A369689 a(n) is the least positive number k such that k^2 is the concatenation of m and m + n for some positive number m, or -1 if there is no such k.

36363636364, 428, 8874, 5, 310, 4, 39, -1, 7747, 465

Offset

0,1

Comments

a(n) is the least number k > 0, if it exists, such that k^2 = (10^d + 1) m + n for some m > 0 where 10^(d-1) <= m + n < 10^d.

The attached file a369689.txt has lines n k m where k = a(n) and k^2 is the concatenation of m and m + n, n -1 where a(n) can be proved to be -1, and n -1 ? where I have not found a k that works but I have not been able to prove that a(n) = -1.

Links

Table of n, a(n) for n=0..9.

Robert Israel, Proving a(n) = -1

Robert Israel, Table of n, a(n), m for n = 0 .. 300 with some conjectured entries.

Example

a(3) = 5 because 5^2 = 25 is the concatenation of 2 and 2 + 3 = 5, and 5 is the least m that works.
a(7) = -1 because it can be proven that 7 is not a square mod (10^d + 1) for any d, and therefore there are no k and m such that k^2 is the concatenation of m and m + 7.

Prog

(Python)
from itertools import count
from sympy import sqrt_mod
def A369689(n):
    for j in count(0):
        b = 10**j
        a = b*10+1
        for k in sorted(sqrt_mod(n, a, all_roots=True)):
            m = (k**2-n)//a
            if m>0 and b <= m+n < a-1:
                return k # Chai Wah Wu, Feb 18 2024

Crossrefs

Cf. A106497.

Sequence in context: A233849 A370486 A116279 * A106497 A204097 A295355

Adjacent sequences: A369686 A369687 A369688 * A369690 A369691 A369692

Keyword

sign,more

Author

Robert Israel, Jan 28 2024

Status

approved