A018796 Smallest square that begins with n.

0, 1, 25, 36, 4, 529, 64, 729, 81, 9, 100, 1156, 121, 1369, 144, 1521, 16, 1764, 1849, 196, 2025, 2116, 225, 2304, 2401, 25, 2601, 2704, 289, 2916, 3025, 3136, 324, 3364, 3481, 35344, 36, 3721, 3844, 3969, 400, 41209, 4225, 4356, 441, 45369, 4624, 4761, 484, 49

Offset: 0

Author

David W. Wilson

Keywords

• nonn
• base
• easy

Comments

If 4*(n+1) < 10^d then a(n) < (n+1)*10^d. - Robert Israel, Aug 01 2014

Examples

			Among the first 100001 terms, the largest is a(99999) = 99999515529 = 316227^2. - _Zak Seidov_, May 22 2016
		

Links

• Zak Seidov, Table of n, a(n) for n = 0..100000 (first 10000 terms from Chai Wah Wu)

Crossrefs

Cf. A018851, A077502.

Programs

Maple

a:= proc(n) local k,d,x;
  if issqr(n) then return n
  else for d from 1 do
    for k from 0 to 10^d-1 do
    x:= 10^d*n+k;
    if issqr(x) then return x fi
    od
    od
  fi
end proc:
seq(a(n),n=1..100); # Robert Israel, Jul 31 2014

Mathematica

Table[With[{d = IntegerDigits@ n}, k = 1; While[Or[IntegerLength[k^2] < Length@ d, Take[IntegerDigits[k^2], Length@ d] != d], k++]; k^2], {n, 49}] (* Michael De Vlieger, May 23 2016 *)

PARI

\\Set precision high enough (for the cases where n+1 is a square)!
a(n) = {my(v=vector(2));if(issquare(n),return(n), v=[sqrt(n),sqrt(n+1-(10^-((#digits(n)+7))))]; while(ceil(v[1])>floor(v[2]),v*=sqrt(10)));ceil(v[1])^2
} \\ David A. Corneth, May 22 2016

Python

n = 1
while n < 100:
    for k in range(10**3):
        if str(k**2).startswith(str(n)):
            print(k**2,end=', ')
            break
    n += 1 # Derek Orr, Jul 31 2014

Python

from gmpy2 import isqrt
def A018796(n):
    if n == 0:
        return 0
    else:
        d, nd = 1, n
        while True:
            x = (isqrt(nd-1)+1)**2
            if x < nd+d:
                return int(x)
            d *= 10
            nd *= 10 # Chai Wah Wu, May 23 2016

Extensions

a(0)=0 prepended by Zak Seidov, May 22 2016