cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A039770 Numbers k such that phi(k) is a perfect square.

Original entry on oeis.org

1, 2, 5, 8, 10, 12, 17, 32, 34, 37, 40, 48, 57, 60, 63, 74, 76, 85, 101, 108, 114, 125, 126, 128, 136, 160, 170, 185, 192, 197, 202, 204, 219, 240, 250, 257, 273, 285, 292, 296, 304, 315, 364, 370, 380, 394, 401, 432, 438, 444, 451, 456, 468, 489, 504, 505
Offset: 1

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Author

Keywords

Comments

A004171 is a subsequence because phi(2^(2k+1)) = (2^k)^2. - Enrique Pérez Herrero, Aug 25 2011
Subsequence of primes is A002496 since in this case phi(k^2+1) = k^2. - Bernard Schott, Mar 06 2023
Products of distinct terms of A002496 form a subsequence. - Chai Wah Wu, Aug 22 2025

Examples

			phi(34) = 16 = 4*4.
		

References

  • D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc., Boston MA, 1976, p. 141.

Crossrefs

Cf. A000010, A007614. A062732 gives the squares. A306882 (squares not totient).

Programs

  • Maple
    with(numtheory); isA039770 := proc (n) return issqr(phi(n)) end proc; seq(`if`(isA039770(n), n, NULL), n = 1 .. 505); # Nathaniel Johnston, Oct 09 2013
  • Mathematica
    Select[ Range[ 600 ], IntegerQ[ Sqrt[ EulerPhi[ # ] ] ]& ]
  • PARI
    for(n=1, 120, if (issquare(eulerphi(n)), print1(n, ", ")))
    
  • Python
    from math import isqrt
    from sympy import totient as phi
    def ok(n): return isqrt(p:=phi(n))**2 == p
    print([k for k in range(1, 506) if ok(k)]) # Michael S. Branicky, Aug 17 2025

Formula

a(n) seems to be asymptotic to c*n^(3/2) with 1 < c < 1.3. - Benoit Cloitre, Sep 08 2002
Banks, Friedlander, Pomerance, and Shparlinski show that a(n) = O(n^1.421). - Charles R Greathouse IV, Aug 24 2009

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

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Author

Giovanni Resta, Feb 13 2006

Keywords

Comments

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

Examples

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

Crossrefs

Programs

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

Views

Author

Benoit Cloitre, Mar 25 2002

Keywords

Comments

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

Examples

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

Crossrefs

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

Programs

  • Mathematica
    Select[Range[132000],EulerPhi[#]==DivisorSigma[0,#]^3&]  (* Harvey P. Dale, Dec 28 2022 *)
  • PARI
    isok(m) = eulerphi(m) == numdiv(m)^3; \\ Michel Marcus, Oct 18 2019

A289276 Numbers k such that phi(k) (the totient function A000010) is a power of the number of divisors of k (A000005).

Original entry on oeis.org

1, 2, 3, 5, 8, 10, 17, 18, 24, 30, 34, 63, 76, 85, 128, 136, 170, 257, 315, 333, 364, 380, 436, 444, 514, 640, 680, 972, 1285, 1542, 1820, 1824, 1836, 1875, 2142, 2220, 2907, 3285, 3488, 3796, 4369, 4788, 4860
Offset: 1

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Author

Keywords

Comments

A019434 is a subsequence. - David A. Corneth, Jun 30 2017
Is the frequency of e such that A000005(a(n))^e = A000010(a(n)) finite? - David A. Corneth, Jul 01 2017

Crossrefs

Programs

  • Mathematica
    Join[{1},Select[Range[2,5000],IntegerQ[Log[DivisorSigma[0,#],EulerPhi[#]]]&]] (* Harvey P. Dale, Aug 06 2017 *)
  • PARI
    ispowerof(n, k)= if(k==1, return(n==1)); while(n>=k, if(n%k!=0, return(0)); n\=k); n==1
    isa(n) = ispowerof(eulerphi(n),numdiv(n)) \\ Quick program, fast enough for early values.
    
  • PARI
    is(n) = if(n==1, return(1)); my(f = factor(n); phi = eulerphi(f), ndiv = numdiv(f), e = logint(phi, ndiv)); ndiv^e == phi \\ David A. Corneth, Jun 30 2017, changed per suggestion of Charles R Greathouse IV
    
  • PARI
    isA289276(n)= if(n==1, return(1)); my(phi = eulerphi(n), ndiv = numdiv(n), v = valuation(phi, ndiv)); ndiv^v == phi; \\ (A variant of above program). - Antti Karttunen, Jun 30 2017
    
  • PARI
    list(lim)=my(v=List([1])); forfactored(n=2,lim\1, my(phi = eulerphi(n), ndiv = numdiv(n)); if(ndiv^valuation(phi,ndiv) == phi, listput(v,n[1]))); Vec(v) \\ Charles R Greathouse IV, Jul 01 2017

A262406 Squarefree k such that phi(k) is a perfect square.

Original entry on oeis.org

1, 2, 5, 10, 17, 34, 37, 57, 74, 85, 101, 114, 170, 185, 197, 202, 219, 257, 273, 285, 370, 394, 401, 438, 451, 489, 505, 514, 546, 570, 577, 629, 677, 679, 802, 902, 969, 978, 985, 1010, 1057, 1095, 1154, 1258, 1285, 1297, 1354, 1358, 1365
Offset: 1

Views

Author

Keywords

Comments

The subsequence of primes is A002496 (primes of the form k^2+1). - Michel Marcus, Oct 14 2015

Crossrefs

Intersection of A039770 and A005117.

Programs

  • Magma
    [n: n in [1..1400] | IsSquarefree(n) and IsSquare(EulerPhi(n))]; // Vincenzo Librandi, May 05 2016
  • Mathematica
    Select[Range[1500], SquareFreeQ[#] && IntegerQ @ Sqrt @ EulerPhi[#] &] (* Amiram Eldar, Jul 16 2022 *)
  • PARI
    is(n)=my(f=factor(n)); issquare(eulerphi(f)) && (n==1 || vecmax(f[,2])==1)
    

Formula

Banks, Friedlander, Pomerance, and Shparlinski show that a(n) = O(n^1.421).
Showing 1-5 of 5 results.