cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A135217 a(n) = smallest number with n+1 digits and without zero digits whose squares have the maximal number of zero digits = A135215(n+1).

Original entry on oeis.org

32, 448, 3747, 62498, 248998, 6244998, 31623251, 498998998, 2449489753, 28284271249
Offset: 2

Views

Author

Artur Jasinski, Nov 23 2007

Keywords

Crossrefs

Extensions

a(8)-a(11) from Lars Blomberg, Jun 26 2011

A135216 a(n)= number of numbers with n+1 digits and without zero digits whose squares have maximal number of zero digits = A135215(n+1).

Original entry on oeis.org

18, 3, 13, 1, 7, 1
Offset: 1

Views

Author

Artur Jasinski, Nov 23 2007

Keywords

Examples

			a(1)=18 because we have 18 numbers with 2 digits and without zero digit whose square have maximal possible value 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 and without zero digit whose square have maximal possible value 3 zeros: 448, 548, 949.
a(3)=13 because we have 13 numbers with 4 digits and without zero digit whose square have maximal possible value 4 zeros: 3747, 3751, 4899, 6245, 6249, 6253, 7746, 7747, 7749, 7751, 7753, 9747, 9798.
a(4)=1 because we have only one number with 5 digits and without zero digit whose square have maximal possible value 6 zeros: 62498.
a(5)=7 because we have 7 numbers with 6 digits and without zero digit whose square have maximal possible value 7 zeros: 248998, 316245, 489898, 498999, 781249, 948951, 997998.
a(6)=1 because we have only one number with 7 digits and without zero digit whose square have maximal possible value 10 zeros: 6244998.
		

Crossrefs

Programs

  • Mathematica
    (*For a(7) *) 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], {m, 1, 9}], {n, 1, 9}], {s, 1, 9}], {p, 1, 9}], {r, 1, 9}], {q, 1, 9}], {b, 1, 9}]; c (*Artur Jasinski*)

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

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Nov 24 2007

Keywords

Crossrefs

Programs

  • Mathematica
    (*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

Formula

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

Extensions

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

Views

Author

Artur Jasinski, Nov 24 2007

Keywords

Examples

			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
		

Crossrefs

Programs

  • Mathematica
    (* 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
Showing 1-4 of 4 results.