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A237350 a(n) = the smallest number k such that Sum_{d|k} 1/tau(d) >= n. 4

%I #84 Sep 08 2022 08:46:06

%S 1,6,24,60,180,420,840,2520,4620,9240,13860,27720,55440,55440,110880,

%T 166320,180180,360360,360360,720720,720720,1441440,1801800,2162160,

%U 3063060,4084080,6126120,6126120,6126120,12252240,12252240,18378360,24504480,24504480,30630600,36756720

%N a(n) = the smallest number k such that Sum_{d|k} 1/tau(d) >= n.

%C Are there numbers n > 1 such that Sum_{d|n} 1/tau(d) is an integer?

%C 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, ...

%C 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

%C From _David A. Corneth_, Nov 05 2019: (Start)

%C Instead of checking all divisors of A025487(n), one could use A318277 to see how often each prime signature occurs as a divisor.

%C 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)

%H David A. Corneth, <a href="/A237350/b237350.txt">Table of n, a(n) for n = 1..3338</a> (first 131 terms from Robert Israel, terms <= 10^25)

%H David A. Corneth, <a href="/A237350/a237350.gp.txt">m, Sum_{d|a(m)} 1/tau(d) and the prime signature of a(m)</a>

%e 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.

%p N:= 10^9: # to get all entries <= N

%p Primorials:= NULL:

%p p:= 2: P:= p:

%p while P <= N do

%p Primorials:= Primorials, P;

%p p:= nextprime(p);

%p P:= P*p;

%p od:

%p Primorials:= [Primorials]:

%p S:= {1}:

%p for i from 1 to nops(Primorials) do

%p S:= {seq(seq(s*Primorials[i]^j,

%p j = 0 .. floor(log[Primorials[i]](N/s))),s=S)}

%p od:

%p A:= NULL:

%p S:= sort(convert(S,list)):

%p xmax:= 0:

%p for s in S do

%p x:= floor(add(1/numtheory:-tau(d),d=numtheory:-divisors(s)));

%p if x > xmax then

%p A:= A, s$(x-xmax);

%p xmax:= x

%p fi

%p od:

%p A; # _Robert Israel_, Dec 21 2015

%t 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 *)

%o (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;

%o (PARI) a(n) = {my(k=1); while(sumdiv(k, d, 1/numdiv(d)) < n, k++); k;} \\ _Michel Marcus_, Dec 20 2015

%Y Cf. A000005, A002110, A025487, A253139, A265390, A265391, A265392, A318277.

%Y Cf. A265393 (a(n) = the smallest number k such that floor(Sum_{d|k} 1/tau(d)) = n).

%K nonn

%O 1,2

%A _Jaroslav Krizek_, Dec 13 2015

%E a(24)-a(30) from _Michel Marcus_, Dec 20 2015

%E a(31)-a(35) from _Robert Israel_, Dec 21 2015

%E Missing a(31) = 12252240 inserted in data section by _Georg Fischer_, Nov 05 2019

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Last modified April 16 17:00 EDT 2024. Contains 371749 sequences. (Running on oeis4.)