OFFSET
1,3
COMMENTS
Since multiplication and multiset union are commutative operations, we count unordered pairs, i.e., we can assume that k1 <= k2.
The sequence could be redefined in terms of the number of distinct n-digit numbers that could be factorized into such pairs.
From David A. Corneth, Feb 27 2024: (Start)
a(n) >= 2*a(n-1).
As we need k1 + k2 == k1 * k2 (mod 9) there are two possible pairs of residues (k1, k2) mod 3, namely, (0, 0) and (2, 2), and six possible residues mod 9, namely, (0, 0), (2, 2), (3, 6), (5, 8), (6, 3), (8, 5). (End)
EXAMPLE
For n=3 the a(3)=3 solutions are:
3 * 51 = 153
6 * 21 = 126
8 * 86 = 688
For n=4 the a(4)=15 solutions are:
3 * 501 = 1503
3 * 510 = 1530
5 * 251 = 1255
6 * 201 = 1206
6 * 210 = 1260
8 * 473 = 3784
8 * 860 = 6880
9 * 351 = 3159
15 * 93 = 1395
21 * 60 = 1260
21 * 87 = 1827
27 * 81 = 2187
30 * 51 = 1530
35 * 41 = 1435
80 * 86 = 6880
PROG
(Python)
def a(n):
count = 0
for i in range(1, 10**(n-1)):
for j in range(i, 10**n//i+1):
if len(str(i*j)) == n and sorted(str(i)+str(j)) == sorted(str(i*j)):
count += 1
print(n, count)
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Danila Potapov, Feb 26 2024
EXTENSIONS
a(9)-a(10) from Michael S. Branicky, Feb 26 2024
a(11) from Chai Wah Wu, Feb 27 2024
a(12)-a(13) from Martin Ehrenstein, Mar 02 2024
STATUS
approved