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 1-10 of 10 results.

A052041 Squares lacking the digit zero in their decimal expansion.

Original entry on oeis.org

1, 4, 9, 16, 25, 36, 49, 64, 81, 121, 144, 169, 196, 225, 256, 289, 324, 361, 441, 484, 529, 576, 625, 676, 729, 784, 841, 961, 1156, 1225, 1296, 1369, 1444, 1521, 1681, 1764, 1849, 1936, 2116, 2916, 3136, 3249, 3364, 3481, 3721, 3844, 3969, 4225, 4356
Offset: 1

Views

Author

Patrick De Geest, Dec 15 1999

Keywords

Comments

This sequence is infinite: see A075415 or A102807 for a constructive proof.
Intersection of A052382 and A000290; A168046(a(n))*A010052(a(n))=1. - Reinhard Zumkeller, Dec 01 2009

Crossrefs

Programs

  • Mathematica
    Select[Range[66]^2, FreeQ[IntegerDigits[#],0]==True &] (* Jayanta Basu, May 25 2013 *)

Formula

a(n) = A052040(n)^2. - R. J. Mathar, Jul 23 2025

A052040 Numbers whose square is zeroless.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 34, 35, 36, 37, 38, 39, 41, 42, 43, 44, 46, 54, 56, 57, 58, 59, 61, 62, 63, 65, 66, 67, 68, 69, 72, 73, 74, 75, 76, 77, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92, 93, 94, 96
Offset: 1

Views

Author

Patrick De Geest, Dec 15 1999

Keywords

Comments

This sequence is infinite, since 33...334^2 = 11...11155...556, for example. This answers an open problem stated in HAKMEM. - Karl W. Heuer, Aug 19 2015

Examples

			From _Jon E. Schoenfield_, Aug 16 2021: (Start)
31 is a term: 31^2 = 961 has no 0's among its digits.
32 is not a term, because 32^2 = 1024. (End)
		

Crossrefs

Programs

A052043 Squares of primes lacking the digit zero in their decimal expansion.

Original entry on oeis.org

4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 3481, 3721, 4489, 5329, 6241, 6889, 7921, 11449, 11881, 12769, 16129, 17161, 18769, 19321, 24649, 26569, 27889, 29929, 32761, 36481, 37249, 44521, 49729, 51529, 52441, 54289
Offset: 1

Views

Author

Patrick De Geest, Dec 15 1999

Keywords

Comments

Intersection of A052382 and A001248; A168046(a(n))*A064911(a(n))*A010052(a(n)) = 1. - Reinhard Zumkeller, Dec 01 2009

Crossrefs

Programs

  • Mathematica
    Select[Prime[Range[100]]^2,DigitCount[#,10,0]==0&] (* Harvey P. Dale, Mar 18 2012 *)
  • PARI
    is(n)=my(d=digits(n));vecsort(d)[1]&&issquare(n,&n)&&isprime(n) \\ Charles R Greathouse IV, Jun 05 2013

Formula

a(n) = A052042(n)^2. - R. J. Mathar, Jul 23 2025

A052042 Primes that lack the digit zero in the decimal expansion of their squares.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 59, 61, 67, 73, 79, 83, 89, 107, 109, 113, 127, 131, 137, 139, 157, 163, 167, 173, 181, 191, 193, 211, 223, 227, 229, 233, 239, 263, 269, 271, 277, 281, 293, 307, 311, 313, 337, 359, 367, 373, 379, 383, 389, 409, 419, 421, 431
Offset: 1

Views

Author

Patrick De Geest, Dec 15 1999

Keywords

Examples

			The primes 47, 53 and 71 are not in the sequence because 47^2=2209, 53^2=2809 and 71^2=5041 contain zeros in their decimal representation.
		

Crossrefs

Programs

  • Mathematica
    fQ[n_] := DigitCount[n^2][[-1]] == 0; Select[Prime@ Range@ 80, fQ] (* Robert G. Wilson v, Aug 22 2012 *)
  • PARI
    {p=2;for(k=1,10^2,if(vecmin(digits(p^2))>0,
    print1(p", "));p=nextprime(1+p))}\\ Zak Seidov, Dec 24 2014

Formula

a(n) = sqrt(A052043(n)). - Zak Seidov, Dec 27 2014

A051750 Primes whose cubes lack zeros.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 41, 53, 61, 71, 83, 97, 113, 137, 139, 151, 157, 167, 173, 179, 181, 191, 197, 211, 233, 239, 241, 251, 257, 263, 277, 283, 293, 307, 331, 337, 347, 353, 359, 373, 379, 383, 389, 409, 421, 433, 457, 461, 463, 499, 503
Offset: 1

Views

Author

G. L. Honaker, Jr., Dec 07 1999

Keywords

Crossrefs

Programs

  • Mathematica
    Select[Prime[Range[100]],FreeQ[IntegerDigits[#^3],0]&] (* Harvey P. Dale, Jan 06 2017 *)
  • PARI
    isok(n) = isprime(n) && vecmin(digits(n^3)); \\ Michel Marcus, Jan 06 2014

Extensions

More terms from Michel Marcus, Jan 06 2014

A051832 Numbers k such that (2*10^(5*k) - 10^(4*k) + 2*10^(3*k) + 10^(2*k) + 10^k + 1)/3 is prime.

