A272235 In base 2, number of steps before n1(i) = n2(i) when n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1)) and n1(1) = 2^(n-1), n2(1) = 0.

1, 3, 5, 8, 1204, 1205, 1199, 1191, 19536395, 19536233, 19535912, 19535673, 19519159

Offset: 0

Author

Anthony Sand, Apr 23 2016

Keywords

• nonn
• base
• more

Comments

The sequence takes two different binary numbers, n1 and n2, and simultaneously adds the digit sum of n1 to n2 and the digit sum of n2 to n1. This process continues until n1 = n2. The two numbers are initialized with n1 = 2^(n-1) and n2 = 0.

Examples

			In base 2: 1000 > 0, 1000 > 1, 1001 > 10, 1010 > 100, 1011 > 110, 11111 > 1100, 10001 > 10000, 10010 = 10010
In base 10: 8 > 0, 8 > 1, 9 > 2, 10 > 4, 11 > 6, 13 > 9, 15 > 12, 17 > 16, 18 = 18
		

Crossrefs

Cf. A004207, A272233.

Programs

PARI

digsum(num) = d=digits(num,2); return(sum(i=1,#d,d[i]));
doubledigsum() = b=2; nnx=5; for(n=1,amx, n1=b^(n-1); n2=0; c=0; until(n1==n2, s1=digsum(n1); s2=digsum(n2); n1+=s2; n2+=s1; c++); print1(c,", "); );

Formula

n1(i) = n1(i-1) + digsum(n2(i-1),base=2), n2(i) = n2(i-1) + digsum(n1(i-1),base=2)