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 12 results. Next

A103670 Smallest m such that A102730(m) = n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 23, 117, 155, 1410, 3702, 29406, 104312, 680407
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

Reinhard Zumkeller conjectures (at A102730) that this sequence is finite. I conjecture the contrary, that a(n) exists for every n. Further, I expect a(n) << n^n. - Charles R Greathouse IV, Aug 21 2011
a(14) > 41000, if it exists. - Amiram Eldar, Apr 03 2025
The number of times that 1! through 10! appear in a(11)! is: 17417, 8777, 4361, 1118, 265, 31, 5, 1, 1, 1. It suggests factorials are normal in base 2 except for their trailing zeros. On this assumption, a(16) is expected around 3-4 million, and a(17) around 20-30 million. - Martin Fuller, Apr 14 2025

Examples

			a(6) = 5: A102730(5) = #{0,1,2,3,4,5} = 6.
a(7) = 8: A102730(8) = #{0,1,2,3,4,7,8} = 7.
a(8) = 23: A102730(23) = #{0,1,2,3,4,5,6,23} = 8.
a(9) = 117: A102730(117) = #{0,1,2,3,4,5,6,7,117} = 9.
a(10) = 155: A102730(155) = #{0,1,2,3,4,5,6,7,8,155} = 10.
		

Crossrefs

Cf. A102730.

Programs

  • Mathematica
    A102730[n_] := Sum[Boole[StringContainsQ[IntegerString[n!, 2], IntegerString[k!, 2]]], {k, 0, n}];
    seq[len_] := Module[{s = Table[0, {len}], c = 0, m = 0, i}, While[c < len, i = A102730[m]; If[i <= len && s[[i]] == 0, c++; s[[i]] = m]; m++]; s]; seq[10] (* Amiram Eldar, Apr 03 2025 *)

Extensions

a(11)-a(12) from D. S. McNeil, Aug 21 2011
a(13) from Amiram Eldar, Apr 03 2025
a(14)-a(15) from Martin Fuller, Apr 14 2025

A103679 Numbers m such that the binary representation of m! does not contain 6!.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 54, 56, 57, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 72, 73, 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 87
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

Last term is probably 802. No numbers between 803 and 500000 belong to the sequence. - Giovanni Resta, Apr 07 2013

Crossrefs

Programs

  • Mathematica
    Select[Range[0,100],SequenceCount[IntegerDigits[#!,2],{1,0,1,1,0,1,0,0,0,0}]==0&] (* Harvey P. Dale, Oct 12 2024 *)
  • PARI
    is(n)=n=n!; while(n>719, my(e=valuation(n, 2), e1=valuation((n>>=e)+1, 2)); n>>=e1; if(e>3 && e1==1 && bitand(n, 31)==22, return(0))); 1 \\ Charles R Greathouse IV, Apr 07 2013

Formula

A103674(a(n)) = 0, A103674(A103678(n)) = 1.

A103680 Numbers m such that the binary representation of m! contains 7!.

Original entry on oeis.org

7, 8, 100, 113, 117, 123, 136, 145, 155, 156, 163, 165, 168, 179, 186, 203, 245, 247, 248, 259, 265, 275, 283, 289, 293, 294, 296, 305, 307, 309, 311, 318, 328, 330, 341, 342, 343, 346, 352, 355, 360, 375, 384, 386, 399, 402, 405, 408, 410, 413, 415, 419
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

All the numbers from 5154 to 5*10^5 belong to the sequence. - Giovanni Resta, Apr 07 2013

Crossrefs

Programs

  • Mathematica
    Select[Range[0, 420], StringContainsQ[IntegerString[#!, 2], IntegerString[7!, 2]] &] (* Amiram Eldar, Apr 03 2025 *)
  • PARI
    is(n)=n=n!; while(n>5039, my(e=valuation(n, 2), e1=valuation((n>>=e)+1, 2)); n>>=e1; if(e>3 && e1==2 && bitand(n, 127)==78, return(1))); 0 \\ Charles R Greathouse IV, Apr 07 2013

Formula

A103675(a(n)) = 1, A103675(A103681(n)) = 0.

A103674 If in binary representation n! contains 6! then 1 else 0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0
Offset: 0

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

Conjecture: a(n) = 1 for n > 802. - Charles R Greathouse IV, Apr 07 2013
Conjecture checked up to n <= 5*10^5. - Giovanni Resta, Apr 07 2013

Crossrefs

Programs

  • PARI
    a(n)=n=n!;while(n>719, my(e=valuation(n,2),e1=valuation((n>>=e)+1,2)); n>>=e1; if(e>3 && e1==1 && bitand(n,31)==22, return(1))); 0 \\ Charles R Greathouse IV, Apr 07 2013

Formula

a(A103678(n)) = 1, a(A103679(n)) = 0.

