A378771 a(n) is the least k such that the last k digits of m = A020666(n)^n contain all 10 possible digits (0 through 9).

10, 10, 10, 11, 13, 11, 13, 17, 15, 16, 15, 15, 16, 18, 17, 15, 17, 15, 13, 16, 17, 15, 24, 17, 23, 16, 19, 20, 20, 22, 22, 25, 32, 17, 20, 23, 20, 19, 19, 23, 19, 14, 21, 19, 17, 25, 22, 17, 27, 19, 24, 13, 20, 28, 18, 33, 26, 18, 33, 22, 23, 20, 25, 25, 25, 22

Offset: 1

Author

David A. Corneth, Dec 06 2024

Keywords

• nonn
• easy
• base

Comments

A conjecture over at A020666 says that A020666(n) = 2 for n >= 189. Perhaps looking at last digits and some modular arithmetic leads to a proof.

A020666(189) = 2 and a(189) = 33 so the last 33 digits of 2^189 contain all 10 digits. Therefore for k = 189 + 4*5^(33-1) the last 33 digits of 2^k contain all 10 possible digits.

Examples

			a(4) = 11 since m = A020666(4)^4 = 763^4 = 338920744561. The last 11 digits contain all 10 possible digits (0 through 9) but the last 10 digits do not; 3 is missing in the last 10 digits.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005054, A020666.

Programs

PARI

a(n) = {
	if(n == 1, return(10));
	my(i, todo = 10, v = vector(10), d);
	for(i = 2, oo,
		d = digits(i^n);
		if(#Set(d) == 10,
			forstep(i = #d, 1, -1,
				if(v[d[i]+1] == 0,
					v[d[i]+1] = 1;
					todo--;
					if(todo == 0,
						return(#d - i + 1))))));
}