Digit pick-up

Consider the multiplication table for the number 125. For each product, we will pick up any digits thereof we have not already collected and add them to a growing list. When will we have collected all ten digits?

 1  125  +1 +2 +5 => {1,2,5}

 2  250        +0 => {0,1,2,5}

 3  375     +3 +7 => {0,1,2,3,5,7}

 4  500

 5  625        +6 => {0,1,2,3,5,6,7}

 6  750

 7  875        +8 => {0,1,2,3,5,6,7,8}

 8 1000

 9 1125

10 1250

11 1375

12 1500

13 1625

14 1750

15 1875

16 2000

17 2125

18 2250

19 2375

20 2500

21 2625

22 2750

23 2875

24 3000

25 3125

26 3250

27 3375

28 3500

29 3625

30 3750

31 3875

32 4000        +4 => {0,1,2,3,4,5,6,7,8}

33 4125

34 4250

35 4375

36 4500

37 4625

38 4750

39 4875

40 5000

41 5125

42 5250

43 5375

44 5500

45 5625

46 5750

47 5875

48 6000

49 6125

50 6250

51 6375

52 6500

53 6625

54 6750

55 6875

56 7000

57 7125

58 7250

59 7375

60 7500

61 7625

62 7750

63 7875

64 8000

65 8125

66 8250

67 8375

68 8500

69 8625

70 8750

71 8875

72 9000        +9 => {0,1,2,3,4,5,6,7,8,9}

So, it takes 72 multiplications before we've bagged all ten digits. In base ten, it turns out that 72 is the largest number of multiplications required for any given starting number.

What is the largest number of multiplications required in base b? Tomas Rokicki has worked them out for thousands of bases: It's the second number on each line in his list. I felt that these maxima (k) deserved to be graphed. Noting that in composite bases k is divisible by (b-1), I decided to do another graph of k/(b-1). Then, to explicitly list the k/(b-1) straight-line points, I did the families, with a question at the end.