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.

A232616 Least positive integer m such that {2^k - k: k = 1,...,m} contains a complete system of residues modulo n.

Original entry on oeis.org

1, 2, 4, 5, 10, 6, 14, 10, 12, 18, 29, 13, 33, 22, 40, 19, 38, 18, 58, 21, 36, 58, 75, 26, 60, 66, 40, 64, 195, 53, 87, 36, 158, 67, 130, 37, 133, 94, 90, 42, 95, 42, 105, 112, 112, 140, 247, 51, 122, 94, 119, 120, 311, 54, 126, 90, 184, 223, 264, 61
Offset: 1

Views

Author

Zhi-Wei Sun, Nov 26 2013

Keywords

Comments

By a result of the author (see arXiv:1312.1166), for any integers a and n > 0, the set {a^k - k: k = 1, ..., n^2} contains a complete system of residues modulo n. (We may also replace a^k - k by a^k + k.) Thus a(n) always exists and it does not exceed n^2.
Conjectures:
(i) a(n) < 2*(prime(n)-1) for all n > 0.
(ii) The Diophantine equation x^n - n = y^m with m, n, x, y > 1 only has two integral solutions: 2^5 - 5 = 3^3 and 2^7 - 7 = 11^2. Also, the Diophantine equation x^n + n = y^m with m, n, x, y > 1 only has two integral solutions: 5^2 + 2 = 3^3 and 5^3 + 3 = 2^7.

Examples

			a(3) = 4 since {2 - 1, 2^2 - 2, 2^3 - 3} = {1, 2, 5} does not contain a complete system of residues mod 3, but {2 - 1, 2^2 - 2, 2^3 - 3, 2^4 - 4} = {1, 2, 5, 12} does.
		

Crossrefs

Programs

  • Mathematica
    L[m_,n_]:=Length[Union[Table[Mod[2^k-k,n],{k,1,m}]]]
    Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2}];
    Print[n," ",0];Label[aa];Continue,{n,1,60}]

A232894 Least positive integer m such that {Catalan(k) - k: k = 1, ..., m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

Original entry on oeis.org

1, 5, 4, 11, 16, 13, 31, 27, 18, 22, 34, 52, 45, 45, 31, 112, 57, 73, 113, 99, 64, 77, 114, 215, 134, 106, 89, 99, 127, 209, 161, 239, 135, 178, 96, 207, 185, 172, 157, 231, 174, 195, 309, 115, 274, 309, 386, 239, 200, 336, 188, 199, 181, 181, 116, 311, 229, 290, 663, 239
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 02 2013

Keywords

Comments

Conjecture: (i) Let n be any positive integer. Then 0 < a(n) <= n^2/2 + 7. Also, {Catalan(k) + k: k = 1, ..., [n^2/2] + 23} contains a complete system of residues modulo n, where [.] is the floor function.
(ii) For any integer n > 3, neither Catalan(n) - n nor Catalan(n) + n has the form x^m with m > 1 and x > 1.

Examples

			a(2) = 5 since Catalan(k) - k is even for each k = 1, 2, 3, 4, and Catalan(5) - 5 = 37 is odd.
		

Crossrefs

Programs

  • Mathematica
    L[m_,n_]:=Length[Union[Table[Mod[CatalanNumber[k]-k,n],{k,1,m}]]]
    Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+7}];
    Print[n," ",counterexample];Label[aa];Continue,{n,1,60}]

A232898 Least positive integer m such that {C(2k,k) + k: k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

Original entry on oeis.org

1, 2, 7, 5, 10, 12, 9, 24, 31, 22, 59, 25, 27, 30, 42, 56, 123, 66, 57, 72, 84, 78, 73, 132, 136, 57, 99, 80, 129, 211, 170, 226, 121, 170, 126, 129, 238, 218, 157, 132, 348, 198, 388, 103, 171, 166, 247, 181, 205, 352, 194, 136, 430, 226, 117, 224, 237, 292, 364, 241
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 02 2013

Keywords

Comments

Conjecture: (i) Let n be any positive integer. Then 0 < a(n) <= n^2/2 + 3. Also, {C(2k,k) - k: k = 1, ..., [n^2/2] + 15} contains a complete system of residues modulo n, where [.] is the floor function.
(ii) For any integer n > 2, neither C(2n,n) + n nor C(2n,n) - n has the form x^m with m > 1.

Examples

			a(2) = 2 since C(2*1,1) + 1 = 3 is odd and C(2*2,2) + 2 = 8 is even.
		

Crossrefs

Programs

  • Mathematica
    L[m_,n_]:=Length[Union[Table[Mod[Binomial[2k,k]+k,n],{k,1,m}]]]
    Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+3}];
    Print[n," ",counterexample];Label[aa];Continue,{n,1,60}]

A232891 Least positive integer m <= n^2/2 + 3 such that {k*prime(k): k = 1,...,m} contains a complete system of residues modulo n, or 0 if such a number m does not exist.

Original entry on oeis.org

1, 3, 4, 11, 7, 7, 10, 17, 43, 13, 51, 22, 51, 36, 31, 49, 64, 71, 119, 73, 86, 68, 141, 110, 153, 85, 83, 86, 144, 81, 174, 127, 115, 87, 122, 138, 143, 134, 133, 142, 211, 229, 152, 104, 109, 177, 259, 142, 194, 176, 196, 311, 312, 193, 243, 197, 396, 169, 156, 171
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 02 2013

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) For any positive integer n not equal to 3, the number n*prime(n) + 1 cannot be a power x^m with m > 1.
(iii) There are infinitely many positive integers n with n - 1, n + 1, n + prime(n), n + prime(n)^2, n^2 + prime(n), n^2 + prime(n)^2 all prime.

Examples

			a(3) = 4 since 1*prime(1) = 2, 2*prime(2) === 3*prime(3) == 0 (mod 3), and 4*prime(4) = 28 == 1 (mod 3).
		

Crossrefs

Programs

  • Mathematica
    L[m_,n_]:=Length[Union[Table[Mod[k*Prime[k],n],{k,1,m}]]]
    Do[Do[If[L[m,n]==n,Print[n," ",m];Goto[aa]],{m,1,n^2/2+3}];
    Print[n," ",counterexample];Label[aa];Continue,{n,1,60}]
Showing 1-4 of 4 results.