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

1,1

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.

Links

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

Example

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.

Prog

(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))))));
}

Crossrefs

Cf. A005054, A020666.

Sequence in context: A166710 A211872 A178166 * A003855 A245431 A377216

Adjacent sequences: A378763 A378767 A378768 * A378772 A378773 A378777

Keyword

nonn,easy,base,new

Author

David A. Corneth, Dec 06 2024

Status

approved