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A350182
Numbers of multiplicative persistence 3 which are themselves the product of digits of a number.
8
49, 75, 96, 98, 147, 168, 175, 189, 196, 288, 294, 336, 343, 392, 448, 486, 648, 672, 729, 784, 864, 882, 896, 972, 1344, 1715, 1792, 1944, 2268, 2744, 3136, 3375, 3888, 3969, 7938, 8192, 9375, 11664, 12288, 12348, 13824, 14336, 16384, 16464, 17496, 18144
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 one-to-one correspondence between values in this sequence and equivalence classes of numbers with multiplicative persistence 4.
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 the numbers A007954(k) such that A031346(k) = 4,
These are the numbers k in A002473 such that A031346(k) = 3,
Or:
- they factor into powers of 2, 3, 5 and 7 exclusively.
- p(n) goes to a single digit in 3 steps.
Postulated to be finite and complete.
Let p(n) be the product of all the digits of n.
The multiplicative persistence of a number mp(n) is the number of times you need to apply p() to get to a single digit.
For example:
mp(1) is 0 since 1 is already a single-digit number.
mp(10) is 1 since p(10) = 0, and 0 is a single digit, 1 step.
mp(25) is 2 since p(25) = 10, p(10) = 0, 2 steps.
mp(96) is 3 since p(96) = 54, p(54) = 20, p(20) = 0, 3 steps.
mp(378) is 4 since p(378) = 168, p(168) = 48, p(48) = 32, p(32) = 6, 4 steps.
There are infinitely many numbers n such that mp(n)=4. But for each n with mp(n)=4, p(n) is a number included in this sequence, and this sequence is likely finite.
This sequence lists p(n) such that mp(n) = 4, or mp(p(n)) = 3.
LINKS
EXAMPLE
49 is in this sequence because:
- 49 goes to a single digit in 3 steps: p(49) = 36, p(36) = 18, p(18) = 8.
- p(77) = p(177) = p(717) = p(771) = 49, etc.
75 is in this sequence because:
- 75 goes to a single digit in 3 steps: p(75) = 35, p(35) = 15, p(15) = 5.
- p(355) = p(535) = p(1553) = 75, etc.
CROSSREFS
Cf. A002473, A003001 (smallest number with multiplicative persistence n), A031346 (multiplicative persistence), A031347 (multiplicative digital root), A046512 (all numbers with mp of 3).
Cf. A350180, A350181, A350183, A350184, A350185, A350186, A350187 (numbers with mp 0, 1 and 3 to 10 that are themselves 7-smooth numbers).
Sequence in context: A155713 A112074 A112057 * A260571 A038510 A063163
KEYWORD
base,nonn
AUTHOR
Daniel Mondot, Dec 18 2021
STATUS
approved