A343727 Numbers with all digits odd whose squares have only one odd digit.

1, 3, 5, 7, 9, 15, 17, 51, 53, 79, 91, 93, 151, 155, 157, 533, 535, 775, 779, 793, 917, 1557, 1571, 1575, 5179, 5333, 5335, 7759, 7799, 9317, 9393, 9395, 15557, 15559, 15755, 51595, 53179, 53333, 53335, 77759, 79151, 79175, 93917, 151151, 155135, 155191

Offset: 1

Author

Jon E. Schoenfield, May 20 2021

Keywords

• nonn
• base

Comments

Of course, the one odd digit in the square is always the last digit.

The sequence is infinite because it contains the family of numbers 5, 53, 533, 5333, ....... with squares 25, 2809, 284089, 28440889, 2844408889. .... and respectively 535, 5335, 53335, ... with squares 286225, 28462225, 2844622225, 284446222225, ... - Marius A. Burtea, May 21 2021

Examples

			53179 is a term: all its digits are odd, and 53179^2 = 2828006041 has only one odd digit.
15113133375599 is a term: all its digits are odd, and 15113133375599^2 = 228406800428644424408608801 has only one odd digit.
		

Crossrefs

Cf. A014261, A030098, A343726, A343728.

Programs

Magma

[n:n in [1..160000 by 2]|Set(Intseq(n)) subset {1,3,5,7,9} and Set(Intseq(n*n div 10)) subset {0,2,4,6,8}]; // Marius A. Burtea, May 21 2021

Mathematica

Select[Range[160000], AllTrue[IntegerDigits[#], OddQ] && AllTrue[Most @ IntegerDigits[#^2], EvenQ] &] (* Amiram Eldar, May 20 2021 *)

Python

def ok(n):
  r, s = str(n), str(n*n)
  return all(d in "13579" for d in r) and all(d in "02468" for d in s[:-1])
print(list(filter(ok, range(1, 155192, 2)))) # Michael S. Branicky, May 20 2021

Python

from itertools import product
A343727_list = [n for n in (int(''.join(d)) for l in range(1,6) for d in product('13579',repeat=l)) if set(str(n**2)[:-1]) <= set('02468')] # Chai Wah Wu, May 21 2021