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A291899
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Numbers n such that (pod(n)/tau(n)) > (pod(k)/tau(k)) for all k < n.
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3
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1, 3, 4, 6, 8, 10, 12, 18, 20, 24, 30, 36, 48, 60, 72, 84, 90, 96, 108, 120, 168, 180, 240, 336, 360, 420, 480, 504, 540, 600, 630, 660, 672, 720, 840, 1080, 1260, 1440, 1680, 2160, 2520, 3360, 3780, 3960, 4200, 4320, 4620, 4680, 5040, 7560, 9240, 10080, 12600
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OFFSET
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1,2
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COMMENTS
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pod(n) = the product of the divisors of n (A007955), tau(n) = the number of the divisors of n (A000005).
Various methods exist to find terms for this sequence, possibly combinable:
- Brute force; checking every positive integer up to some bound.
- Finding terms based on the prime signature.
- Relating to that, the number of divisors.
- Finding terms based on the GCD of some earlier found terms.
- ... (?)
There seems to be a method that helps finding terms < 10^150 for the similar A034287. (End)
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LINKS
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FORMULA
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EXAMPLE
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6 is a term because pod(6)/tau(6) = 36/4 = 9 > pod(k)/tau(k) for all k < 6.
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MAPLE
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f:= proc(n) local t; t:= numtheory:-tau(n); simplify(n^(t/2))/t end proc:
N:= 20000: # to get all terms <= N
Res:= NULL: m:= 0:
for n from 1 to N do
v:= f(n);
if v > m then Res:= Res, n; m:= v fi
od:
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MATHEMATICA
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With[{s = Array[Times @@ Divisors@ # &, 12600]}, Select[Range@ Length@ s, Function[m, AllTrue[Range[# - 1], m > s[[#]]/DivisorSigma[0, #] &]][s[[#]]/DivisorSigma[0, #]] &]] (* Michael De Vlieger, Oct 10 2017 *)
DeleteDuplicates[Table[{n, Times@@Divisors[n]/DivisorSigma[0, n]}, {n, 13000}], GreaterEqual[ #1[[2]], #2[[2]]]&][[;; , 1]] (* Harvey P. Dale, Mar 03 2024 *)
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PROG
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(Magma) a:=1; S:=[a]; for n in [2..60] do k:=0; flag:= true; while flag do k+:=1; if &*[d: d in Divisors(a)] / #[d: d in Divisors(a)] lt &*[d: d in Divisors(k)] / #[d: d in Divisors(k)] then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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