A023396 If any odd power of 2 ends with k 1's and 2's, they must be the first k terms of this sequence in reverse order.

2, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1

Offset: 1

Author

David W. Wilson

Keywords

• nonn
• easy
• base

Examples

			2^1 ends in 2;
2^5 ends in 32;
2^9 ends in 512;
2^13 ends in 8192;
2^89 ends in ...562112.
There exists a power of two ending in 12, so for n = 3 the choice for a(3) = 1 or a(3) = 2 comes from the existence of a power of two ending in either 112 or 212. As 112 is divisible by 2^n = 8 (and 212 is not) a(3) = 1. - _David A. Corneth_, Jun 11 2020
		

Links

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

Crossrefs

Cf. A053312, A126933.

Programs

PARI

first(n) = my(f = 2, pow10 = 1, pow2 = 2); { for(i = 2, n, pow10*=10; pow2<<=1; c1 = pow10 + f; if(c1 % pow2 == 0, f = c1, c2 = 2*pow10 + f; if(c2 % pow2 == 0, f = c2 ) ) ); Vecrev(digits(f)) } \\ David A. Corneth, Jun 11 2020

Extensions

Definition corrected by Gerry Leversha, Mar 17 2007