A362001 Numbers k such that the digits of k^2 are a subsequence of the digits of 2^k.

2, 4, 26, 52, 58, 64, 73, 76, 86, 87, 89, 95, 96, 98, 100, 101, 104, 106, 110, 111, 112, 113, 121, 122, 126, 129, 133, 134, 135, 136, 138, 139, 140, 141, 144, 146, 148, 152, 156, 157, 161, 162, 163, 164, 165, 167, 169, 170, 171, 172, 173, 175, 176, 177, 178, 180, 182, 183, 184, 186, 187, 189, 190

Offset

1,1

Comments

Does the sequence contain all numbers > 327?

Conjecture is true up to 10^6. - Michael S. Branicky, Apr 03 2023

Links

Robert Israel, Table of n, a(n) for n = 1..1000

Example

a(3) = 26 is a term because the digits of 26^2 = 676 form a subsequence of those of 2^26 = 67108864.

Maple

filter:= proc(n) local L, nL, M, nM, i, j, k;
  L:= convert(n^2, base, 10); nL:= nops(L);
  M:= convert(2^n, base, 10); nM:= nops(M);
  j:= 1:
  for i from 1 to nL do
    if not member(L[i], M[j..nM], 'k') then return false fi;
    j:= j+k;
  od;
  true
end proc:
select(filter, [$0..200]);

Prog

(Python)
from itertools import count, islice, product
def ok(n):
    s, p = n**2, 2**n
    while s and p:
        if p%10 == s%10:
            s //= 10
        p //= 10
    return s == 0 and n > 0
print([k for k in range(200) if ok(k)]) # Michael S. Branicky, Apr 03 2023
(Python)
from itertools import count, islice
def A362001_gen(startvalue=1): # generator of terms >= startvalue
    for k in count(max(startvalue, 1)):
        c = iter(str(1<<k))
        if all(map(lambda b:any(map(lambda a:a==b, c)), str(k**2))):
            yield k
A362001_list = list(islice(A362001_gen(), 20)) # Chai Wah Wu, Apr 03 2023

Crossrefs

Cf. A046829.

Sequence in context: A186431 A129894 A028386 * A259374 A155120 A356442

Adjacent sequences: A361998 A361999 A362000 * A362002 A362003 A362004

Keyword

nonn,base

Author

Robert Israel, Apr 02 2023

Status

approved