A103675 a(n) = 1 if the binary representation of n! contains 7! (bit string "1001110110000"), otherwise a(n) = 0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0
Offset: 0

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

a(A103680(n)) = 1, a(A103681(n)) = 0.
Probably a(5153) is the last zero term. a(n) = 1 for n from 5154 to 5*10^5. - Giovanni Resta, Apr 07 2013

Crossrefs

Programs

  • PARI
    a(n)=n=n!; while(n>5039, my(e=valuation(n, 2), e1=valuation((n>>=e)+1, 2)); n>>=e1; if(e>3 && e1==2 && bitand(n, 127)==78, return(1))); 0 \\ Charles R Greathouse IV, Apr 07 2013

Extensions

Name edited by Antti Karttunen, Dec 24 2018

A103678 Numbers m such that in binary representation m! contains 6!.

Original entry on oeis.org

6, 20, 23, 30, 31, 32, 37, 43, 53, 55, 58, 65, 71, 74, 81, 86, 91, 94, 95, 99, 102, 106, 108, 111, 115, 116, 117, 118, 122, 123, 124, 127, 128, 129, 133, 134, 135, 139, 141, 143, 144, 145, 149, 153, 155, 157, 158, 159, 160, 162, 163, 166, 167, 168, 169, 170, 171
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

Complement of A103679: A103674(a(n))=1, A103674(A103679(n))=0.
All the numbers 803 <= n <= 500000 are in the sequence. - Giovanni Resta, Apr 07 2013

Crossrefs

Programs

  • Mathematica
    With[{c=IntegerDigits[6!,2]},Select[Range[180],SequenceCount[ IntegerDigits[ #!,2], c]>0&]] (* The program uses the SequenceCount function from Mathematica version 10 *) (* Harvey P. Dale, Jul 26 2016 *)
  • PARI
    is(n)=n=n!; while(n>719, my(e=valuation(n, 2), e1=valuation((n>>=e)+1, 2)); n>>=e1; if(e>3 && e1==1 && bitand(n, 31)==22, return(1))); 0 \\ Charles R Greathouse IV, Apr 07 2013

A103681 Numbers m such that in binary representation m! does not contain 7!.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

Last term is probably 5153, since all numbers from 5154 to 5*10^5 do not belong to the sequence. - Giovanni Resta, Apr 07 2013

Crossrefs

Programs

  • Mathematica
    Select[Range[0, 100], !StringContainsQ[IntegerString[#!, 2], IntegerString[7!, 2]] &] (* Amiram Eldar, Apr 03 2025 *)
  • PARI
    is(n)=n=n!; while(n>5039, my(e=valuation(n, 2), e1=valuation((n>>=e)+1, 2)); n>>=e1; if(e>3 && e1==2 && bitand(n, 127)==78, return(0))); 1 \\ Charles R Greathouse IV, Apr 07 2013

Formula

A103675(a(n)) = 0, A103675(A103680(n)) = 1.

A103673 If in binary representation n! contains 5! then 1 else 0.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

a(A103676(n)) = 1, a(A103677(n)) = 0.
Conjecture: a(n) = 1 for n > 65. - Charles R Greathouse IV, Apr 07 2013

Crossrefs

Programs

  • PARI
    a(n)=n=n!;while(n>119, my(e=valuation(n,2),e1=valuation((n>>=e)+1,2)); n>>=e1; if(e>2 && e1>3, return(1))); 0 \\ Charles R Greathouse IV, Apr 07 2013

A103676 Numbers m such that in binary representation m! contains 5!.

Original entry on oeis.org

5, 10, 12, 22, 23, 25, 27, 29, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 54, 55, 56, 57, 59, 60, 61, 62, 63, 64, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

Complement of A103677: A103673(a(n))=1, A103673(A103677(n))=0.

Crossrefs

Programs

  • Mathematica
    Select[Range[100],MemberQ[Partition[IntegerDigits[#!,2],7,1],{1,1,1,1,0,0,0}]&] (* Harvey P. Dale, Apr 09 2012 *)
  • PARI
    is(n)=n=n!; while(n>119, my(e=valuation(n, 2), e1=valuation((n>>=e)+1, 2)); n>>=e1; if(e>2 && e1>3, return(1))); 2 \\ Charles R Greathouse IV, Apr 07 2013

A103677 Numbers m such that in binary representation m! doesn't contain 5!.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 9, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 24, 26, 28, 30, 31, 32, 36, 46, 53, 58, 65
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 12 2005

Keywords

Comments

Conjecture: there are no further terms;
complement of A103676: A103673(a(n))=0, A103673(A103676(n))=1.

Crossrefs

Programs

  • PARI
    is(n)=n=n!; while(n>119, my(e=valuation(n, 2), e1=valuation((n>>=e)+1, 2)); n>>=e1; if(e>2 && e1>3, return(0))); 1 \\ Charles R Greathouse IV, Apr 07 2013
Showing 1-10 of 12 results. Next