A190146 Decimal expansion of Sum_{k>=2} (1/Sum_{j=2..k} j'), where n' is the arithmetic derivative of n.

2, 3, 3, 0, 0, 9

Offset

1,1

Comments

Slow convergence.

a(7) is likely either 3 or 4. Is there a simple proof that this sum converges? - Nathaniel Johnston, May 24 2011

From Husnain Raza, Aug 29 2023: (Start)

The series indeed converges: we have that the series is C = Sum_{k>=2} (1/Sum_{j=2..k} A003415(j)).

Let s_k = Sum_{j=2..k} A003415(j) be the inner sum.

It is known that s_k = (1/2)*T_0*k^2 + O(k^(1+n)) for all real n > 0 where T_0 = A136141.

Therefore, 1/s_k = (2/T_0)*k^(-2) + O(k^(-3+n)) = (2/T_0)*k^(-2) + O(k^(-3)).

Summing both sides from k=2 to infinity, we have that:

C = Sum_{k >= 2} 1/s_k = Sum_{k >= 2} ((2/T_0)*k^(-2) + O(k^(-3))), which converges. (End)

Links

Table of n, a(n) for n=1..6.

Example

1/2' + 1/(2'+3') + 1/(2'+3'+4') + 1/(2'+3'+4'+5') + ... = 1 + 1/2 + 1/6 + 1/7 + ... = 2.33009...

Maple

with(numtheory);
P:=proc(i)
local a, b, f, n, p, pfs;
a:=0; b:=0;
for n from 2 to i do
  pfs:=ifactors(n)[2];
  f:=n*add(op(2, p)/op(1, p), p=pfs);
  b:=b+f; a:=a+1/b;
od;
print(evalf(a, 300));
end:
P(1000);

Mathematica

digits = 5; d[0] = d[1] = 0; d[n_] := d[n] = n*Total[Apply[#2/#1 &, FactorInteger[n], {1}]]; p[m_] := p[m] = Sum[1/Sum[d[j], {j, 2, k}], {k, 2, m}] // RealDigits[#, 10, digits]& // First; p[digits]; p[m = 2*digits]; While[Print["p(", m, ") = ", p[m]]; p[m] != p[m/2], m = 2*m]; p[m] (* Jean-François Alcover, Feb 21 2014 *)

Crossrefs

Cf. A003415, A190144, A190145, A190147.

Sequence in context: A260208 A242457 A340261 * A026932 A087401 A332915

Adjacent sequences: A190143 A190144 A190145 * A190147 A190148 A190149

Keyword

nonn,more,cons

Author

Paolo P. Lava, May 05 2011

Extensions

a(6) corrected and a(7) removed by Nathaniel Johnston, May 24 2011

Status

approved