A194235 a(n) = [Sum_{k=1..n} (k/8)], where [ ]=floor, ( )=fractional part.
0, 0, 0, 1, 1, 2, 3, 3, 3, 3, 4, 4, 5, 6, 7, 7, 7, 7, 7, 8, 8, 9, 10, 10, 10, 10, 11, 11, 12, 13, 14, 14, 14, 14, 14, 15, 15, 16, 17, 17, 17, 17, 18, 18, 19, 20, 21, 21, 21, 21, 21, 22, 22, 23, 24, 24, 24, 24, 25, 25, 26, 27, 28, 28, 28, 28, 28, 29, 29, 30, 31, 31, 31, 31
Offset: 1
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 1..5000
- Index entries for linear recurrences with constant coefficients, signature (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1).
Crossrefs
Cf. A194236.
Programs
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Mathematica
r = 1/8; a[n_] := Floor[Sum[FractionalPart[k*r], {k, 1, n}]] Table[a[n], {n, 1, 90}] (* A194235 *) s[n_] := Sum[a[k], {k, 1, n}] Table[s[n], {n, 1, 100}] (* A194236 *)
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PARI
a(n) = floor(sum(k=1, n, frac(k/8))); \\ Michel Marcus, Nov 03 2017
Formula
From Chai Wah Wu, Jun 10 2020: (Start)
a(n) = a(n-1) + a(n-16) - a(n-17) for n > 17.
G.f.: x*(x^14 + x^13 + x^12 + x^10 + x^6 + x^5 + x^3)/(x^17 - x^16 - x + 1). (End)
Comments