A227013 -id:A227013 - cp's OEIS frontend

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

Clark Kimberling, Jul 01 2013

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.

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

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

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

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 *)

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

A227016 Floor(M(g(n-1)+1,..,g(n))), where M = harmonic mean and g(n) = n(n + 1)(n + 2)/6.

Original entry on oeis.org

1, 2, 7, 14, 27, 45, 69, 101, 141, 191, 252, 323, 408, 506, 618, 746, 890, 1052, 1233, 1432, 1653, 1895, 2159, 2447, 2759, 3097, 3462, 3853, 4274, 4724, 5204, 5716, 6260, 6838, 7451, 8098, 8783, 9505, 10265, 11065, 11905, 12787, 13712, 14679, 15692, 16750

Offset: 1

Clark Kimberling, Jul 01 2013

See A227012.

			a(1) = floor(1/(1/1)) = 1; a(2) = floor(3/(1/2 + 1/3 + 1/4)) = 2; a(3) = floor(6/(1/5 + 1/6 + ... + 1/10)) = 7.

Clark Kimberling, Table of n, a(n) for n = 1..200

Cf. A227012, A227013.

z = 200; f[x_] := f[x] = 1/x; g[n_] := g[n] = n (n + 1) (n + 2)/6; s[n_] := s[n] = Sum[f[k], {k, g[n - 1] + 1, g[n]}]; v[n_] := v[n] = (g[n] - g[n - 1])/s[n]; Table[g[n], {n, 1, z}];
Table[Floor[v[n]], {n, 1, z}]

a(n) + 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-8) - 3*a(n-9) + 3*a(n-10) - a(n-11) for n > 1 (conjectured).

G.f.: (1 - x + 4*x^2 - 2*x^3 + 4*x^4 - x^5 + x^6 + 2*x^7 - x^8 + 3*x^9 - 3*x^10 + x^11)/((x - 1)^4 (1+x) (1+x^2) (1+x^4)) (conjectured). (G.f. found by Peter J. C. Moses, Jul 01 2013)

A227014 a(n) = floor(M(g(n-1)+1,..,g(n))), where M = harmonic mean and g(n) = n^5.

Original entry on oeis.org

1, 10, 104, 543, 1883, 5102, 11717, 23906, 44626, 77735, 128110, 201769, 305989, 449428, 642243, 896212, 1224852, 1643541, 2169636, 2822595, 3624095, 4598154, 5771249, 7172438, 8833478, 10788947, 13076362, 15736301, 18812521, 22352080

Offset: 1

Clark Kimberling, Jul 01 2013

See A227012. It is conjectured that A227014 is a linear recurrence sequence with signature (5,-10,10,-5,1,...Z...,1,-5,-10,-10,-1,0,0), where ...Z... represents a string of 138 zeros; has been confirmed for a(1), a(2),..., a(150000).

			a(1) = floor(1/(1/1)) = 1.
a(2) = floor(31/(1/2 + 1/3 + ... + 1/32)) = 10.

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

Cf. A227012, A227013.

Clear[g]; g[n_] := N[n^5, 100]; a = {1}; Do[AppendTo[a, Floor[(#2 - #1 + 1)/(HarmonicNumber[#2]-HarmonicNumber[#1 - 1])] &[g[k - 1] + 1, g[k]]], {k, 2, 200}]; a (* Peter J. C. Moses, Jul 05 2012 *)
(* confirm generating function *)
p = {1, -4, 5, 9, 54, 117, 117, 122, 118, 122, 118, 122, 118, 122,
   118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118,
   122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122,
   118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118,
   122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122,
   119, 117, 129, 107, 134, 106, 134, 106, 134, 106, 134, 106, 134,
   106, 134, 106, 134, 107, 129, 117, 119, 122, 118, 122, 118, 122,
   118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118,
   122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122,
   118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118,
   122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122, 118, 122,
   117, 126, 113, 113, 64, 5, 1};
q = {0, 0, 1, -5, 10, -10, 5, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
   0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 5, -10, 10, -5,
    1}; gf = Fold[x #1 + #2 &, 0, p]/Fold[x #1 + #2 &, 0, q]; CoefficientList[Series[Factor[gf], {x, 0, 100}], x] (* Peter J. C. Moses, Jul 08 2012 *)

A227015 a(n) = floor(M(g(n-1)+1, ..., g(n))), where M = harmonic mean and g(n) = n^3 + n^2 + n + 1.

Original entry on oeis.org

2, 8, 26, 60, 117, 203, 324, 487, 696, 958, 1279, 1666, 2123, 2657, 3274, 3981, 4782, 5684, 6693, 7816, 9057, 10423, 11920, 13555, 15332, 17258, 19339, 21582, 23991, 26573, 29334, 32281, 35418, 38752, 42289, 46036, 49997, 54179, 58588, 63231, 68112, 73238

Offset: 1

Clark Kimberling, Jul 01 2013

nonn

See A227012.

Clark Kimberling, Table of n, a(n) for n = 1..100

Cf. A227012, A227013.

z = 100; f[x_] := f[x] = 1/x; g[n_] := g[n] = n^3 + n^2 + n + 1; s[n_] := s[n] = Sum[f[k], {k, g[n - 1] + 1, g[n]}]; v[n_] := v[n] = (g[n] - g[n - 1])/s[n]; Table[g[n], {n, 1, z}];
Table[Floor[v[n]], {n, 1, z}]

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 > 2 (conjectured).
G.f.: x*(2 + 2*x + 8*x^2 + 4*x^3 + 5*x^4 + 4*x^5 - 2*x^6 + 3*x^7 - 3*x^8 + x^9)/((x - 1)^4*(1 + x + x^2 + x^3)) (conjectured).
From Franck Maminirina Ramaharo, Apr 16 2018: (Start)
a(n) = (1/2)*((-1)^(n - 1)! + 2*n^3 - n^2 + n + 3 + 2*floor(max(0, n - 4)/4)) (conjectured).
E.g.f.: (1/24)*exp(-x)*(exp(x)*(6*sin(x) + 6*cos(x) + 4*x^3 - 24) + exp(2*x)*(24*x^3 + 60*x^2 + 30*x + 15) + 3) (conjectured).
(End)

cp's OEIS Frontend

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

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A227016 Floor(M(g(n-1)+1,..,g(n))), where M = harmonic mean and g(n) = n(n + 1)(n + 2)/6.

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A227014 a(n) = floor(M(g(n-1)+1,..,g(n))), where M = harmonic mean and g(n) = n^5.

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A227015 a(n) = floor(M(g(n-1)+1, ..., g(n))), where M = harmonic mean and g(n) = n^3 + n^2 + n + 1.

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