A339999 Squares that are divisible by both the sum of their digits and the product of their nonzero digits.
1, 4, 9, 36, 100, 144, 400, 900, 1296, 2304, 2916, 3600, 10000, 11664, 12100, 14400, 22500, 32400, 40000, 41616, 82944, 90000, 121104, 122500, 129600, 152100, 176400, 186624, 202500, 219024, 230400, 260100, 291600, 360000, 419904, 435600, 504100
Offset: 1
Examples
For the perfect square 144 = 12^2, the sum of its digits is 9, which divides 144, and the product of its nonzero digits is 16, which also divides 144 so 144 is a term of the sequence.
Links
- H. G. Grundman, Sequences of consecutive n-Niven numbers, Fibonacci Quarterly (1994) 32 (2): 174-175.
- Jean-Marie De Koninck and Florian Luca, Positive integers divisible by the product of their nonzero digits, Port. Math. 64 (2007) 75-85. (This proof for upper bounds contains an error. See the paper below.)
- Jean-Marie De Koninck and Florian Luca, Corrigendum to "Positive integers divisible by the product of their nonzero digits", Portugaliae Math. 64 (2007), 1: 75-85, Port. Math. 74 (2017), 169-170.
Programs
-
Mathematica
Select[Range[720]^2, And @@ Divisible[#, {Plus @@ (d = IntegerDigits[#]), Times @@ Select[d, #1 > 0 &]}] &] (* Amiram Eldar, Jul 23 2021 *) -
Python
from math import prod def sumd(n): return sum(map(int, str(n))) def nzpd(n): return prod([int(d) for d in str(n) if d != '0']) def ok(sqr): return sqr > 0 and sqr%sumd(sqr) == 0 and sqr%nzpd(sqr) == 0 print(list(filter(ok, (i*i for i in range(1001))))) # Michael S. Branicky, Jul 23 2021