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A344214
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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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1
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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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 -> ... .
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MATHEMATICA
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Select[Range@100, Last@NestWhileList[Total[IntegerDigits@#]^2+1&, #, #>10&&#!=26&]==5&] (* Giorgos Kalogeropoulos, May 12 2021 *)
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PROG
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(Python)
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 = ", ")
(Python)
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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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