%I #24 Oct 07 2023 11:22:14
%S 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,
%T 4,3,3,10,1,2,3,1,1,1,7,1,1
%N 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.
%C 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
%e n a(n) 10^(2n-1)-a(n) j m
%e 1 1 9 1 9
%e 2 1 999 27 37
%e 3 1 99999 123 813
%e 4 1 9999999 2151 4649
%e 5 10 999999990 10001 99990
%e 6 1 99999999999 194841 513239
%e 7 3 9999999999997 2769823 3610339
%e More than one pair (j,m) may exist, e.g., 9 = 1*9 = 3*3.
%o (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)))))};
%o (Python)
%o from itertools import count, takewhile
%o from sympy import divisors
%o def A366165(n):
%o a, l1, l2 = 10**((n<<1)-1), 10**(n-1), 10**n
%o for k in count(1):
%o b = a-k
%o if any(l1<=d<l2 and d*l2>b for d in takewhile(lambda m:m*m<=b, divisors(b))):
%o return k # _Chai Wah Wu_, Oct 05 2023
%Y Cf. A002275, A003020, A004022, A057951, A327435.
%Y A067272 are the solutions for even exponents of 10, corresponding to (j,m) = (9,9), (99,99), (999,999), ... .
%K nonn,base,more
%O 1,5
%A _Hugo Pfoertner_, Oct 04 2023
%E a(33)-a(35) from _Chai Wah Wu_, Oct 05 2023
%E a(36)-a(46) from _Chai Wah Wu_, Oct 07 2023