A100800 Let f(n) = n + sum of the digits of n. If f(n) is multiple of n then a(n)= f(n) else a(n) = f(f(f(n)))... until one gets a multiple of n; a(n) = 0 if no such number exists.

2, 4, 6, 8, 10, 12, 14, 16, 18, 130, 341, 24, 130, 392, 30, 320, 119, 36, 950, 80, 84, 88, 115, 96, 950, 104, 54, 392, 406, 120, 341, 736, 231, 578, 455, 72, 851, 950, 507, 320, 328, 210, 559, 440, 90, 184, 658, 480, 392, 950, 204, 416, 530, 162, 1430, 2128, 114

Offset: 1

Author

Amarnath Murthy, Dec 17 2004

Keywords

• base
• easy
• nonn

Comments

Conjecture: No term is zero.

Examples

			a(10) = 130, f(10) = 10 + 1 = 11, f(f(10)) = f(11) = 13,... we get the sequence 10,11,13,17,25,32,37,47,58,71,79,95,109,119,130,...
		

Crossrefs

Cf. A100801, A101183.

Extensions

Extended by Ray Chandler, Dec 19 2004