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A325237
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Squarefree k such that 1/2 - phi(k)/k is positive and minimal for k with gpf(k) = prime(n).
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2
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2, 6, 10, 105, 165, 195, 4641, 5187, 5313, 266133, 8870433, 3068957045, 11063481, 10164297, 667797009, 909411789, 32221169781185, 1963007211216415, 421522466365, 3012887561310445
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OFFSET
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1,1
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COMMENTS
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Let gpf(k) = A006530(k) and let phi(n) = A000010(n) for k in A005117. There are 2^(n-1) numbers k with gpf(k) = n, since we can only either have p_i^0 or p_i^1 where p_i | k and i <= n. For example, for n = 2, there are only 2 squarefree numbers k with prime(2) = 3 as greatest prime factor. These are 3 = 2^0 * 3^1, and 6 = 2^1 * 3^1. We observe that we can write multiplicities of the primes as A067255(k), and thus for the example derive 3 = "0,1" and 6 = "1,1". Thus for n = 3, we have 5 = "0,0,1", 15 = "0,1,1", 10 = "1,0,1", and 30 = "1,1,1". This establishes the possible values of k with respect to n. We choose the value of k in n for which 1/2 - phi(k)/k is positive and minimal.
We know that prime k (in A000040) have phi(k)/k = A006093(n)/A000040(n) and represent maxima in n. We likewise know primorials k (in A002110) have phi(k)/k = A038110(n)/A060753(n) and represent minima in n. This sequence shows squarefree numbers k with gpf(k) = n such that their value phi(k)/k is closest to but less than 1/2.
Conjecture: for n > 3, k is always odd. This assertion is reliant upon phi(2 prime(n))/2 prime(n) = phi(2)/2 * phi(prime(n))/prime(n) = 1/2 * (prime(n) - 1)/prime(n), and it is clear that 1/2 is an asymptote for even k.
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LINKS
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EXAMPLE
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First terms of this sequence appear in the chart below between asterisks.
The values of n appear in the header, values of k followed parenthetically by phi(k)/k appear in column n. The x axis plots k according to primepi(gpf(k)), while the y axis plots k according to phi(k)/k:
0 1 2 3 4
. . . . .
--- 1 ------------------------------------------------
(1/1) . . . .
. . . . .
. . . . .
. . . . 7
. . . 5 (6/7)
. . . (4/5) .
. . . . .
. . . . 35
. . 3 . (24/35)
. . (2/3) . .
. . . . .
. . . . .
. . . . 21
. . . . (4/7)
. . . 15 .
. . . (8/15) .
. *2* . . .
----------(1/2)---------------------------------------
. . . . .
. . . . *105*
. . . . (16/35)
. . . . 14
. . . *10* (3/7)
. . . (2/5) .
. . . . .
. . . . 70
. . *6* . (12/35)
. . (1/3) . .
. . . . 42
. . . 30 (2/7)
. . . (4/15) .
. . . . 210
. . . . (8/35)
...
a(3) = 10 for the following reasons. There are 4 possible values of k with n = 3. These are 5, 15, 10, and 30 with phi(k)/k = 4/5, 8/15, 2/5, and 4/15, respectively. Subtracting each of the latter values from 1/2, we derive -3/10, -1/30, 1/10, and 7/30 respectively. Since the smallest of these differences is 1/10 pertaining to k = 10, a(3) = 10.
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MATHEMATICA
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With[{e = 20}, Map[MinimalBy[#, If[# > 0, # + 1, Abs@ #] &[#[[2]] - 1/2] &] &, SplitBy[#, Last]] &@ Array[{#2, EulerPhi[#2]/#2, If[! IntegerQ@ #, 0, #] &[1 + Floor@ Log2@ #1]} & @@ {#, Times @@ MapIndexed[Prime[First@ #2]^#1 &, Reverse@ IntegerDigits[#, 2]]} &, 2^e - 1]][[All, 1, 1]]
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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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