A349744 a(n) is the number of distinct numbers of steps required for the last n digits of integers to repeat themselves by iterating the map m -> m^3.
2, 5, 7, 12, 15, 18, 29, 42, 57, 87, 108
Offset: 1
Examples
a(1) = 2. The paths of the last digit of integers resulted from iterating the map, m -> m^3, are: 0->0; 1->1; 2->8->2; 3->7->3; 4->4; 5->5; 6->6; 7->3->7; 8->2->8; 9->9. Integers ending with 0, 1, 4, 5, 6 or 9 take 1 step to repeat the last digit. Integers ending with 2, 3, 7 or 8 takes 2 steps to repeat the last digit. Therefore, for n = 1, the distinct numbers of steps s(1) = {1, 2} and a(1) = 2.
a(2) = 5 because the distinct steps for the last two digits of integers to repeat themselves by iterating the map, m -> m^3, is s(2) = {1, 2, 3, 4, 5}.
a(3) = 7: s(3) = s(2) + {20, 21}.
a(4) = 12: s(4) = s(3) + {6, 22, 100..102}.
a(5) = 15: s(5) = s(4) + {500..502}.
a(6) = 18: s(6) = s(5) + {2500..2502}.
a(7) = 29: s(7) = s(6) + {8..10, 40, 200, 1000, 5000, 12500..12502, 25000}
a(8) = 42: s(8) = s(7) + {16..18, 80, 400, 2000, 10000, 50000, 62500..62502, 125000, 250000}.
a(9) = 57: s(9) = s(8) + {32..34, 160, 800, 4000, 20000, 100000, 312500..312502, 500000, 625000, 1250000, 2500000}.
a(10)= 87: s(10)= s(9) + {7, 11, 19, 23, 35, 64..67, 103, 320, 503, 1600, 2503, 8000, 12503, 40000, 62503, 200000, 312503, 1000000, 1562500..1562503, 3125000, 5000000, 6250000, 12500000, 25000000}.
a(11)=108: s(11)=s(10) + {128..131, 640, 3200, 16000, 80000, 400000, 2000000, 7812500..7812503, 10000000, 15625000, 31250000, 50000000, 62500000, 125000000, 250000000}.
Programs
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Python
for n in range(1, 12): b = 10**n; M = set() for i in range(b): t = i; L = set() while t not in L: L.add(t); t = (t**3)%b d = len(L) if d not in M: M.add(d) print(len(M), end = ', ')