A100801 -id:A100801 - cp's OEIS frontend

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

Amarnath Murthy, Dec 17 2004

Conjecture: No term is zero.

			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,...

Cf. A100801, A101183.

Extended by Ray Chandler, Dec 19 2004

A101183 Number of times (>0) function f must be applied in A100800 to arrive at multiple of n, or 0 if multiple of n is never reached.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 14, 32, 3, 12, 36, 3, 31, 10, 2, 71, 8, 8, 8, 11, 8, 67, 8, 3, 35, 35, 10, 31, 53, 22, 45, 38, 4, 59, 70, 44, 27, 27, 19, 44, 36, 5, 14, 47, 42, 33, 66, 16, 33, 42, 11, 105, 151, 5, 92, 69, 7, 7, 48, 6, 23, 20, 22, 7, 62, 22, 145, 7, 7, 20, 7, 58, 7, 184, 44

Offset: 1

Ray Chandler, Dec 19 2004

Conjecture: No term is zero.

			a(10) = 14 since f has to be applied 14 times to 10 to arrive at 130, a multiple of 10.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A100800, A100801.

f:= n -> n + convert(convert(n,base,10),`+`):
g:= proc(n) local j,t;
  t:= n;
  for j from 1 do
    t:= f(t);
    if t mod n = 0 then return j fi
  od
end proc:
map(g, [$1..100]); # Robert Israel, May 14 2019

cp's OEIS Frontend

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.

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