A259497 a(n) = number of steps before A055641(num(i)) = 0, when num(i) = num(i-1) + A055641(num(i-1)) and num(0) = 10^n, where A055641(n) is the number of zero digits in n.

1, 10, 96, 918, 8778, 83970, 803652, 7695702, 73736351, 706940003, 6782083197, 65107856455, 625462980243, 6012764576492, 57843691102715, 556865542063090, 5364870125881211, 51722954280818076, 499024949301954326, 4818086348226292202, 46551954003050282966

Offset: 1

Author

Anthony Sand, Jun 29 2015

Keywords

• nonn
• base

Examples

			For example, when num(0) = 10^1 = 10, A055641(10) = 1. 10 + 1 = 11 and A055641(11) = 0. Therefore a(1) = 1, because the procedure has taken one step before A055641(num(i)) = 0.
a(2) begins with num(0) = 10^2 = 100.
1: 100 + A055641(100) = 100 + 2 = 102.
2: 102 + A055641(102) = 102 + 1 = 103.
.
.
9: 109 + A055641(109) = 109 + 1 = 110.
10: 110 + A055641(110) = 110 + 1 = 111.
At the next step, A055641(111) = 0, so the procedure takes ten steps before A055641(num(i)) = 0 and a(2) = 10.
		

Links

• Hiroaki Yamanouchi, Table of n, a(n) for n = 1..100

Crossrefs

Cf. A055641.

Programs

PARI

{ b=10; digmx=10; for(ni=2,digmx, n=b^(ni-1); s=0; until(z==0, d=digits(n,b); z=sum(i=2,ni,if(d[i]==0,1,0)); n+=z; s++; ); print1(s-1,", "); ); }

Extensions

a(12)-a(21) from Hiroaki Yamanouchi, Aug 02 2015