

A346630


Smallest number which reaches the narcissistic number 153 after n steps when repeatedly summing the cubes of its digits.


2



153, 135, 18, 3, 9, 12, 33, 114, 78, 126, 6, 117, 669, 177, 12558, 44499999999999999999
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OFFSET

0,1


COMMENTS

All the terms a(n) as well as the intermediate results will be multiples of 3:
x^3 mod 3 = x mod 3 [0^3 = 0; 1^3 = 1; (1)^3 = 1].
Therefore (sum of cubes of digits) mod 3 = (sum of digits) mod 3.
Because the only multiple of 3 in A046197 is 153, every number which is a multiple of 3 will end up at 153.
Some other terms (not dealt with here) may reach a cycle of length > 1:
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.
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.
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:
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).
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 61042524005486968digit number consisting of the digit 5 followed by 753611407475147 copies of the 81digit string 182441700960219478737997256515775034293552812071330589849108367626886145404663923 followed by a single instance of the 60digit string 182441700960219478737997256515775034293552812071330589849106.


LINKS

Table of n, a(n) for n = 0..15.
Shyam Sunder Gupta, Curious Properties of 153.
Elizabeth Todd, Happy numbers, p. 11.


EXAMPLE

a(3) = 3, for 3^3 = 27, 2^3 + 7^3 = 351, 3^3 + 5^3 + 1^3 = 153.
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).
The process ends because 153 = 1^3 + 5^3 + 3^3.


MATHEMATICA

Table[k=0; While[Last[s=NestList[Total[IntegerDigits@#^3]&, k, n]]!=153Count[s, 153]!=1, k=k+3]; k, {n, 0, 14}] (* Giorgos Kalogeropoulos, Jul 30 2021 *)


CROSSREFS

Cf. A055012 (sum of cubes of digits), A182111 (number of steps to a cycle), A165330 (cycle end), A046156.
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).
Cf. A346789 (concluding the number from the sum of the cubes of its digits).
Sequence in context: A109778 A156740 A095226 * A165340 A183985 A184045
Adjacent sequences: A346627 A346628 A346629 * A346631 A346632 A346633


KEYWORD

nonn,base


AUTHOR

Jörg Zurkirchen, Jul 25 2021


STATUS

approved



