WPCL 2BPZCourier 10cpi#|xx6X@8;X@HP LaserJet Series IIHPLASEII.PRSx  @,t0.jX@#|x2 ZB$XF`HP LaserJet Series IIHPLASEII.PRSx  @,t0.jX@Courier 10cpiLine Printer 16.67cpi?xxx,x6X@8;X@tk+HHH,H H@;@u\jTH\T<u3Th/ B, then gk(n) < n, where B = [(1 + 9logk)/(1 c)] (*) with the brackets representing the greatest integer function and c = (loge)/e. We are interested in the eventual cycle destination of the  ?@ sequence gk(n), gk2(n), gk3(n), ... . Proposition 1 shows that  ? eventually this sequence will reach a point where gki(n) will have no more than B digits, where B is defined in (*). Consequently, we will only need to know the effect of repeated  ?`" iteration of gk on integers with no more than B digits.  ?# COROLLARY 1. To find the cycles of the digit functions gk, it is only necessary to find the cycle destinations of integers n with no more than B digits, where B is defined at (*). In particular, we have: k 1 2 3 4 5 6 7 8 9 10 B 1 4 6 7 8 9 10 10 11 11h)0*0*0*ԌTHE COMPUTER PROGRAM Since the order of the digits of n does not affect the value of  ?X the digit function gk, one only needs to consider unordered sets of digits with repetitions allowed. In an 11 digit search, this reduces the number of cases from 100,000,000,000 to 352,715 a critical savings. The function is iterated on a starting set of digits, extracting a new set of digits at each step and checking to see if one of the known cycle destinations has been reached. After 200 iterations, if no known cycle destination is reached, the current value is printed out for further investigation. If necessary, new cycle destinations are added to the program. PROPOSITION 2. If 1 <= k <= 10, the ultimate destination of any  ?H base10 integer n under repeated iteration of the function gk is one of the cycles represented below. The complete cycle can be  ? obtained by iterating gk on the given cycle representative (rep). Cycle Cycle Cycle Cycle k Length Rep k Length Rep 1 1 1 7 1 13,177,388 7 4 6,002,858 2 2 148 7 44 19,260 2 5 98 2 6 70 8 1 1,033 2 15 5 8 2 41,498 8 3 19,402,896 3 1 12 8 4 299,593 3 2 8,757 8 9 41,561 3 13 54 8 18 66,250 3 13 118 8 22 4,747 3 14 32 8 88 5,696 4 1 4,624 9 1 10 4 1 595,968 9 4 96,788,601 4 4 5,268 9 10 6,669 4 15 36,929 9 13 43,119,713 9 23 185,897 5 1 3,909,511 9 24 597,080 5 2 1,959,655 9 61 605,161 5 3 81,886 9 91 60,185 5 3 1,954,031 5 6 17,031 10 2 21 5 20 81,902 10 6 22 6 2 49,473 6 2 1,967,408 6 14 55,800 6 50 9,338 h)0*0*0*ԌFREQUENCY OF CYCLE OCCURRENCE  During the execution of the computer program, one can keep track of the "frequency of occurrence" of each cycle destination in the following sense. The program is executed on unordered sets of up to B digits (with repeated digits permitted), so it can keep track of how many of these sets reach each cycle destination  ?x under repeated iteration of gk. Some examples: Cycle Cycle k B Length Rep Frequency 4 7 1 4,624 12 1 595,968 2 4 5,268 7,610 15 36,929 11,824 19,448 5 8 1 3,909,511 11 2 1,959,655 704 3 81,886 118 3 1,954,031 23,898 6 17,031 366 20 81,902 18,661 43,758 8 10 1 1,033 1 2 41,498 29 3 19,402,896 54 4 299,593 1,612 9 41,561 186 18 66,250 2,304 22 4,747 1,002 88 5,696 179,568 184,756 OTHER DIGIT FUNCTIONS WHOSE CYCLES HAVE BEEN INVESTIGATED p(n) = the product of the digits of n e.g. p(432) = 4x3x2 = 24  ?(# fk(n) = the sum of the kth powers of the digits of n  ?$ e.g. f3(432) = 43 + 33 + 23 = 99 f(n) = the sum of the factorials of the digits of n e.g. f(432) = 4! + 3! + 2! = 32 h)0*0*0*Ԍg(n) = the sum of the digittodigit powers of n  ? e.g. g(432) = 44 + 33 + 22 = 287 Note: These investigations can be (and some have been) done in other bases using baseb digits of the integer in the various functions and extracting the baseb digits of the result. References: These papers contain other related references: D.C. Morrow, Variations on Perfect Digital Invariants, J. Recreational Mathematics, 27:1, pp. 912, 1995. D.C. Morrow, Recurring DigittoDigit Invariants, J. Recreational Mathematics, 27:2, pp. 154156, 1995.