Problems & Puzzles: Puzzles

Puzzle 328. Is this the largest? It has been told that 12157692622039623539 is the largest number n such that n equals the sum of its digits raised to the consecutive powers 1, 2, 3, ... 12157692622039623539 = 1^1 + 2^2 + ... + 9^20 See the following 'fini & full' sequence A032799 Questions: 1. Demonstrate the above claim or get a larger example. 2. Suppose you don't know the largest known term of this sequence. Explain a method to get it.

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Contribution came from Faride Firoozbakht:

Faride wrote:

The sequence A032799 (0,1,2,3,4,5,6,7,8,9,89,135,175,518,
598,1306,1676,2427,2646798,12157692622039623539) is finite.

Proof: If m is a n-digit term of this sequence then

10^(n-1) <= m <= 9^1+9^2+...+9^n = 9/8*(9^n-1) < 9/8*9^n

so (10/9)^n < 45/4 and n <= [log(45/4)/log(10/9)]=22.

Hence this sequence is finite and the largest term is less than 10^22.

Surely Macsy Zhang who has found the nice and large 20-digit number 12157692622039623539, has searched until the bound 10^22 and he hasn't found further terms..

I was curious about the way of finding this large number and I wanted to contact with Macsy Zhang one year ago but I didn't get any reply. I also contacted with Patrick De Geest about it that time, but he didn't know anything more.
 

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