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User: Joseph Brown

Joseph Brown has authored 2 sequences.

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

Joseph Brown, May 11 2021

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)

			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 -> ... .

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

Cf. A007953, A344214.

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

A344214 Numbers k such that repeated iterations of f(m) = (digsum(f(m-1)))^2 + 1 starting from f(1) = k will eventually yield 5 before any other single-digit number.

Original entry on oeis.org

5, 11, 15, 18, 19, 20, 24, 27, 28, 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, 101, 105, 108, 109, 110, 114, 117, 118, 123, 126, 127, 129, 132, 135, 136, 138, 141, 144, 145, 147, 150, 153, 154

Offset: 1

Joseph Brown, May 11 2021

f(x) = digsum(x)^2 + 1 < x for x >= 400, and all iterations terminate in a single digit or lead to the cycle 65 -> 122 -> 26. - Michael S. Branicky, May 14 2021

			11 is in the list because (1+1)^2 + 1 = 5.
12 is not in the list because repeatedly iterating the function starting with f(1) = 12 will yield 2 before 5.
13 is not in the list because it will never yield 5. Specifically, 13 -> 17 -> 65 -> 122 -> 26 -> 65 -> ... .

Subsequence of A344208.

Cf. A007953, A052216, A052224, A344208, A118881.

Select[Range@100,Last@NestWhileList[Total[IntegerDigits@#]^2+1&,#,#>10&&#!=26&]==5&] (* Giorgos Kalogeropoulos, May 12 2021 *)

def f(n):
    s = 0
    while n > 0:
        s, n = s+n%10, n//10
    return s*s+1
n, pota = 0, 0
while n < 62:
    a, repf, i, ii = pota, 0, 0, 4
    while a > 9 and a != repf:
        a, i = f(a), i+1
        if i == ii:
            repf, ii = a, 2*ii
    if a == 5:
        n = n+1
        print(pota, end = ", ")
    pota = pota+1 # A.H.M. Smeets, May 13 2021

def f(x): return sum(map(int, str(x)))**2 + 1
def ok(n):
  iter = n  # set to f(n) if number of iterations must be >= 1
  while iter > 9:
    if iter in {65, 122, 26}: return False
    iter = f(iter)
  return iter == 5
print(list(filter(ok, range(1, 155)))) # Michael S. Branicky, May 19 2021

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User: Joseph Brown

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

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A344214 Numbers k such that repeated iterations of f(m) = (digsum(f(m-1)))^2 + 1 starting from f(1) = k will eventually yield 5 before any other single-digit number.

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Python

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