A134845 -id:A134845 - cp's OEIS frontend

A135251 Maximal number of zero digits in square of number with n digits not divisible by 10.

0, 1, 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124

Offset: 1

Artur Jasinski, Nov 24 2007

Cf. A134843, A134844, A134845, A134846, A134847, A134848, A134849, A135215, A135217, A135219, A135252, A135253.

(*For a(7)*) mx = 0; Do[Do[Do[Do[Do[Do[Do[k = 10^6b + 10^5q + 10^4r + 10^3p + 10^2s + 10n + m; w = IntegerDigits[k^2]; ile = 0; Do[If[w[[t]] == 0, ile = ile + 1; If[ile > mx, mx = ile]], {t, 1, Length[w]}], {m, 1, 9}], {n, 0, 9}], {s, 0, 9}], {p, 0, 9}], {r, 0, 9}], {q, 0, 9}], {b, 1, 9}]; mx

2*n-4 <= a(n) <= 2*n-2 since, if k is an n-digit number not divisible by 10, then k^2 has at most 2*n digits of which the first and last are nonzero; and for n >= 2, the square of the n-digit number 10^(n-1)+1 contains 2*n-4 zeros. It seems likely that a(n) = 2*n-4 for all n >= 4. - Pontus von Brömssen, Jun 09 2025

a(8)-a(64) from Pontus von Brömssen, Jun 09 2025

A135252 a(n) = number of numbers with n+1 digits and not divisible by 10 whose squares have maximal number of zero digits = A135251(n+1).

Original entry on oeis.org

18, 3, 24, 11, 10, 11

Offset: 1

Artur Jasinski, Nov 24 2007

			a(1)=18 because we have 18 numbers with 2 digits not divisible by 10 whose squares have maximal possible number of zero digits, namely 1 zero: 32, 33, 45, 47, 48, 49, 51, 52, 53, 55, 64, 71, 78, 84, 95, 97, 98, 99
a(2)=3 because we have 3 numbers with 3 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 3 zeros: 448, 548, 949
a(3)=24 because we have 24 numbers with 4 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 4 zeros: 1001, 1002, 1003, 2001, 2002, 3001, 3747, 3751, 4001, 4899, 5001, 5002, 5003, 6245, 6249, 6253, 7746, 7747, 7749, 7751, 7753, 9503, 9747, 9798
a(4)=11 because we have 11 numbers with 5 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 6 zeros: 10001, 10002, 10003, 20001, 20002, 30001, 40001, 50001, 50002, 50003, 62498
a(5)=10 because we have 10 numbers with 6 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 8 zeros: 100001, 100002, 100003, 200001, 200002, 300001, 400001, 500001, 500002, 500003
a(6)=11 because we have 11 numbers with 7 digits not divisible by 10 whose square have maximal possible number of zero digits, namely 10 zeros: 1000001, 1000002, 1000003, 2000001, 2000002, 3000001, 4000001, 5000001, 5000002, 5000003, 6244998

Cf. A134843, A134844, A134845, A134846, A134847, A134848, A134849, A135215, A135217, A135219, A135251, A135253.

(* For a(6) *) a = {}; c = 0; mx = 10; Do[Do[Do[Do[Do[Do[Do[k = 10^6b + 10^5q + 10^4r + 10^3p + 10^2s + 10n + m; w = IntegerDigits[k^2]; ile = 0; Do[If[w[[t]] == 0, ile = ile + 1], {t, 1, Length[w]}]; If[ile == mx, c = c + 1; AppendTo[a, k]], {m, 1, 9}], {n, 0, 9}], {s, 0, 9}], {p, 0, 9}], {r, 0, 9}], {q, 0, 9}], {b, 1, 9}]; c

A241489 Least number k not divisible by 10 such that k^3 contains n zeros.

Original entry on oeis.org

16, 52, 101, 252, 1002, 1001, 10003, 10002, 10001, 100003, 100002, 100001, 1000003, 1000002, 1000001, 10000003, 10000002, 10000001, 100000003, 100000002, 100000001, 1000000003, 1000000002, 1000000001, 10000000003, 10000000002, 10000000001, 100000000003, 100000000002

Offset: 1

Derek Orr, Apr 23 2014

It is believed that a(n) will have the pattern 1000...0003, 1000...0002, 1000...0001 after a(6).

			16 is not divisible by 10 and 16^3 = 4096, has 1 zero. So a(1) = 16.
52 is not divisible by 10 and 52^3 = 140608, has 2 zeros. So a(2) = 52.

Cf. A134845.

a(n) = {k = 1; while ((d = digits(k^3)) && (((k % 10) == 0) || (sum(i=1, #d, d[i] == 0) != n)), k++); k;} \\ Michel Marcus, Apr 30 2014

def Cu(n):
  for k in range(10**100):
    if k % 10 != 0:
      if str(k**3).count("0") == n:
        return k
n = 1
while n < 100:
  print(Cu(n))
  n += 1

For n > 6, a(n) = 10^(ceiling(n/3) + 1) + 3 - (n+2) mod 3.

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A135251 Maximal number of zero digits in square of number with n digits not divisible by 10.

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A135252 a(n) = number of numbers with n+1 digits and not divisible by 10 whose squares have maximal number of zero digits = A135251(n+1).

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A241489 Least number k not divisible by 10 such that k^3 contains n zeros.

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