A381699 - OEIS

A381699

a(n) is the least nontrivial multiple of 2*n with the least possible number of even digits.

4, 8, 12, 16, 30, 36, 56, 32, 36, 40, 110, 72, 52, 56, 90, 96, 136, 72, 76, 80, 336, 132, 92, 96, 150, 156, 378, 112, 116, 120, 310, 192, 132, 136, 350, 576, 370, 152, 156, 160, 574, 336, 172, 176, 990, 552, 376, 192, 196, 300, 510, 312, 318, 756, 330, 336, 570, 1392, 354, 360, 732, 372, 378

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OFFSET

1,1

COMMENTS

By nontrivial multiple, we mean a multiple strictly larger than the number.

For even numbers, the last digit of any multiple will always be even. Also, for multiples of 10^k, all multiples will always have at least k even digits, namely k trailing '0's. Thus, if the number is a multiple of 2*10^k, there will be at least k+1 trailing even digits.

Question: if n is even, but not a multiple of 10, is there always a multiple k*n for which the last digit is the only even digit? If not, what is the smallest counterexample?

Records values of k(n) = a(n)/2n are k(1) = 2, k(5) = 3, k(7) = 4, k(11) = 5, k(21) = 8, k(45) = 11, k(58) = 12, k(101) = 55, k(182) = 108, k(1001) = 555, k(2001) = 778, k(3996) = 1001, k(7992) = 3253, k(9091) = 21545, k(9901) = 161155, ...

LINKS

Table of n, a(n) for n=1..63.

EXAMPLE

a(4) = 16 is the least nontrivial multiple of 8 with only one even digit.

a(5) = 30 is the least nontrivial multiple of 10 with only one even digit.

a(10) = 40 because 40 is the least nontrivial multiple of 20, and all multiples of 20 will always have (at least) the last two digits even.

a(41) = 574 is the least positive multiple of 82 that has only one even digit.

PROG

(PARI) apply( {A381699(n, o=valuation(5*n*=2, 10))=for(k=2, oo, #[0|d<-digits(n*k)%2, !d]>o|| return(k*n))}, [1..99]) \\ M. F. Hasler, Mar 04 2025

CROSSREFS

Cf. A061808 (similar for odd numbers), A061807.

Cf. A118070 (numbers with exactly one even decimal digit).