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 1-3 of 3 results.

A232861 Numbers k with k - 1, k + 1, prime(k) - k, prime(k) + k, k*prime(k) - 1, k*prime(k) + 1 all prime.

Original entry on oeis.org

22110, 23742, 128238, 275592, 346560, 1061910, 1281522, 1339002, 1378188, 1461600, 1850130, 2064150, 2354952, 2478270, 2523708, 2689260, 2694300, 3916638, 4422618, 4933530, 6179082, 6541080, 6641562, 6740478, 6759030, 7315812, 8484798, 8711010, 9133308, 9687720
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 01 2013

Keywords

Comments

Obviously, each term of the sequence is a multiple of 6.
Conjecture: (i) This sequence contains infinitely many terms.
(ii) Let P(x) be a non-constant integer-valued polynomial with positive leading coefficient. Then, there are infinitely many positive integers k with prime(k) - k in the range P(Z) = {P(m): m is an integer}, if and only if the degree of P(x) is at most 3. We may also replace prime(k) - k by prime(k) + k.

Examples

			a(1) = 22110 with the six numbers 22110 - 1 = 22109, 22110 + 1 = 22111, prime(22110) - 22110 = 228841, prime(22110) + 22110 = 273061, 22110*prime(22110) - 1 = 5548526609, 22110*prime(22110) + 1 = 5548526611 all prime.
		

Crossrefs

Programs

  • Mathematica
    n=0
    Do[If[PrimeQ[k-1]&&PrimeQ[k+1]&&PrimeQ[Prime[k]-k]&& PrimeQ[Prime[k]+k]&& PrimeQ[k*Prime[k]-1]&& PrimeQ[k*Prime[k]+1],n=n+1;Print[n," ",k]],{k,1,9700000}]

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}]

A232862 Least positive integer m <= n^2/2 + 3 such that the set {prime(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, 3, 4, 11, 9, 8, 10, 15, 29, 13, 23, 22, 23, 37, 32, 28, 48, 44, 53, 41, 45, 67, 76, 117, 119, 91, 121, 88, 89, 101, 72, 88, 100, 143, 144, 185, 145, 104, 176, 141, 144, 175, 187, 213, 121, 255, 128, 129, 189, 243, 122, 267, 275, 242, 209, 205, 130, 153, 263, 335
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 01 2013

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
Note that a(4) = 11 = 4^2/2 + 3.

Examples

			a(3) = 4 since prime(1) - 1 = prime(2) - 2 = 1, prime(3) - 3 = 2, prime(4) - 4 = 3, and {1,2,3} is a complete system of residues modulo 3.
		

Crossrefs

Programs

  • Mathematica
      L[m_,n_]:=Length[Union[Table[Mod[Prime[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," ",0];Label[aa];Continue,{n,1,60}]
Showing 1-3 of 3 results.