A337230 Numbers that are a divisor of the product of the digits of its divisors.

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 14, 15, 16, 18, 21, 24, 25, 27, 28, 32, 35, 36, 42, 45, 48, 54, 56, 63, 64, 72, 75, 84, 96, 112, 125, 126, 128, 135, 144, 147, 162, 168, 175, 189, 192, 224, 225, 252, 256, 288, 294, 336, 375, 378, 384, 441, 448, 486, 512, 567, 576, 588, 625, 672, 675, 729

Offset: 1

Author

Scott R. Shannon, Aug 20 2020

Keywords

• nonn
• fini
• full
• base

Comments

In the first 130 million numbers there are 117 terms, the last being 234375. It is possible no more terms exist, although this is unknown.

The number with the largest ratio of it divisors' digit product to the number itself is 5376, with a ratio of ~2.2*10^43.

From David A. Corneth, Aug 22 2020: (Start)

The product of the digits of divisors must be nonzero. For example, 10 isn't a term as the product of digits of divisors is 0.

Terms are 7-smooth.

The largest term is 234375. Proof:

2^k1 has a 0 for k1 = 10, 3^k2 has a 0 for k2 = 10, 5^k3 has a 0 for k3 = 8 and 7^k4 has a 0 for k4 = 4.

Checking all numbers of the form 2^e1 * 3^e2 * 5^e3 * 7^e4 with ei bounded by the respective ki gives the 117 terms mentioned above with the largest of them being 234375. (End)

Examples

			a(4) = 4 is a term as the divisors of 4 are 1,2,4 and 1*2*4 = 8 which is a divisible by 4.
a(10) = 12 is a term as the divisors of 12 are 1,2,3,4,6,12 and 1*2*3*4*6*1*2 = 288 which is divisible by 12.
a(19) = 28 is a term as the divisors of 28 are 1,2,4,7,14,28 and 1*2*4*7*1*4*2*8 = 3584 which is divisible by 28.
		

Links

• Scott R. Shannon, Table of n, a(n) for n = 1..117

Crossrefs

Cf. A002473, A027750, A000005, A007954, A007602, A005349, A038186, A333617.

Programs

Mathematica

Select[Range[3^6], (prod = Times @@ (Times @@@ IntegerDigits @ Divisors[#])) > 0 && Divisible[prod, #] &] (* Amiram Eldar, Aug 22 2020 *)

PARI

is(n) = {if(n < 10, return(n > 0)); f = factor(n); if(f[#f~, 1] > 7, return(0)); my(d = divisors(n), p = 1); for(i = 2, #d, dd = digits(d[i]); for(j = 1, #dd, p *= dd[j]); if(p == 0, return(0))); p % n == 0} \\ David A. Corneth, Aug 22 2020