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.

A282423 a(n) = smallest k such that A282026(k) = n, or 0 if no such k exists.

Original entry on oeis.org

3, 2, 0, 13, 19, 0, 427, 4, 0, 0, 1, 0, 802, 99412, 0, 3097, 7, 0, 637, 0, 0, 7225627, 30898822, 0, 0, 280134277, 0, 31705902442, 43190647, 0, 965577112
Offset: 1

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Author

Andrey Zabolotskiy and Altug Alkan, Feb 14 2017, following a suggestion from N. J. A. Sloane

Keywords

Comments

a(n) is nonzero if n is in A282429.
For n>4 and nonzero a(n), 2*a(n)+3 is in A022004. For n>8 and nonzero a(n), 2*a(n)+3 is also in A153417. For n>16 and nonzero a(n), 2*a(n)+3 is also in A049481.

Examples

			a(10) = 0. Proof: Suppose 10 is a term of A282026. For the corresponding n, 2*n + 1 cannot be divisible by 5 because of A282026’s definition (gcd(10, 2*n + 1) = 1). So 2*n + 1 can be only of the form 10*k + 1, 10*k + 3, 10*k + 7, 10*k + 9. But 10*k + 1 + 2*2, 10*k + 3 + 2*1, 10*k + 7 + 2*4, 10*k + 9 + 2*8 are all composite and 1, 2, 4, 8 are relatively prime to any odd number. Since all of them are smaller than 10, this is the contradiction to the assumption that 10 is the term which is the smallest number for corresponding n. This also proves that a(5*k) = 0 for any k > 1.
		

Crossrefs

A282429 List of distinct terms of A282026.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 11, 13, 14, 16, 17, 19, 22, 23, 26, 28, 29, 31
Offset: 1

Views

Author

Altug Alkan and Andrey Zabolotskiy, Feb 15 2017, following a suggestion from N. J. A. Sloane

Keywords

Comments

a(n) occurs in A282026 for the first time at the position A282423(a(n)).

Examples

			3 is not a term. Proof: Suppose 3 is a term of A282026. For the corresponding n, 2*n + 1 cannot be divisible by 3 because of A282026’s definition (gcd(3, 2*n + 1) = 1). So 2*n + 1 can be only of the form 6*k + 1 or 6*k + 5. But 6*k + 1 + 2*1 and 6*k + 5 + 2*2 are both composite numbers and 1, 2 are relatively prime to any odd number. Since they are smaller than 3, this is the contradiction to the assumption that 3 is the term which is the smallest number for corresponding n. This also proves that 3*k cannot be a term of this sequence for any k >= 1.
		

Crossrefs

Programs

  • Mathematica
    Union@ Table[m = 1; While[Nand[CoprimeQ[m, 2 n + 1], CompositeQ[2 (n + m) + 1]], m++]; m, {n, 0, 10^7}] (* Michael De Vlieger, Feb 18 2017 *)

A282194 a(n) = smallest positive k such that 2*n + 2^k + 1 is composite.

Original entry on oeis.org

3, 5, 2, 1, 4, 2, 1, 7, 2, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 4, 2, 1, 2, 1, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 3, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 4, 2, 1, 4, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 3, 2, 1, 1, 1, 1, 4, 2, 1, 3, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1
Offset: 0

Views

Author

Altug Alkan, Feb 15 2017

Keywords

Comments

Least k such that a(k) = n are 3, 2, 0, 4, 1, 112, 7, 32917, 802, 9712, 1198673602 for the initial terms.

Examples

			a(1) = 5 because 3 + 2^k is prime for 0 < k < 5 and 3 + 2^5 = 35 is composite.
		

Crossrefs

Programs

  • Mathematica
    spk[n_]:=Module[{k=1},While[!CompositeQ[2n+2^k+1],k++];k]; Array[spk,110,0] (* Harvey P. Dale, Apr 26 2017 *)
  • PARI
    a(n) = my(k=1); while(isprime(2*n+2^k+1), k++); k;
Showing 1-3 of 3 results.