A061051 Smallest square of the form [n digits][same n digits][further digits].

0, 225, 40401, 2042041, 317231721, 16198161984, 3921203921209, 400000040000001, 23391004233910041, 1100298301100298304, 141162631214116263129, 1322314049613223140496, 3171326702963171326702969, 107786983188610778698318864, 29726516052320297265160523209, 1003781781031081003781781031081

Offset: 0

Author

Erich Friedman, May 26 2001

Keywords

• base
• nonn

Comments

a(n) <= (2*10^n+1)^2. This bound is tight for n = 2, 7. Are there other values of n for which this bound is tight? For n = 11, there are no [further digits] block, i.e. the smallest square has 2n digits. This is true for all n in A086982. For instance, a(21) = 183673469387755102041183673469387755102041, a(33) = 132231404958677685950413223140496132231404958677685950413223140496. - Chai Wah Wu, Mar 25 2020

Examples

			40401 is the first square to have the first two digits the same as the next two digits
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..37

Crossrefs

Cf. A086982.

Programs

Python

from sympy import integer_nthroot
def A061051(n):
    if n == 0:
        return 0
    nstart = 10**(n-1)
    nend = 10*nstart
    for i in range(nstart,nend):
        k = int(str(i)*2)
        if integer_nthroot(k,2)[1]:
            return k
    for i in range(nstart,nend):
        si = str(i)*2
        for sj in '014569':
            k = int(si+sj)
            if integer_nthroot(k,2)[1]:
                return k # Chai Wah Wu, Mar 25 2020

Extensions

One more term from Vladeta Jovovic, Jun 02 2001

a(7)-a(15) from Chai Wah Wu, Mar 25 2020