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.

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A065389 Numbers m such that sigma(phi(m)) sets a new record, i.e., sigma(phi(m)) > sigma(phi(k)) for all k < m numbers.

Original entry on oeis.org

1, 3, 5, 7, 11, 13, 17, 19, 25, 29, 31, 37, 43, 53, 61, 73, 97, 109, 127, 143, 151, 157, 181, 211, 241, 313, 331, 337, 397, 403, 421, 527, 541, 601, 631, 661, 757, 779, 899, 1009, 1147, 1201, 1321, 1333, 1517, 1621, 1763, 1801, 2017, 2161, 2341, 2501, 2521
Offset: 1

Views

Author

Labos Elemer, Nov 05 2001

Keywords

Comments

Numbers m at which A000203(A000010(m)) = A062402(m) reaches a new maximal value.

Examples

			A062402 begins {1, 1, 3, 3, 7, 3, 12, 7, 12, 7, 18, 7, 28, 12, 15, 15, 31, 12, 39, 15, 28, 18, 36, 15, 42}. New peaks are reached at positions 3, 5, 7, 11, 13, 17, 19, 25. These peak values are 3, 7, 12, 18, 28, 31, 39, 42, respectively.
		

Crossrefs

Programs

  • Mathematica
    a=0; s=0; Do[s=DivisorSigma[1, EulerPhi[n]]; If[s>a, a=s; Print[n]], {n, 1, 10000}];
    DeleteDuplicates[Table[{n,DivisorSigma[1,EulerPhi[n]]},{n,2600}],GreaterEqual[#1[[2]],#2[[2]]]&][[;;,1]] (* Harvey P. Dale, Jul 20 2023 *)
  • PARI
    { n=r=0; for (m=1, 10^9, x=sigma(eulerphi(m)); if (x > r, r=x; write("b065389.txt", n++, " ", m); if (n==500, return)) ) } \\ Harry J. Smith, Oct 18 2009

A065293 Number of values of s, 0 <= s <= n-1, such that 2^s mod n = s.

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 2, 0, 0, 1, 0, 2, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 2, 1, 1, 1, 0, 0, 3, 0, 0, 1, 1, 2, 0, 1, 2, 1, 0, 2, 0, 2, 0, 1, 1, 1, 1, 0, 2, 1, 0, 0, 0, 2, 1, 1, 1, 1, 1, 0, 1, 2, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 0, 2, 2, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1
Offset: 1

Views

Author

Jonathan Ayres (jonathan.ayres(AT)btinternet.com), Oct 28 2001

Keywords

Examples

			For n=5 we have (2^0) mod 5 = 1, (2^1) mod 5 = 2, (2^2) mod 5 = 4, (2^3) mod 5 = 3, (2^4) mod 5 = 1. Only for s=3 does (2^s) mod 5=s, so a(5)=1
		

Crossrefs

Cf. A065294.

Programs

Extensions

a(1) corrected by Michel Marcus, Jun 20 2018
Showing 1-2 of 2 results.