A344208 Numbers k such that iterating x -> digsum(x)^2 + 1 from k one or more times results in a number < 10.

1, 2, 3, 6, 9, 10, 11, 12, 15, 18, 19, 20, 21, 24, 27, 28, 30, 33, 36, 37, 39, 42, 45, 46, 48, 51, 54, 55, 57, 60, 63, 64, 66, 69, 72, 73, 75, 78, 81, 82, 84, 87, 90, 91, 93, 96, 99, 100, 101, 102, 105, 108, 109, 110, 111, 114, 117, 118, 120, 123, 126, 127

Offset: 1

Author

Joseph Brown, May 11 2021

Keywords

• nonn
• base

Comments

The number of iterations must be nonzero.

From Michael S. Branicky, May 15 2021: (Start)

f(x) = digsum(x)^2 + 1 < x for x >= 400.

All iterations terminate or lead to the cycle 65 -> 122 -> 26.

There are 5, 47, 395, 3213, 27724, 253490, 2362998, 22649995, 224689951, 2236788357 terms with 1..10 digits, resp. (End)

Examples

			15 is a term because (1+5)^2 + 1 = 37, (3+7)^2 + 1 = 101, (1+0+1)^2 + 1 = 5.
13 is not a term in this sequence because iterating 13 through this function will never yield a single-digit number.  Specifically, 13 -> 17 -> 65 -> 122 -> 26 -> 65 -> ... .
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007953, A344214.

Programs

Python

def f(x): return sum(map(int, str(x)))**2 + 1
def ok(n):
  iter = f(n)  # set to n for number of iterations >= 0
  while iter > 9:
    if iter in {65, 122, 26}: return False
    iter = f(iter)
  return True
print(list(filter(ok, range(1, 128)))) # Michael S. Branicky, May 14 2021