[seqfan] Re: Last term?

Hans Havermann gladhobo at teksavvy.com
Mon Oct 26 20:12:54 CET 2015


No easier than Eric's forward sequence. The first 79 terms are: 999999999, 999999998, 999999997, 999999996, 999999995, 999999994, 999999993, 999999992, 999999991, 999999989, 999999988, 999999987, 999999986, 999999985, 999999984, 999999983, 999999982, 99999999, 999999981, 999999979, 999999978, 999999977, 999999976, 999999975, 999999974, 999999973, 9999999, 999999972, 999999971, 999999969, 999999968, 999999967, 999999966, 999999965, 999999964, 999999, 999999963, 999999962, 999999961, 999999959, 999999958, 999999957, 999999956, 999999955, 99999, 999999954, 999999953, 999999952, 999999951, 999999949, 999999948, 999999947, 999999946, 9999, 999999945, 999999944, 999999943, 999999942, 999999941, 999999939, 999999938, 999999937, 999, 999999936, 999999935, 999999934, 999999933, 999999932, 999999931, 999999929, 999999928, 99, 999999927, 999999926, 999999925, 999999924, 999999923, 999999922, 999999921. So next is a nine-digit number followed by a one-digit number. We've only used eight 1s thus far, so there's still one 1 left to be used. So term #80 is 999999919. Well, no! The eight 1s we've already used will be assigned to the numbers 9 to 2. Using a ninth 1 will assign that 1 to the number 1. But we can't have any more 1s! So term #80 will be 999999899 (followed by 9). That kind of reasoning gets even more slippery when one tackles the end of two-digit numbers and the end of three-digit numbers. And it gets impossible when one contemplates the end of nine-digit numbers: The total number of digits is greater than the total number of terms (regardless of where one stops) and there will (therefore) be predicted terms long after we have run out of terms.


> On Oct 26, 2015, at 2:48 AM, Andrew Weimholt <andrew.weimholt at gmail.com> wrote:
> 
> 999999999, 999999998, 999999997, 999999996, 999999995,
> 999999994, 999999993, 999999992, 999999991, 999999989,
> 999999988, 999999987, 999999986, 999999985, 999999984,
> 999999983, 999999982, 99999999, 999999981, ...
> 
> Perhaps someone would like to figure out when this one ends?



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