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A237350
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a(n) = the smallest number k such that Sum_{d|k} 1/tau(d) >= n.
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4
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1, 6, 24, 60, 180, 420, 840, 2520, 4620, 9240, 13860, 27720, 55440, 55440, 110880, 166320, 180180, 360360, 360360, 720720, 720720, 1441440, 1801800, 2162160, 3063060, 4084080, 6126120, 6126120, 6126120, 12252240, 12252240, 18378360, 24504480, 24504480, 30630600, 36756720
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OFFSET
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1,2
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COMMENTS
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Are there numbers n > 1 such that Sum_{d|n} 1/tau(d) is an integer?
Values of function F = Sum_{d|n} 1/tau(d) for some numbers according to their prime signature: F{} = 1; F{1} = 3/2; F{2} = 11/6; F{1, 1} = 9/4; F{3} = 25/12; F{2, 1} = 11/4; F{4} = 137/60; F{3, 1} = 25/8, ...
All terms are of the form Product_{j=1..k} prime(j)^e(j) where e(j+1)<= e(j), and thus products of (not necessarily distinct) primorials. - Robert Israel, Dec 21 2015
Instead of checking all divisors of A025487(n), one could use A318277 to see how often each prime signature occurs as a divisor.
Knowing the lcm of the terms below some m drastically improves the possibility of finding terms. In hindsight, knowing the lcm of the terms below 10^25 yields having to consider 1056 terms of A025487 instead of 222124. Is there some way to accurately predict the lcm to improve computation? (End)
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LINKS
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EXAMPLE
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For n = 2; a(2) = 6 because 6 is the smallest number with Sum_{d|6} 1/tau(d) = 1/1 + 1/2 + 1/2 + 1/4 = 9/4 >= 2.
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MAPLE
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N:= 10^9: # to get all entries <= N
Primorials:= NULL:
p:= 2: P:= p:
while P <= N do
Primorials:= Primorials, P;
p:= nextprime(p);
P:= P*p;
od:
Primorials:= [Primorials]:
S:= {1}:
for i from 1 to nops(Primorials) do
S:= {seq(seq(s*Primorials[i]^j,
j = 0 .. floor(log[Primorials[i]](N/s))), s=S)}
od:
A:= NULL:
S:= sort(convert(S, list)):
xmax:= 0:
for s in S do
x:= floor(add(1/numtheory:-tau(d), d=numtheory:-divisors(s)));
if x > xmax then
A:= A, s$(x-xmax);
xmax:= x
fi
od:
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MATHEMATICA
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s[1] = 1; s[n_] := DivisorSum[n, 1/DivisorSigma[0, #] &]; n = 1; k = 1; seq = {}; Do[While[s[k] < n, k++]; AppendTo[seq, k]; n++, {j, 1, 20}]; seq (* Amiram Eldar, Jan 30 2019 *)
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PROG
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(Magma) a:=1; S:=[a]; for n in [2..14] do k:=0; flag:= true; while flag do k+:=1; if &+[1/NumberOfDivisors(d): d in Divisors(k)] gt n then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;
(PARI) a(n) = {my(k=1); while(sumdiv(k, d, 1/numdiv(d)) < n, k++); k; } \\ Michel Marcus, Dec 20 2015
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CROSSREFS
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Cf. A265393 (a(n) = the smallest number k such that floor(Sum_{d|k} 1/tau(d)) = n).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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Missing a(31) = 12252240 inserted in data section by Georg Fischer, Nov 05 2019
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STATUS
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approved
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