A351877 Nonnegative integers whose trajectory under iteration of taking the absolute value of the alternating sum of the cubes of the digits includes zero.

0, 11, 22, 33, 44, 55, 58, 66, 77, 85, 88, 99, 110, 135, 138, 142, 179, 220, 232, 241, 256, 267, 284, 328, 330, 345, 346, 387, 396, 429, 440, 464, 482, 486, 531, 543, 550, 580, 587, 643, 652, 660, 684, 693, 762, 770, 783, 785, 808, 823, 831, 849, 850, 868, 880, 924, 948, 971, 990

Offset: 1

Author

Luca Onnis, Feb 23 2022

Keywords

• nonn
• base

Comments

The sequence is infinite. Any number which is formed by concatenating two-digit multiples of 11 is a term.

To determine whether a given number k is a term of this sequence, start with k, take the cube of each digit of k, sum them together with alternating signs and take the absolute value of the result, apply the same process to the result, and continue until 0 is reached or a loop is entered. If 0 is reached, k is a term of this sequence. If not, k is not a term of this sequence.

Examples

			346 is a term of the sequence since: 346->179->387->142->55->0.
8 is not a term since: 8->512->132->18->511->125->118->512 (we reached a loop of length 6 starting with 512).
		

Crossrefs

Cf. A345680, A351985.

Programs

Mathematica

Select[Range[10000],FixedPoint[Abs[Sum[(-1)^(n + 1)*Part[IntegerDigits[#]^3, n], {n, 1,Length[IntegerDigits[#]]}]] &, #, 30] == 0 &]

PARI

f(n) = my(d=digits(n)); abs(sum(k=1, #d, (-1)^k*d[k]^3)); \\ A351985
already(m, v) = {for (i=1, #v, if (v[i] == m, return (1)););}
isok(m) = {my(v=[]); while (m=f(m), if (already(m, v), return(0)); v = concat(v, m);); return(1);} \\ Michel Marcus, Feb 27 2022

Python

def happyish_function(number, base: int = 10):
  total = 0
  times = 0
  while number > 0:
    total += pow(-1, times) * pow(abs(number) % base, 3)
    number = abs(number) // base
    times += 1
  return abs(total)
def is_happyish(number: int) -> bool:
  seen_numbers = set()
  while number > 0 and number not in seen_numbers:
    seen_numbers.add(number)
    number = happyish_function(number)
  return number == 0
print([k for k in range(1000) if is_happyish(k)])