1,1

Josiah H. Drummond, Problem 57, Amer. Math. Monthly, Vol. 5 (1898), p. 26; reprinted on p. 906 of Vol. 105 (1998).

Any square ends with one of [ 0 ], 1, 4, 5, 6, 9, so a(1) = 5.

a(3) = A000993(3) - a(2) - #{100, 104, 201, 204, 209, 304, 400, 401, 404, 409, 500, 504, 600, 601, 604, 609, 704, 801, 804, 809, 900, 904} = 159 - 18 - 22 = 119, cf. A122986. - Reinhard Zumkeller, Mar 21 2010

(* A partly empirical script *) a[n_] := (Clear[qr]; qr[_] = False; For[k = 1, k <= 10^n/4, k++, m = PowerMod[k, 2, 10^n]; If[m > 10^(n-1) && FreeQ[IntegerDigits[m], 0], qr[m] = True]]; For[cnt = 0; k = 10^(n-1)+1, k <= 10^n-1, k++, If[qr[k], cnt++]]; cnt); a[1] = 5; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 10}] (* Jean-François Alcover, Jul 31 2015 *)

(Python)

from math import isqrt

def a(n):

suffixes = set()

for k in range(isqrt(10 ** (n - 1)) + 1, 10 ** n):

kk = k * k

s = str(kk)[-n:]

if "0" not in s and len(s) >= n:

suffixes.add(s)

return len(suffixes)

print([a(n) for n in range(1, 8)]) # Michael S. Branicky, May 18 2021

Cf. A036788.

Sequence in context: A332784 A302435 A332467 * A009348 A009365 A140296

Adjacent sequences: A036685 A036686 A036687 * A036689 A036690 A036691

base,nonn,nice,more

Explanation and more terms from David W. Wilson

a(11)-a(12) from Bert Dobbelaere, Mar 10 2021

approved