A107655 -id:A107655 - cp's OEIS frontend

A033844 a(n) = prime(2^n).

2, 3, 7, 19, 53, 131, 311, 719, 1619, 3671, 8161, 17863, 38873, 84017, 180503, 386093, 821641, 1742537, 3681131, 7754077, 16290047, 34136029, 71378569, 148948139, 310248241, 645155197, 1339484197, 2777105129, 5750079047, 11891268401, 24563311309, 50685770167, 104484802057, 215187847711

Offset: 0

Vasiliy Danilov (danilovv(AT)usa.net), Jun 15 1998

a(n) is the smallest number m such that pi(m)=d(m)^n, where d(m) is number of positive divisors of m (the proof is easy). - Farideh Firoozbakht, Jun 06 2005

David Baugh, Table of n, a(n) for n = 0..78 (terms n = 0..57 from Charles R Greathouse IV, terms n = 58..78 found using Kim Walisch's primecount program).

Cf. A000040, A018249, A051438, A051440, A051439, A006988, A107655.

Mathematica
```
Table[Prime[2^n], {n, 0, 32}]
```

a(n)=prime(2^n) \\ Charles R Greathouse IV, Nov 02 2014

More terms from Robert G. Wilson v, Jun 09 2000

A114063 Numbers k such that phi(k) = tau(k)^4, where tau(k) = A000005(k).

Original entry on oeis.org

1, 17, 514, 8738, 32301, 33003, 36351, 41504, 42292, 43852, 51860, 62226, 549117, 561051, 571311, 599067, 617967, 629811, 634005, 657495, 673184, 674505, 683168, 701024, 705568, 718964, 722684, 732628, 745484, 759772, 774368

Offset: 1

Giovanni Resta, Feb 13 2006

For all large enough k, we have tau(k) < k^(1/5) and phi(k) > k^(4/5). Hence, tau(k)^4 < k^(4/5) < phi(k), implying that this sequence is finite. - Max Alekseyev, Mar 10 2016

Sequence is composed of 94030 terms. - Max Alekseyev, Jun 01 2024

			phi(33003) = 20736. tau(33003) = 12, 20736 = 12^4.
a(2) = A107655(4) = 17.

Max Alekseyev, Table of n, a(n) for n = 1..94030 (first 660 terms from Enrique Pérez Herrero; terms 661..7000 from Giovanni Resta)

Subsequence of A078164.

Cf. A020488, A062516, A063469, A063470, A107655, A112954, A112955, A175667, A068560, A068559.

Select[Range[10^6], EulerPhi[#] == DivisorSigma[0, #]^4 &] (* Paolo Xausa, May 31 2024 *)

isok(n) = eulerphi(n) == numdiv(n)^4; \\ Michel Marcus, Jan 22 2014

A068559 Numbers m such that phi(m) = tau(m)^3.

Original entry on oeis.org

1, 85, 333, 436, 1542, 1875, 2907, 3285, 3488, 3796, 5196, 10280, 17532, 17776, 20080, 21250, 28305, 30368, 30555, 32708, 34748, 35308, 36860, 37060, 41544, 41568, 43068, 44004, 45162, 48468, 51930, 81324, 98304, 98688, 104856, 131070

Offset: 1

Benoit Cloitre, Mar 25 2002

For all large enough k, we have tau(k) < k^(1/4) and phi(k) > k^(3/4). Hence, tau(k)^3 < k^(3/4) < phi(k), implying that this sequence is finite. In fact, the sequence consists of 614 terms. - Max Alekseyev, May 30 2024

			a(2) = A107655(3) = 85.

Max Alekseyev, Table of n, a(n) for n = 1..614 (first 470 terms from Enrique Pérez Herrero)

Subsequence of A039771. - Enrique Pérez Herrero, Aug 29 2010

Cf. A020488, A068560, A107655, A114063.

Select[Range[132000],EulerPhi[#]==DivisorSigma[0,#]^3&]  (* Harvey P. Dale, Dec 28 2022 *)

isok(m) = eulerphi(m) == numdiv(m)^3; \\ Michel Marcus, Oct 18 2019

A068560 Numbers k such that phi(k) = tau(k)^2.

Original entry on oeis.org

1, 5, 34, 63, 76, 128, 136, 170, 315, 364, 380, 444, 640, 680, 972, 1820, 1824, 1836, 2142, 2220, 4788, 4860, 6000, 8064, 8568, 8736, 9120, 9180, 10710, 23940, 40320, 42840, 43680

Offset: 1

Benoit Cloitre, Mar 25 2002

This sequence is finite because phi(k) >= sqrt(k) for all k >= 6, and for any e > 0, tau(k) < k^e for k large enough. Choosing e=1/4 gives tau(k)^2 < sqrt(k) <= phi(k). It remains unknown, however, if this sequence is full. - Nathaniel Johnston, Apr 28 2011

It can be shown that tau(k) <= 120 and the sequence is complete. - Max Alekseyev, May 30 2024

			a(2) = A107655(2) = 5.

Cf. A039770, A068559, A107655, A114063.

Do[If[EulerPhi[n] == DivisorSigma[0, n]^2, Print[n]], {n, 10^5}] (* Ryan Propper, Jun 09 2006 *)
Select[Range[10^5], EulerPhi[#] == DivisorSigma[0, #]^2 &] (* Alonso del Arte, Aug 25 2011 *)

More terms from Ryan Propper, Jun 09 2006

"full" keyword added by Max Alekseyev, May 30 2024

cp's OEIS Frontend

A033844 a(n) = prime(2^n).

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A114063 Numbers k such that phi(k) = tau(k)^4, where tau(k) = A000005(k).

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A068559 Numbers m such that phi(m) = tau(m)^3.

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A068560 Numbers k such that phi(k) = tau(k)^2.

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