A309883 Numbers k such that A003132(k^2) = A003132(k), where A003132(n) is the sum of the squares of the digits of n.

0, 1, 10, 35, 100, 152, 350, 377, 452, 539, 709, 1000, 1299, 1398, 1439, 1519, 1520, 1569, 1591, 1679, 1965, 2599, 2838, 3332, 3500, 3598, 3770, 4520, 4586, 4754, 4854, 5390, 5501, 5835, 5857, 6388, 6595, 6735, 6861, 6951, 7090, 7349, 7887, 8395, 9795, 10000, 10056, 10159, 10389, 11055, 11091, 12990, 12999

Offset: 1

Author

Antonio Roldán, Aug 21 2019

Keywords

• nonn
• base

Comments

If k is in the sequence, then so are k*10^r, r >= 1.

Examples

			377^2 = 142129, A003132(377) = 3^2 + 7^2 + 7^2 = 107, A003132(142129) = 1^2 + 4^2 + 2^2 + 1^2 + 2^2 + 9^2 = 107.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A003132, A058369, A070276, A174460, A176670, A178213, A307735, A309884.

Programs

Magma

[0] cat [k:k in [1..13000]| &+[c^2: c in Intseq(k)] eq &+[c^2: c in Intseq(k^2)]]; // Marius A. Burtea, Aug 24 2019

Maple

filter:= proc(n) local t;
  add(t^2, t = convert(n,base,10)) = add(t^2, t = convert(n^2,base,10))
end proc:
select(filter, [$0..20000]); # Robert Israel, Apr 30 2023

Mathematica

digSum[n_] := Total[IntegerDigits[n]^2]; Select[Range[0, 13000], digSum[#] == digSum[#^2] &] (* Amiram Eldar, Aug 22 2019 *)

PARI

for(i = 0, 30000, if(norml2(digits(i^2)) == norml2(digits(i)), print1(i, ", ")))

Python

def A003132(n):
    s = 0
    while n > 0:
        s, n = s+(n%10)**2, n//10
    return s
n, a = 0, 0
while n < 50:
    if A003132(a) == A003132(a*a):
        n = n+1
        print(n,a)
    a = a+1 # A.H.M. Smeets, Aug 23 2019