cp's OEIS Frontend

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).

%I A378771 #16 Dec 12 2024 18:27:25
%S A378771 10,10,10,11,13,11,13,17,15,16,15,15,16,18,17,15,17,15,13,16,17,15,24,
%T A378771 17,23,16,19,20,20,22,22,25,32,17,20,23,20,19,19,23,19,14,21,19,17,25,
%U A378771 22,17,27,19,24,13,20,28,18,33,26,18,33,22,23,20,25,25,25,22
%N 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).
%C A378771 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.
%C A378771 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.
%H A378771 David A. Corneth, <a href="/A378771/b378771.txt">Table of n, a(n) for n = 1..10000</a>
%e A378771 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.
%o A378771 (PARI) a(n) = {
%o A378771 	if(n == 1, return(10));
%o A378771 	my(i, todo = 10, v = vector(10), d);
%o A378771 	for(i = 2, oo,
%o A378771 		d = digits(i^n);
%o A378771 		if(#Set(d) == 10,
%o A378771 			forstep(i = #d, 1, -1,
%o A378771 				if(v[d[i]+1] == 0,
%o A378771 					v[d[i]+1] = 1;
%o A378771 					todo--;
%o A378771 					if(todo == 0,
%o A378771 						return(#d - i + 1))))));
%o A378771 }
%Y A378771 Cf. A005054, A020666.
%K A378771 nonn,easy,base
%O A378771 1,1
%A A378771 _David A. Corneth_, Dec 06 2024