BMAD 62 Report post Posted January 28, 2014 A sequence of natural numbers is determined by the following formula, A[n+1] = a[n] + f(n) Where f(n) is the product of digits in a[n]. Is there an a[1] such that the above sequence is unbounded? Share this post Link to post Share on other sites

0 bonanova 77 Report post Posted February 2, 2014 Apparently not. The series clearly ceases to increase as soon as any term contains a zero. If we start with 1, the series goes eleven terms before hitting the term 102, and that is long compared with other small starting numbers. For the first 100 starting numbers the series seems to stop growing at either 102 or 202: 1 2 4 8 16 22 26 38 62 74 102 For starting values less than 1000 the latest first occurrence of zero is the 28^{th} term. The lowest such starting value is 187. There are two others, namely 248 and 264, and the zero-term in each case is 4506: 187 243 267 351 366 474 586 826 922 958 1318 1342 1366 1474 1586 1826 1922 1958 2318 2366 2582 2742 2854 3174 3258 3498 4362 4506 248 312 318 342 366 474 586 826 922 958 1318 1342 1366 1474 1586 1826 1922 1958 2318 2366 2582 2742 2854 3174 3258 3498 4362 4506 264 312 318 342 366 474 586 826 922 958 1318 1342 1366 1474 1586 1826 1922 1958 2318 2366 2582 2742 2854 3174 3258 3498 4362 4506 These show two different paths to the term 366, after which the term 4506 is inevitably reached. Next in length is the starting value of 3237326, which has 29 terms: 3237326 3241862 3244166 3247622 3251654 3255254 3261254 3262694 3278246 3294374 3312518 3313238 3314534 3316694 3328358 3345638 3371558 3384158 3395678 3531758 3544358 3573158 3585758 3753758 3841958 3876518 3916838 3947942 4002374 Next is 3515987, which goes a whopping 33 terms: 3515987 3553787 3641987 3678275 3748835 3829475 3889955 4278755 4357155 4367655 4443255 4452855 4484855 4587255 4643255 4657655 4783655 4884455 4986855 5332455 5341455 5347455 5389455 5497455 5623455 5641455 5653455 5698455 5914455 5932455 5959455 6161955 6170055 No starting point less than 10 million has a longer sequence, although 5112691 also reaches 33 terms before hitting a zero: 5112691 5113231 5113321 5113411 5113471 5113891 5114971 5116231 5116411 5116531 5116981 5119141 5119321 5119591 5121616 5121976 5125756 5136256 5141656 5145256 5151256 5152756 5163256 5168656 5211856 5214256 5216656 5227456 5244256 5253856 5289856 5462656 5505856 There may be a proof that a zero must be reached -- and the OP suggests there is one -- other than the intuitive observation that unless some mechanism actively prevents zeros, and it's hard to imagine such a mechanism exists, eventually the terms have so many digits that the probability that none of them is zero becomes vanishingly small. That is, for the series to be unbounded, the terms will reach one million digits in length. For an endless number of such terms not to contain a zero is unimaginable. Share this post Link to post Share on other sites

0 BMAD 62 Report post Posted February 2, 2014 your logic is sound and the last of your explanation is on the correct track Share this post Link to post Share on other sites

A sequence of natural numbers is determined by the following formula, A[n+1] = a[n] + f(n) Where f(n) is the product of digits in a[n]. Is there an a[1] such that the above sequence is unbounded?

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