A359347 -id:A359347 - cp's OEIS frontend

1234549876609, 9066789454321, 123452587690084, 123454387666009, 123454987660900, 123456987654400, 123458987664100, 123478988652100, 125688987432100, 146678985432100, 445678965432100, 480096785254321, 900666783454321, 906678945432100, 10223418547690084

Offset: 1

Martin Renner, Dec 27 2022

These are perfect squares containing each digit from 0 to 9 at least once and still remain square numbers (not necessarily of the same length) when reversing the digits.

In 1905, inspired by a question about all pandigital square numbers containing each digit from 0 to 9 exactly once (cf. A036745, A156977), the British mathematician Allan Cunningham (1842-1928) asked for reversible and palindromic pandigital square numbers. In his answer, he gives possible solutions, but actually not the least possible numbers he was asking for in his question.

			Sequence starts with 1111103^2 = 1234549876609 <~> 9066789454321 = 3011111^2, which is the smallest possible such number.

Martin Renner, Table of n, a(n) for n = 1..458
Allan Cunningham, Question 15789, The Educational Times, and Journal of the College of Preceptors 58 (1905), nr. 530 (June 1), p. 273; Solution 15789, Ibid., 59 (1906), nr. 537 (Jan. 1), p. 39.

Cf. A036745, A061457, A156977, A225218, A359347.

isok(k) = if (issquare(k), my(d=digits(k)); (#Set(d) == 10) && issquare(fromdigits(Vecrev(d)));); \\ Michel Marcus, Dec 31 2022

from math import isqrt
from itertools import count, islice
def c(n): return len(set(s:=str(n)))==10 and isqrt(r:=int(s[::-1]))**2==r
def agen(): yield from (k*k for k in count(10**6) if c(k*k))
print(list(islice(agen(), 15))) # Michael S. Branicky, Dec 27 2022

cp's OEIS Frontend

A359346 Reversible pandigital square numbers.

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