A339999 - OEIS

%I #118 Aug 16 2021 22:08:32

%S 1,4,9,36,100,144,400,900,1296,2304,2916,3600,10000,11664,12100,14400,

%T 22500,32400,40000,41616,82944,90000,121104,122500,129600,152100,

%U 176400,186624,202500,219024,230400,260100,291600,360000,419904,435600,504100

%N Squares that are divisible by both the sum of their digits and the product of their nonzero digits.

%H H. G. Grundman, <a href="http://www.fq.math.ca/Scanned/32-2/grundman.pdf">Sequences of consecutive n-Niven numbers</a>, Fibonacci Quarterly (1994) 32 (2): 174-175.

%H Jean-Marie De Koninck and Florian Luca, <a href="https://doi.org/10.4171/PM/1777">Positive integers divisible by the product of their nonzero digits</a>, Port. Math. 64 (2007) 75-85. (This proof for upper bounds contains an error. See the paper below.)

%H Jean-Marie De Koninck and Florian Luca, <a href="https://doi.org/10.4171/PM/1999">Corrigendum to "Positive integers divisible by the product of their nonzero digits", Portugaliae Math. 64 (2007), 1: 75-85</a>, Port. Math. 74 (2017), 169-170.

%e 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.

%t Select[Range[720]^2, And @@ Divisible[#, {Plus @@ (d = IntegerDigits[#]), Times @@ Select[d, #1 > 0 &]}] &] (* _Amiram Eldar_, Jul 23 2021 *)

%o (Python)

%o from math import prod

%o def sumd(n): return sum(map(int, str(n)))

%o def nzpd(n): return prod([int(d) for d in str(n) if d != '0'])

%o def ok(sqr): return sqr > 0 and sqr%sumd(sqr) == 0 and sqr%nzpd(sqr) == 0

%o print(list(filter(ok, (i*i for i in range(1001)))))

%o # _Michael S. Branicky_, Jul 23 2021

%Y Intersection of A000290, A005349 and A055471.

%Y Cf. A118547, A342262, A342650.

%K nonn,base

%O 1,2

%A _Michael Gohn_, _Joshua Harrington_, _Sophia Lebiere_, _Hani Samamah_, _Kyla Shappell_, _Wing Hong Tony Wong_, Jul 23 2021