A366165 a(n) is the least k > 0 such that 10^(2*n-1) - k can be written as a product j*m, where j and m have an equal number of decimal digits.

1, 1, 1, 1, 10, 1, 3, 1, 5, 3, 1, 6, 1, 7, 1, 2, 2, 1, 4, 7, 5, 1, 1, 3, 2, 1, 1, 1, 1, 2, 1, 1, 10, 4, 3, 3, 10, 1, 2, 3, 1, 1, 1, 7, 1, 1

Offset: 1

Author

Hugo Pfoertner, Oct 04 2023

Keywords

• nonn
• base
• more

Comments

a(n) <= 10 since 10^(2n-1)-10 = (10^(n-1)+1)(10^n-10). A consequence is that j and m in the product both have n decimal digits. - Chai Wah Wu, Oct 05 2023

Examples

			n a(n) 10^(2n-1)-a(n)       j       m
1  1   9                    1       9
2  1   999                 27      37
3  1   99999              123     813
4  1   9999999           2151    4649
5 10   999999990        10001   99990
6  1   99999999999     194841  513239
7  3   9999999999997  2769823 3610339
More than one pair (j,m) may exist, e.g., 9 = 1*9 = 3*3.
		

Crossrefs

Cf. A002275, A003020, A004022, A057951, A327435.

A067272 are the solutions for even exponents of 10, corresponding to (j,m) = (9,9), (99,99), (999,999), ... .

Programs

PARI

a366165(n)={my (p10=10^(2*n-1)); for (dd=1, p10, my (d=p10-dd); fordiv (d, x, fordiv (d, y, if (x*y==d && #digits(x)==#digits(y), return(dd)))))};

Python

from itertools import count, takewhile
from sympy import divisors
def A366165(n):
    a, l1, l2 = 10**((n<<1)-1), 10**(n-1), 10**n
    for k in count(1):
        b = a-k
        if any(l1<=db for d in takewhile(lambda m:m*m<=b, divisors(b))):
            return k # Chai Wah Wu, Oct 05 2023

Extensions

a(33)-a(35) from Chai Wah Wu, Oct 05 2023

a(36)-a(46) from Chai Wah Wu, Oct 07 2023