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.
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
Examples
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 -> ... .
Programs
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Mathematica
Select[Range@100,Last@NestWhileList[Total[IntegerDigits@#]^2+1&,#,#>10&&#!=26&]==5&] (* Giorgos Kalogeropoulos, May 12 2021 *)
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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 = ", ") pota = pota+1 # A.H.M. Smeets, May 13 2021 -
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 print(list(filter(ok, range(1, 155)))) # Michael S. Branicky, May 19 2021
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