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 1-4 of 4 results.

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

Views

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.

Links

Crossrefs

Subsequence of A078164.
Cf. A020488, A062516, A063469, A063470, A107655, A112954, A112955, A175667, A068560, A068559.

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

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

Views

Author

Benoit Cloitre, Mar 25 2002

Keywords

Comments

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

Examples

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

Crossrefs

Cf. A039770, A068559, A107655, A114063.

Programs

  • Mathematica
    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 *)

Extensions

More terms from Ryan Propper, Jun 09 2006
"full" keyword added by Max Alekseyev, May 30 2024

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

Views

Author

Franklin T. Adams-Watters, Jun 30 2017

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

Links

Crossrefs

Cf. A000005, A000010, A019434, A020488, A032447, A036913, A051281, A068559, A068560, A114063, A286627.

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

A087334 a(1) = 3, then smallest number such that every partial product + 1 is a distinct square.

Original entry on oeis.org

3, 5, 8, 3, 14, 40, 1270, 15874, 251984126, 28673777890680889, 2806729404119595479093401735, 15973219322678152520589944038429546981629762353084607
Offset: 1

Views

Author

Amarnath Murthy, Sep 06 2003, May 22 2004

Keywords

Examples

			3 + 1 = 4, 3*5 + 1 = 16, 3*5*8 + 1 = 121 etc. are squares.

Crossrefs

Cf. A068559, A087335.

Programs

  • Mathematica
    p = 1; Do[k = 2; While[ !IntegerQ[Sqrt[p*k + 1]], k++ ]; Print[k]; p *= k, {n, 1, 9}] (* Ryan Propper *)

Extensions

Edited by Don Reble, Sep 12 2003
Entry revised by N. J. A. Sloane, Aug 29 2006
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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