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 1-3 of 3 results.

A338189 Integers m such that A337454(m) = 1.

Original entry on oeis.org

1, 77, 161, 209, 231, 253, 329, 377, 385, 413, 437, 473, 483, 497, 517, 539, 581, 589, 627, 649, 689, 713, 737, 749, 759, 805, 817, 869, 893, 913, 917, 987, 989, 1045, 1073, 1081, 1121, 1127, 1133, 1155, 1169, 1177, 1197, 1239, 1253, 1265, 1273, 1309, 1311, 1313, 1333
Offset: 1

Views

Author

Michel Marcus, Oct 15 2020

Keywords

Crossrefs

Programs

  • PARI
    f(n) = sumdiv(n, d, n % (sum(m=0, n-1, Mod(m, n)^d == m)/sum(m=0, n-1, -Mod(m, n)^d == m)) == 0); \\ A337454
    isok(n) = f(n) == 1;

A338190 Integers m such that A337454(m) = tau(m), the number of divisors of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 22, 23, 25, 26, 27, 29, 30, 31, 32, 34, 36, 37, 38, 39, 41, 43, 45, 46, 47, 49, 50, 51, 53, 54, 58, 59, 61, 62, 64, 67, 71, 73, 74, 75, 78, 79, 81, 82, 83, 85, 86, 87, 89, 90, 91, 94, 97, 98, 99, 101, 102
Offset: 1

Views

Author

Michel Marcus, Oct 15 2020

Keywords

Crossrefs

Programs

  • PARI
    f(n) = sumdiv(n, d, n % (sum(m=0, n-1, Mod(m, n)^d == m)/sum(m=0, n-1, -Mod(m, n)^d == m)) == 0); \\ A337454
    isok(n) = f(n) == numdiv(n);

A334006 Triangle read by rows: T(n,k) = (the number of nonnegative bases m < n such that m^k == m (mod n))/(the number of nonnegative bases m < n such that -m^k == m (mod n)) for nonnegative k < n, n >= 1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 5, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 7, 1, 3, 1, 3, 1, 1, 4, 1, 5, 1, 5, 1, 5, 1, 9, 1, 3, 1, 3, 1, 7, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 5, 1, 11, 1, 3, 1, 3, 1, 3, 1, 3, 1, 1, 6, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 13, 1, 1, 1, 5, 1, 1, 1, 5, 1, 1, 1, 1, 7, 1, 3, 1, 3
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Apr 12 2020

Keywords

Comments

If the sum of proper divisors of q in row q <= q, then q are 1, 2, 3, 4, 5, 8, 16, 17, 32, 64, 128, 256, 257, ...(union of Fermat primes and powers of 2).

Examples

			Triangle T(n,k) begins:
  n\k| 0   1  2  3  4   5  6  7  8   9 10 11 12  13 14 15 16
  ---+------------------------------------------------------
   1 | 1;
   2 | 1,  1;
   3 | 1,  3, 1;
   4 | 1,  2, 1, 3;
   5 | 1,  5, 1, 1, 1;
   6 | 1,  3, 1, 3, 1,  3;
   7 | 1,  7, 1, 3, 1,  3, 1;
   8 | 1,  4, 1, 5, 1,  5, 1, 5;
   9 | 1,  9, 1, 3, 1,  3, 1, 7, 1;
  10 | 1,  5, 1, 1, 1,  5, 1, 1, 1,  5;
  11 | 1, 11, 1, 3, 1,  3, 1, 3, 1,  3, 1;
  12 | 1,  6, 1, 9, 1,  9, 1, 9, 1,  9, 1, 9;
  13 | 1, 13, 1, 1, 1,  5, 1, 1, 1,  5, 1, 1, 1;
  14 | 1,  7, 1, 3, 1,  3, 1, 7, 1,  3, 1, 3, 1, 7;
  15 | 1, 15, 1, 3, 1, 15, 1, 3, 1, 15, 1, 3, 1, 15, 1;
  16 | 1,  8, 1, 5, 1,  9, 1, 5, 1,  9, 1, 5, 1,  9, 1, 5;
  17 | 1, 17, 1, 1, 1,  1, 1, 1, 1,  1, 1, 1, 1,  1, 1, 1, 1;
  ...
For (n, k) = (7, 3), there are three nonnegative values of m < n such that m^3 == m (mod 7) (namely 0, 1, and 6) and one nonnegative value of m < n such that -m^3 == m (mod 7) (namely 0), so T(7,3) = 3/1 = 3.
		

Crossrefs

Programs

  • Magma
    [[#[m: m in [0..n-1] | m^k mod n eq m]/#[m: m in [0..n-1] | -m^k mod n eq m]: k in [0..n-1]]: n in [1..17]];
    
  • PARI
    T(n, k) = sum(m=0, n-1, Mod(m, n)^k == m)/sum(m=0, n-1, -Mod(m, n)^k == m);
    matrix(7, 7, n, k, k--; if (k>=n, 0, T(n,k))) \\ to see the triangle \\ Michel Marcus, Apr 17 2020

Extensions

Name corrected by Peter Kagey, Sep 12 2020
Showing 1-3 of 3 results.