

A350184


Numbers of multiplicative persistence 5 which are themselves the product of digits of a number.


8



2688, 18816, 26244, 98784, 222264, 262144, 331776, 333396, 666792, 688128, 1769472, 2939328, 3687936, 4214784, 4917248, 13226976, 19361664, 38118276, 71663616, 111476736, 133413966, 161414428, 169869312, 184473632, 267846264, 368947264, 476171136, 1783627776
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

The multiplicative persistence of a number mp(n) is the number of times the product of digits function p(n) must be applied to reach a single digit, i.e., A031346(n).
The product of digits function partitions all numbers into equivalence classes. There is a onetoone correspondence between values in this sequence and equivalence classes of numbers with multiplicative persistence 5.
There are infinitely many numbers with mp of 1 to 11, but the classes of numbers (p(n)) are postulated to be finite for sequences A350181....
Equivalently:
This sequence consists of all numbers A007954(k) such that A031346(k) = 6.
Or:
 they factor into powers of 2, 3, 5 and 7 exclusively.
 p(n) goes to a single digit in 5 steps.
Postulated to be finite and complete.


LINKS



EXAMPLE

2688 is in this sequence because:
 2688 goes to a single digit in 5 steps: p(2688)=768, p(768)=336, p(336)=54, p(54)=20, p(20)=0.
 p(27648) = p(47628) = 2688, etc.
331776 is in this sequence because:
 331776 goes to a single digit in 5 steps: p(331776)=2646, p(2646)=288, p(288)=128, p(128)=16, p(16)=6.
 p(914838624) = p(888899) = 331776, etc.


MATHEMATICA

mx=10^10; lst=Sort@Flatten@Table[2^i*3^j*5^k*7^l, {i, 0, Log[2, mx]}, {j, 0, Log[3, mx/2^i]}, {k, 0, Log[5, mx/(2^i*3^j)]}, {l, 0, Log[7, mx/(2^i*3^j*5^k)]}];
Select[lst, Length@Most@NestWhileList[Times@@IntegerDigits@#&, #, #>9&]==5&] (* code for 7smooth numbers from A002473.  Giorgos Kalogeropoulos, Jan 16 2022 *)


PROG

(Python)
from math import prod
def hd(n):
while (n&1) == 0: n >>= 1
while (n%3) == 0: n /= 3
while (n%5) == 0: n /= 5
while (n%7) == 0: n /= 7
return(n)
def pd(n): return prod(map(int, str(n)))
def ok(n):
if hd(n) > 9: return False
return (p := pd(n)) > 9 and (q := pd(p)) > 9 and (r := pd(q)) > 9 and (s := pd(r)) > 9 and pd(s) < 10
print([k for k in range(10, 476200000) if ok(k)])


CROSSREFS

Cf. A003001 (smallest number with multiplicative persistence n), A031346 (multiplicative persistence), A031347 (multiplicative digital root).


KEYWORD

nonn,base


AUTHOR



STATUS

approved



