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

1,5

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

Links

Table of n, a(n) for n=1..46.

Example

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.

Prog

(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<=d<l2 and d*l2>b for d in takewhile(lambda m:m*m<=b, divisors(b))):
            return k # Chai Wah Wu, Oct 05 2023

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), ... .

Sequence in context: A359736 A010179 A174209 * A010181 A111525 A138261

Adjacent sequences: A366162 A366163 A366164 * A366166 A366167 A366168

Keyword

nonn,base,more

Author

Hugo Pfoertner, Oct 04 2023

Extensions

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

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

Status

approved