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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. 2
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 (list; graph; refs; listen; history; text; internal format)
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
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
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
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

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Last modified April 16 04:14 EDT 2024. Contains 371696 sequences. (Running on oeis4.)