A046838 Numbers k such that k^2 can be obtained from k by inserting a block of digits.

0, 1, 5, 6, 10, 11, 25, 76, 95, 96, 100, 101, 125, 376, 625, 976, 995, 996, 1000, 1001, 1025, 1376, 9376, 9625, 9976, 9995, 9996, 10000, 10001, 10025, 10376, 10625, 90625, 99376, 99625, 99976, 99995, 99996, 100000, 100001, 100025, 100376, 100625, 109376, 890625

Offset

1,3

Comments

From Robert Israel, Oct 24 2025: (Start)

Every term is of one of the following types:

(a) x with d digits such that x^2 == x (mod 10^d).

(b) 10^r + x, x as in (a).

(c) 10^r - 10^d + x, x as in (a).

(End)

Links

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

Maple

N:= 10: # for terms of <= N digits
R:= 0, seq(10^i, i=0..N-1), seq(10^i+1, i=1..N-1):
for d from 0 to N do
   for b in [chrem([0, 1], [2^d, 5^d]), chrem([1, 0], [2^d, 5^d])] do
     for r from 1 to N-d do
       x:= b + (10^r-1)*10^d;
       if floor(x^2/10^(ilog10(x^2)-r+1)) = 10^r-1 then R:= R, x fi;
     od;
     if b >= 10^(d-1) then
       R:= R, b; # for a = 0
     fi;
     for r from 0 to N-d-1 do
         x:= b + 10^(r+d);
         if floor(x^2/10^(ilog10(x^2)-r)) = 10^r then R:= R, x fi;
       od;
od od:
sort(convert({R}, list));

Prog

(Python)
def ok(n):
    s, t = str(n), str(n**2)
    return any(s == t[:i]+t[j:] for i in range(len(t)) for j in range(i, len(t)+1))
print([k for k in range(900000) if ok(k)]) # Michael S. Branicky, Sep 12 2025

Crossrefs

Sequence in context: A074913 A046829 A052212 * A326418 A037359 A099538

Adjacent sequences: A046835 A046836 A046837 * A046839 A046840 A046841

Keyword

nonn,base

Author

David W. Wilson

Status

approved