A136812 -id:A136812 - cp's OEIS frontend

A136809 Numbers k such that k and k^2 use only the digits 0, 1, 2 and 3.

0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 10000, 10001, 10010, 10011, 10100, 10101, 10110, 10111, 11000, 11001, 11010, 11100, 11101, 100000, 100001, 100010, 100011, 100100, 100101, 100110, 100111, 101000, 101001, 101010, 101011, 101100, 101110, 110000, 110001, 110010, 110100, 110101, 111000, 111001, 111010

Offset: 1

Jonathan Wellons (wellons(AT)gmail.com), Jan 22 2008

Generated with "DrScheme".
Subsequence of A136810-A136815. - M. F. Hasler, Jan 24 2008
From M. F. Hasler, Nov 03 2020: (Start)
A136813(144) = 31733311 is the first term of that sequence which is not in this sequence. All others among A136810-A136815 differ much earlier.
If a(n) is a term then so is 10*a(n). Conjecture: the sequence contains only "binary numbers" (digits 0 or 1) having no more than three 1's in a row, and not more than one run of two or more consecutive 1's. (But not all such numbers, since 101101 for example is not in the sequence.) (End)
Here is a counterexample with two instances of two consecutive ones, i.e., two non-overlapping occurrences of the substring 11: 110010000010100001010000010011^2 = 12102200102222202222322212223022221222322220222220100220121. - Michael S. Branicky, Nov 04 2020
One cannot test candidates digit-by-digit from the right. Specifically, a suffix of the valid counterexample above is invalid: 10100001010000010011^2 = 102010020402001222322220222220100220121. - Michael S. Branicky, Nov 05 2020
With respect to the conjectures and comment above, only digits 0 and 1 occur and no 1111's occur in the first 83990 terms (all with <= 25 digits). These were generated incrementally from the right based on partial screening (see Python program). - Michael S. Branicky, Jul 07 2022

			111^2 = 12321,
11101^2 = 123232201, and
101011^2 = 10203222121,
111001^2 = 12321222001, so 111, 11101, 101011 and 111001 are in the sequence, but:
110011^2 = 12102420121, so 110011 is not in the sequence; also
1100011^2 = 1210024200121, so 1100011 is not in the sequence, and
1010101^2 = 1020304030201, so 1010101 is not in the sequence; but
1110001^2 = 1232102220001, so 1110001 is in the sequence; also
1010100100001^2 = 1020302212022030200200001.

Michael S. Branicky, Table of n, a(n) for n = 1..12670 (all terms with <= 19 digits; terms 1..1380 from Jonathan Wellons)
Jonathan Wellons, Tables of Shared Digits

Cf. A007088 (binary numbers), A136808 (subsequence: only 0, 1, 2), A136810, A136811, A136812, A136813, A136814, A136815.

Select[Range[0,200000],And@@(ContainsAll[{0,1,2,3},Union@IntegerDigits@#]&/@{#,#^2})&] (* Giorgos Kalogeropoulos, May 21 2021 *)
With[{c={0,1,2,3}},Select[FromDigits/@Tuples[c,6],SubsetQ[c,IntegerDigits[ #^2]]&]] (* Harvey P. Dale, Jun 01 2021 *)

select( {is_A136809(n,o(n)=vecmax(digits(n))<4)=o(n^2)&&o(n)}, [fromdigits(binary(n))|n<-[0..99]]) \\ M. F. Hasler, Nov 03 2020

from itertools import count, islice
def agen(only="0123"):
    digset, valid = set(only), set(only)
    for e in count(1):
        found, newvalid = set(), set()
        for tstr in valid:
            t = int(tstr)
            if (tstr == "0" or tstr[0] != "0") and set(str(t**2)) <= digset:
                found.add(t)
            for d in digset:
                dtstr = d + tstr
                dt = int(dtstr)
                remstr = str(dt**2)[-e:]
                if set(remstr) <= digset:
                    newvalid.add(dtstr)
        valid = newvalid
        yield from sorted(found)
print(list(islice(agen(), 50))) # Michael S. Branicky, Jul 07 2022

cp's OEIS Frontend

A136809 Numbers k such that k and k^2 use only the digits 0, 1, 2 and 3.

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