A115530 -id:A115530 - cp's OEIS frontend

A380428 Numbers k for which nonnegative integers x and y exist such that k is the concatenation of x and y as well as k = (x + y)^2.

81, 100, 2025, 3025, 88209, 494209, 4941729, 7441984, 24502500, 25502500, 52881984, 60481729, 300814336, 493817284, 6049417284, 6832014336, 20408122449, 21948126201, 33058148761, 35010152100, 43470165025, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969

Offset: 1

Felix Huber, Jan 25 2025

Subsequence of A000290.
From David A. Corneth, Apr 26 2025: (Start)
If y has q digits then a term m is of the form (x + y) = 10^q * x + y. Choosing some y we can solve for x (the equation is a quadratic with respect to x) and see if it produces a term.
y comes from A238712.
The sequence is infinite; it contains (25*100^i +- 5*10^i)^2 = concat(25*100^i +- 5*10^i, 25*100^i) for all i >= 0.
Neither x nor y can have a leading 0. (End)

			2025 is in the sequence because (20 + 25)^2 = 2025.
100 is in the sequence because (10 + 0)^2 = 100.
88209 is in the sequence because (88 + 209)^2 = 88209.
From _David A. Corneth_, Apr 26 2025: (Start)
9801 is not in the sequence even though (98 + 01)^2 = 9801 but 01 has a leading 0 which is disallowed.
If a term m ends in y = 209 where y has three digits we have 10^3*x + y = (x + y)^2. Solving for x gives x = 88 or x = 494 corresponding to terms 88209 and 494209. (End)

David A. Corneth, Table of n, a(n) for n = 1..96 (terms <= 10^20)

Cf. A000290, A238712.

Cf. A115527, A115528, A115529, A115530, A115531, A115532, A115533, A115534, A115535, A115536, A115537, A115538, A115539, A115540, A115541, A115542, A115543, A115544, A115545, A115546, A115547, A115548, A115549, A115550, A115551, A115552, A115553, A115554, A115555, A115556.

A380428:=proc(n)
    option remember;
    local a,i,k,x,y;
    if n=1 then
        81
    elif n=2 then
        100
    else
        for a from isqrt(procname(n-1))+1 do
            k:=length(a^2);
            for i to k-1 do
                x:=floor(a^2/10^i);
                y:=a^2-x*10^i;
                if x+y=a and length(x)+length(y)=k then
                    return a^2
                fi
            od
        od
    fi;
end proc;
seq(A380428(n),n=1..26);

cp's OEIS Frontend

A380428 Numbers k for which nonnegative integers x and y exist such that k is the concatenation of x and y as well as k = (x + y)^2.

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