Original entry on oeis.org

0, 1, 7, 133
Offset: 1

Views

Author

G. L. Honaker, Jr., Dec 11 1999

Keywords

Comments

The Baxter-Hickerson function provides a number whose cube lacks zeros.
The next term is > 4400. - Jason Earls, Sep 10 2005
The next term is > 20000 (found using pfgw64). - Patrick De Geest, Jul 22 2012

Crossrefs

Programs

  • Maple
    f := n->(2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3;
  • PARI
    is(n)=isprime((2*10^(5*n)-10^(4*n)+2*10^(3*n)+10^(2*n)+10^n+1)/3) \\ Charles R Greathouse IV, Feb 17 2017

A052044 Numbers k such that k^3 lacks the digit zero in its decimal expansion.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 17, 18, 19, 21, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 35, 36, 38, 39, 41, 44, 45, 46, 49, 51, 53, 54, 55, 56, 57, 58, 61, 62, 64, 65, 66, 68, 71, 72, 75, 76, 77, 78, 81, 82, 83, 85, 88, 91, 92, 95, 96, 97, 98, 104, 105, 108, 111
Offset: 1

Views

Author

Patrick De Geest, Dec 15 1999

Keywords

Comments

This sequence is infinite since A052427 is a subsequence. - Amiram Eldar, Nov 23 2020

Crossrefs

Programs

Formula

a(n) = A052045(n)^(1/3). - Amiram Eldar, Nov 23 2020

A051751 Cubes arising in A051750.

Original entry on oeis.org

8, 27, 125, 343, 1331, 2197, 4913, 6859, 12167, 24389, 29791, 68921, 148877, 226981, 357911, 571787, 912673, 1442897, 2571353, 2685619, 3442951, 3869893, 4657463, 5177717, 5735339, 5929741, 6967871, 7645373, 9393931, 12649337
Offset: 1

Views

Author

G. L. Honaker, Jr., Dec 07 1999

Keywords

Crossrefs

Formula

A030078 INTERSECT A052382. - R. J. Mathar, Mar 23 2007

A052045 Cubes lacking the digit zero in their decimal expansion.

Original entry on oeis.org

1, 8, 27, 64, 125, 216, 343, 512, 729, 1331, 1728, 2197, 2744, 3375, 4913, 5832, 6859, 9261, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 29791, 32768, 35937, 42875, 46656, 54872, 59319, 68921, 85184, 91125, 97336, 117649, 132651, 148877
Offset: 1

Views

Author

Patrick De Geest, Dec 15 1999

Keywords

Comments

This sequence is infinite since A052427(n)^3 is a term for all n>=0. - Amiram Eldar, Nov 23 2020

Crossrefs

Programs

  • Maple
    select(t -> not has(convert(t,base,10),0), [seq(m^3,m=1..10^3)]); # Robert Israel, Aug 24 2014
  • Mathematica
    Select[Range[53]^3, DigitCount[#, 10, 0] == 0 &] (* Amiram Eldar, Nov 23 2020 *)
  • PARI
    lista(nn) = {for (n=1, nn, if (vecmin(digits(cub=n^3)), print1(cub, ", ")););} \\ Michel Marcus, Aug 25 2014
  • Python
    A052045 = [n**3 for n in range(1,10**5) if not str(n**3).count('0')]
    # Chai Wah Wu, Aug 24 2014
    

Formula

Intersection of A052382 and A000578; A168046(a(n))*A010057(a(n)) = 1. - Reinhard Zumkeller, Dec 01 2009
a(n) = A052044(n)^3. - Amiram Eldar, Nov 23 2020

A052427 Baxter-Hickerson numbers.

Original entry on oeis.org

2, 64037, 6634003367, 666334000333667, 66663334000033336667, 6666633334000003333366667, 666666333334000000333333666667, 66666663333334000000033333336666667
Offset: 0

Views

Author

Keywords

Comments

From Amiram Eldar, Nov 23 2020: (Start)
Named after Lew Baxter and Dean Hickerson.
Pegg (1999) conjectured that the sequence of zeroless cubes (A052045) is finite. On April 19, 1999, Hickerson gave the counterexample: if n == 2 (mod 3) and n >= 5, then the cube of (2*10^(5*n) - 10^(4*n) + 17*10^(3*n-1) + 10^(2*n) + 10^n - 2)/3 is zeroless. Three days later, Baxter gave a simpler variation which is valid for all n>=0 and is given in the Formula section. (End)

References

  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005. See p. 109.

Crossrefs

Subsequence of A052044.

Programs

  • Maple
    a(0) = 2, and 2^3 = 8 is zeroless.
    a(1) = 64037, and 64037^3 = 262598918898653 is zeroless.
  • Mathematica
    a[n_] := (2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3; Array[a, 10, 0] (* Amiram Eldar, Nov 23 2020 *)

Formula

a(n) = (2*10^(5*n) - 10^(4*n) + 2*10^(3*n) + 10^(2*n) + 10^n + 1)/3 (Baxter, 1999). - Amiram Eldar, Nov 23 2020

Extensions

Offset changed to 0 by Amiram Eldar, Nov 23 2020
Showing 1-10 of 10 results.