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A241846 Numbers for which the cube of the sum of the digits is equal to the square of the product of their digits. 2

%I #21 Jun 04 2017 02:42:59

%S 0,1,88,333,11248,11284,11428,11482,11824,11842,12148,12184,12418,

%T 12481,12814,12841,14128,14182,14218,14281,14812,14821,18124,18142,

%U 18214,18241,18412,18421,21148,21184,21418,21481,21814,21841,24118,24181,24811,28114

%N Numbers for which the cube of the sum of the digits is equal to the square of the product of their digits.

%C Let d_1 d_2... d_q denote the decimal expansion of a number n. The sequence lists the numbers n such that (d_1 + d_2 +...+ d_q)^3 = (d_1 * d_2 *...* d_q)^2.

%C The sequence is finite and contains 1419 terms because the maximum sum of the digits of a(n) is 16, the maximum product of the digits is 64 with 16^3 = 64^2 and the greatest number of the sequence is 2222221111.

%C The primitive values of a(n) (numbers whose decimal digits are not a permutation of another number of the sequence) are 0, 1, 88, 333, 11248, 112228, 1111444, 11112244, 111122224, 1111222222.

%C Nevertheless, the numbers 112228, 1111444, 11112244, 111122224, 1111222222 are not completely independent; for example, a decimal digit 4 of 1111444 becomes 22 and gives the number 11112244.

%H Michel Lagneau, <a href="/A241846/b241846.txt">Table of n, a(n) for n = 1..1419</a>

%e 333 is in the sequence because (3+3+3)^3 = (3*3*3)^2 = 729.

%e 11248 is in the sequence because (1+1+2+4+8)^3 = (1*1*2*4*8)^2 = 4096.

%t Select[Range[30000], (Plus @@ IntegerDigits[ # ]^3) == (Times @@ IntegerDigits[ # ]^2) &]

%Y Cf. A034710, A117720.

%K nonn,base,fini,full

%O 1,3

%A _Michel Lagneau_, Apr 30 2014

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Last modified April 19 08:45 EDT 2024. Contains 371782 sequences. (Running on oeis4.)