A145679 - OEIS

%I #163 Aug 16 2024 20:56:53

%S 2,1,1,0,1,0,0,1,0,1,0,1,1,0,0,1,0,1,0,0,0,1,1,1,1,0,1,1,1,1,0,0,0,0,

%T 1,0,1,1,0,0,1,1,0,0,1,0,1,1,0,1,0,1,0,1,1,1,1,1,1,0,0,0,0,1,0,1,1,1,

%U 0,0,1,1,0,1,1,0,0,1,1,0,0,0,1,0,0,0,0,0

%N Lower limit of backward value of 2^n and n!.

%C For n!, omitting the trailing sequence zeros. - _Simon Plouffe_, Mar 05 2017

%C The terms are deduced from sequence A023415.

%C The sum of constants in this sequence and A023415 is conjectured to be 11 exactly.

%C From _Cezary Glowacz_, Jul 25 2024: (Start)

%C The backward value of 2^k is formed by writing the decimal digits of 2^k from least to most significant, so that for example 2^2489 = ...610112 becomes 2.11016..., and then the present constant is the infimum of all such backward values (excluding 2^0).

%C The sequence is not eventually periodic. Assuming any period results in a condition a(1)=0 mod 10 which contradicts a(1)=2.

%C The above conjecture about the sum of constants in this sequence and A023415 can be proved using the recursion formulas for the sequences.

%C The equality of the backward sequences for n! omitting the trailing sequence zeros and 2^n holds under the conjecture that for each k and each r not divisible by 5 there are infinitely many n for which n! without trailing zeros = r mod 5^k. Actually this conjecture is true because for m > 0 g(k)^m = ((5^k)(5^(4m) - 1)/(5^4 - 1))! without leading zeros mod 5^k, and g(k) is a generator of the multiplicative group mod 5^k for k=3 mod 4 by induction using g(k+4)^l = g(k)^l <> 1 mod 5^k for l = 5^i or 4.

%C (End)

%C From _David A. Corneth_, Jun 15 2024: (Start)

%C a(1) through a(n) describes the smallest number with n digits (in base 10) not ending in 0 such that the number formed by concatenating the last k digits in reverse is a multiple of 2^k for 1 <= k <= n by choosing digits 0 or 1 for n >= 2.

%C If a(1) were 0, then a(n) would be 0 for all n. (End)

%H Cezary Glowacz, <a href="/A145679/a145679.pdf">Two proofs regarding the n! without leading zeros sequence</a>

%F a(n) >= 0 and is the minimum satisfying (Sum_{i=1..n} a(i)*10^(i-1)) == 0 (mod 2^n), for n >= 2. - _Cezary Glowacz_, Jun 25 2024

%e From _David A. Corneth_, Jun 15 2024: (Start)

%e a(1) = 2. a(2) = 0 or 1. If a(2) = 0 then any number ending in 02 (the backwards concatenation of (2, 0)) should be a multiple of 2^2 = 4 but it is not. Any number ending in 12 (the backwards concatenation of (2, 1)) is a multiple of 2^2 = 4 so a(2) = 1.

%e Similarly a(3) = 1 as 112 is a multiple of 2^3 and 012 is not.

%e a(4) = 0 as 0112 is a multiple of 2^4 and 1112 is not.

%e a(5) = 1 as 10112 is a multiple of 2^5 and 00112 is not. (End)

%o (Python)

%o # lower limit of backward value of 2^n

%o a,i=2,0; x=a

%o while 1:

%o      i+=1; print(x, end=', ')

%o      if a%2**(i+1) == 0: x=0

%o      else: x=1; a+=10**i

%o # _Cezary Glowacz_, Jun 15 2024

%o (PARI) first(n) = {

%o 	my(t = 2);

%o 	for(i = 2, n,

%o 		if(t%2^i != 0,

%o 			t = t + 10^(i-1);

%o 		);

%o 	);

%o 	res = Vecrev(digits(t));

%o 	res = concat(res, vector(n - #res));

%o } \\ _David A. Corneth_, Jun 15 2024

%Y Cf. A000079, A004154, A023415, A158624, A158625.

%K nonn,base

%O 1,1

%A _Simon Plouffe_, Mar 23 2009

%E More terms from _Cezary Glowacz_, Feb 26 2017

%E More terms from _Jinyuan Wang_, Mar 01 2020