A343924 a(n) = the maximum number of times n can be multiplied by a number > 1 such that each product has distinct digits.

15, 14, 10, 13, 10, 9, 6, 12, 5, 9, 3, 8, 6, 6, 10, 11, 3, 4, 5, 8, 5, 4, 6, 7, 3, 6, 10, 3, 4, 9, 4, 10, 3, 4, 5, 4, 4, 4, 6, 7, 3, 3, 2, 4, 4, 5, 2, 6, 5, 4, 8, 4, 3, 9, 3, 3, 3, 4, 4, 8, 3, 3, 3, 9, 4, 5, 4, 3, 4, 5, 3, 3, 4, 4, 4, 3, 4, 5, 5, 6, 4, 2, 3, 2, 4, 3, 3, 7, 5, 5, 4, 4, 2, 2, 4, 5

Offset: 1

Author

Scott R. Shannon, May 04 2021

Keywords

• nonn
• base

Comments

See A343925 for the list of numbers for each n which can multiply n to produce the maximum length series of products with distinct digits.

Examples

			a(1) = 15 as 1 can be multiplied by 2 a total of fifteen times with each product containing distinct digits. The products are 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16348, 32768. No other number can multiply 1 to produce a longer series.
a(7) = 6 as 7 can be multiplied by 5 a total of six times with each product containing distinct digits. The products are 35, 175, 875, 4375, 21875, 109375. No other number can multiply 7 to produce a longer series.
a(17) = 3 as 17 can be multiplied by 2, 3, 6, or 17 a total of three times with each product containing distinct digits. For example for 17 the products are 289, 4913, 83521. No other numbers can multiply 17 to produce a longer series.
		

Crossrefs

Cf. A343925, A343921 (addition), A010784, A003991, A043537.

Formula

a(n) = 0 for n > 4938271605 or for any number n ending in two or more 0's.