A371078 Numbers formed by the rightmost decimal digits of n^n that are the same as those n^(n^n), where -1 indicates that n^n <> n^(n^n) (mod 10).

0, 1, -1, 7, 6, 3125, 656, 43, 6, 89, 0, 611, 6, 53, 6, 380859375, 616, 77, -1, 79, 0, 421, -1, 7, 76, 447265625, 203776, 3, 6, 69, 0, 431, 76, 13, 6, 46875, 136, 17, -1, 59, 0, 641, -1, 7, 6, 828125, 696, 3, 6, 449, 0, 15051, 6, 73, 6, 5234375, 456, 688057

Offset: 0

Author

Marco Ripà, Mar 10 2024

Keywords

• sign
• base

Comments

The common digits might include leading 0's (such as at n = 5 or n = 43) and they are discarded (in particular, a(0) = 0 indicates that the corresponding zero digit term results in a 0 integer entry).

For n = k*10^c with c >= 1 and k != 0 (mod 10), a(n) = 0 since n^n has (exactly) c*n rightmost 0's.

a(n) is equal to -1 if and only if n == 2 or 18 (mod 20).

Examples

			For n = 6, 6^6 = 46656 and 6^(6^6) == 8656 (mod 10^4) so there are three common final digits and a(6) = 656.
		

Links

• Jorge Jiménez Urroz and José Luis Andrés Yebra, On the Equation a^x == x (mod b^n), Journal of Integer Sequences, Article 09.8.8, 2009.
• Marco Ripà, Congruence speed of tetration bases ending with 0, arXiv:2402.07929 [math.NT], 2024.

Crossrefs

Cf. A000312, A317905, A349425, A369624, A369826, A370211, A371048, A371074.

Formula

If n <> 2,18 (mod 20), then a(n) = A000312(n) (mod 10^k), where k is such that n^n == n^(n^n) (mod 10^k) and n^n <> n^(n^n) (mod 10^(k+1)), whereas a(n) = -1 otherwise.