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A190146
Decimal expansion of Sum_{k>=2} (1/Sum_{j=2..k} j'), where n' is the arithmetic derivative of n.
5
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)
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
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