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 1-2 of 2 results.

A373205 Numbers m such that m^m == m (mod 10^(len(m) + 1)), where len(m) is the number of digits of m (A055642).

Original entry on oeis.org

1, 51, 57, 101, 151, 176, 201, 301, 351, 401, 501, 551, 576, 601, 625, 701, 751, 801, 901, 951, 976, 1001, 1376, 2001, 2057, 2751, 3001, 4001, 4193, 4751, 5001, 5376, 6001, 6249, 6751, 7001, 8001, 8751, 9001, 9375, 9376, 10001, 10751, 11001, 12001, 13001
Offset: 1

Views

Author

Marco Ripà, May 27 2024

Keywords

Comments

By definition, the present sequence is a subsequence of A082576.
For each integer r >= 2 this sequence contains 10^r + 1.
All terms > 1 end in 01, 25, 49, 51, 57, 75, 76, or 93.

Examples

			51 is a term since 51 is a 2-digit number and 51^51 == 5051 (mod 10^4) and thus 51^51 == 51 (mod 10^(2 + 1)).
		

Crossrefs

Programs

  • PARI
    for (len_m = 1, 5, for (m = 10^(len_m - 1), 10^len_m - 1, if (m == Mod(m, 10^(len_m + 1))^m, print1(m, ", "))))
    
  • Python
    from itertools import count
    def A373205_gen(): # generator of terms
        for i in count(0,100):
            for j in (1, 25, 49, 51, 57, 75, 76, 93):
                m = i+j
                if pow(m,m,10*10**(len(str(m)))) == m:
                    yield m
    A373205_list = list(islice(A373205_gen(),20)) # Chai Wah Wu, Jun 02 2024

A373386 Smallest integer m > 1 such that m == m^m (mod 10^(len(m) + n)), where len(m) is the number of digits of m.

Original entry on oeis.org

5, 51, 751, 10001, 100001, 1000001, 10000001, 100000001, 1000000001
Offset: 0

Views

Author

Marco Ripà, Jun 02 2024

Keywords

Comments

By definition, this sequence is a subsequence of A082576.
It is not known if a(n) = 10^(n + 1) + 1 holds for all n >= 3.

Examples

			a(2) = 751 since m = 751 is the smallest integer satisfying m == m^m (mod 10^(len(m) + 2)), given the fact that 751 is a 3-digit number and 751^751 == 500751 (mod 10^6) and thus 751^751 == 751 (mod 10^(3 + 2)).
		

Crossrefs

Programs

  • PARI
    a(n) = my(im); for (len_m = 1, oo, if (len_m==1, im=2, im=10^(len_m - 1)); for (m = im, 10^len_m - 1, if (m == Mod(m, 10^(len_m + n))^m, return(m)))); \\ Michel Marcus, Jun 03 2024

Extensions

a(7)-a(8) from Michel Marcus, Jun 03 2024
Showing 1-2 of 2 results.