a(n10) + 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/(x1)^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/(1x) = 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


FORMULA

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


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((n1)^2/10)


CROSSREFS

Sequence in context: A004052 A247781 A051742 * A064661 A226982 A280952
Adjacent sequences: A134116 A134117 A134118 * A134120 A134121 A134122


KEYWORD

nonn


AUTHOR

Alexander R. Povolotsky, Jan 12 2008


EXTENSIONS

More terms from N. J. A. Sloane, Jan 22 2008


STATUS

approved



