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%I #102 Aug 15 2021 13:50:10
%S 153,135,18,3,9,12,33,114,78,126,6,117,669,177,12558,
%T 44499999999999999999
%N Smallest number which reaches the narcissistic number 153 after n steps when repeatedly summing the cubes of its digits.
%C All the terms a(n) as well as the intermediate results will be multiples of 3:
%C x^3 mod 3 = x mod 3 [0^3 = 0; 1^3 = 1; (-1)^3 = -1].
%C Therefore (sum of cubes of digits) mod 3 = (sum of digits) mod 3.
%C Because the only multiple of 3 in A046197 is 153, every number which is a multiple of 3 will end up at 153.
%C Some other terms (not dealt with here) may reach a cycle of length > 1:
%C Elizabeth Todd has shown that only numbers (1 mod 3) and (2 mod 3) may reach a cycle, and the only possible cycles are {55, 230, 130}, {136, 244}, {160, 217, 352}, {919, 1459}. That means that numbers (0 mod 3) never reach a cycle but just a single number, namely 153.
%C Shyam Sunder Gupta tested all the multiples of 3 less than 10^5. He found that they all reach 153, in accordance with the above statements.
%C The values a(n) for n>15 are really too big to be fully written out (and so are missing in the list), as Jon E. Schoenfield calculated for n=16 and n=17:
%C a(16) = 3.777999...999*10^61042524005486970; it has one 3, three 7's, and 61042524005486967 9's, so the sum of the cubes of its digits is 1*3^3 + 3*7^3 + 61042524005486967*9^3 = 44499999999999999999 = a(15).
%C a(17) consists of the digit string 45888 followed by a very, very long string of 9's. The number of 9's in that string is (a(16) - 1725)/729, which is a 61042524005486968-digit number consisting of the digit 5 followed by 753611407475147 copies of the 81-digit string 182441700960219478737997256515775034293552812071330589849108367626886145404663923 followed by a single instance of the 60-digit string 182441700960219478737997256515775034293552812071330589849106.
%H <a href="/A346630/b346630.txt">Table of n, a(n) for n = 0..15</a>.
%H Shyam Sunder Gupta, <a href="http://www.shyamsundergupta.com/c153.htm">Curious Properties of 153</a>.
%H Elizabeth Todd, <a href="https://lib.dr.iastate.edu/creativecomponents/109">Happy numbers</a>, p. 11.
%e a(3) = 3, for 3^3 = 27, 2^3 + 7^3 = 351, 3^3 + 5^3 + 1^3 = 153.
%e a(13) = 177, for 177 -> 687 -> 1071 -> 345 -> 216 -> 225 -> 141 -> 66 -> 432 -> 99 -> 1458 -> 702 -> 351 -> 153 (13 = longest chain for numbers up to 10^4).
%e The process ends because 153 = 1^3 + 5^3 + 3^3.
%t Table[k=0;While[Last[s=NestList[Total[IntegerDigits@#^3]&,k,n]]!=153||Count[s,153]!=1,k=k+3];k,{n,0,14}] (* _Giorgos Kalogeropoulos_, Jul 30 2021 *)
%Y Cf. A055012 (sum of cubes of digits), A182111 (number of steps to a cycle), A165330 (cycle end), A046156.
%Y Cf. A046197 (proving that {0, 1, 153, 370, 371, 407} are the only possible fixed points for all numbers of any size when repeatedly summing the cubes of its digits).
%Y Cf. A346789 (concluding the number from the sum of the cubes of its digits).
%K nonn,base
%O 0,1
%A _Jörg Zurkirchen_, Jul 25 2021
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