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
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
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<=d
b 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
Comments