A371048 Numbers formed by the rightmost decimal digits of n that are the same as those n^n, where -1 indicates that n <> n^n (mod 10).

0, 1, -1, -1, -1, 5, 6, -1, -1, 9, 0, 11, -1, 3, -1, 5, 16, 7, -1, 9, 0, 21, -1, -1, -1, 25, 6, -1, -1, 9, 0, 31, -1, 3, -1, 5, 36, 7, -1, 9, 0, 41, -1, -1, -1, 5, 6, -1, -1, 49, 0, 51, -1, 3, -1, 5, 56, 57, -1, 9, 0, 61, -1, -1, 5, 6, -1, -1, 9, 0, 71, -1, 3

Offset

0,6

Comments

The common digits might include leading 0's (such as at n = 51 or n = 57) and they are discarded (in particular, a(0) = 0 indicates that the corresponding zero digit term results in a 0 integer entry).

Assuming that c > 0 is an integer not a multiple of 10, then a(c*10^k) = 0 for every positive integer k, since (c*10^k) and (c*10^k)^(c*10^k) have in common only their rightmost k digits.

a(n) is equal to -1 if and only if n == 2,4,8 (mod 10) or n == 3,7 (mod 20).

A082576 is a subsequence of the present one.

Links

Table of n, a(n) for n=0..72.

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.

Formula

If n <> 2,4,8 (mod 10) or n <> 3,7 (mod 20), then a(n) = n (mod 10^k), where k is such that n == n^n (mod 10^k) and n <> n^n (mod 10^(k+1)), whereas a(n) = -1 otherwise.

Example

For n = 51, 51^51 = 1219211305094648479473193481872927834667576992593770717189298225284399541977208231315051 and 51^51 == 5051 (mod 10^4), so there might be three common final digits by including a leading 0 that should instead be disregarded. Consequently, a(51) = 51.

Crossrefs

Cf. A000312, A082576, A317905, A349425, A369624, A369826, A370211.

Sequence in context: A287610 A298171 A046614 * A080130 A188640 A198419

Adjacent sequences: A371045 A371046 A371047 * A371049 A371050 A371051

Keyword

sign,base

Author

Marco Ripà, Mar 10 2024

Status

approved