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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) = 10^(n-1), n2(1) = 0
1

%I #22 Jul 22 2017 09:16:16

%S 1,12,57,22820,754504

%N 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) = 10^(n-1), n2(1) = 0

%C The sequence takes two different 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 = 10^(n-1) and n2 = 0.

%C a(5) > 10^12. - _Lars Blomberg_, Jul 19 2017

%F n1(i) = n1(i-1) + digsum(n2(i-1)), n2(i) = n2(i-1) + digsum(n1(i-1))

%e 10 > 0, 10 > 1, 11 > 2, 13 > 4, 17 > 8, 25 > 16, 32 > 23, 37 > 28, 47 > 38, 58 > 49, 71 > 62, 79 > 70, 86 = 86

%o (PARI) {digsum(num) = d=digits(num,b); return(sum(i=1,#d,d[i]));} {doubledigsum() = b=10; nmx=5; for(n=1,nmx, n1=b^(a-1); n2=0; c=0; until(n1==n2, s1=digsum(n1); s2=digsum(n2); n1+=s2; n2+=s1; c++); print1(c,", "); ); }

%Y Cf. A004207, A272235

%K nonn,base,more

%O 0,2

%A _Anthony Sand_, Apr 23 2016