A227012 a(n) = floor(M(g(n-1)+1, ..., g(n))), where M = harmonic mean and g(n) = n^3.

1, 4, 16, 43, 91, 166, 275, 422, 614, 857, 1158, 1521, 1953, 2460, 3049, 3724, 4492, 5359, 6332, 7415, 8615, 9938, 11391, 12978, 14706, 16581, 18610, 20797, 23149, 25672, 28373, 31256, 34328, 37595, 41064, 44739, 48627, 52734, 57067, 61630, 66430, 71473

Offset: 1

Author

Clark Kimberling, Jul 01 2013

Keywords

• nonn
• easy

Comments

Suppose that f(k) is a sequence such that f(k) > 0 for k >= 1, the limit of f(k) is 0, and the sum of f(k) as k->oo diverges. Let g(n) be a strictly increasing sequence of positive integers, and s(n) = Sum_{k=g(n-1)+1..g(n)} f(k). If f(k) = 1/k, then M(n) = (g(n) - g(n-1))/s(n) is the harmonic mean of g(n-1),...,g(n).

Conjecture: if f(k) = u/(v*k + w), where u,v,w are integers, and g(n) is a polynomial, then the sequence with n-th term m(n) = floor(M(n)) is linearly recurrent. The conjecture extends to these cases, in which a,b,c,d are integers and a > 0:

(1) if g(n) = a*n^2 + b*n + c, the recurrence has order 2, and the first 3 recurrence coefficients for m(n) are 3, -3, 1; these are followed by some nonnegative number of 0's, a property abbreviated below as "(fbz)"; e.g., A002378.

(2) if g(n) has the form (a*n^2 + b*n + c)/2 where a and b are odd, then the recurrence has order 4, and the first 4 coefficients for m(n) are 2, 0-, -1, 2 (fbz); e.g., A080576.

(3) if g(n) = a*n^3 + b*n^2 + c*n + d, the recurrence has order 7, and the first 7 coefficients for m(n) are 3, -3, 1, 1, -3, 3, -1 (fbz); e.g., A227012.

Examples

			a(1) = floor(1/(1/1)) = 1, a(2) = floor(7/(1/2 + 1/3 + ... + 1/8)).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A227013, A227014, A227015, A227016, A227017.

Programs

Mathematica

Clear[g]; g[n_] := g[n] = n^3; a = {1}; Do[AppendTo[a, Floor[(Last[#] - First[#] + 1)/(HarmonicNumber[Last[#]]-HarmonicNumber[First[#] - 1])] &[
   N[{g[k - 1] + 1, g[k]}, 150]]], {k, 2, 100}]; a (* Peter J. C. Moses, Jul 03 2012 *)

Formula

a(n+2) = (1/8)*(27 - (-1)^n - 2*cos(n*Pi/2) + 2*sin(n*Pi/2) + 2*n*(4*n^2 + 18 n + 27)) for n >= 1 (conjectured).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-4) - 3*a(n-5) + 3*a(n-6) - a(n-7) for n >= 10 (conjectured).

G.f.: x*(1 + x + 7*x^2 + 6*x^3 + 5*x^4 + 5*x^5 - 2*x^7 + x^8)/(((x - 1)^4)*(1 + x + x^2 + x^3)) (conjectured).

a(n) = (2*n^3 - 3*n^2 + n + 2)/2 + floor(max(0, n - 3)/4) (conjectured). - Franck Maminirina Ramaharo, Apr 12 2018