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 1-2 of 2 results.

A169818 Rank of n-th prime as defined in A177854.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 2, 2, 2, 1, 3, 2, 2, 3, 2, 3, 2, 2, 2, 3, 3, 3, 2, 3, 2, 3, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 3, 3, 2, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 3, 3
Offset: 1

Views

Author

T. D. Noe, May 28 2010

Keywords

Comments

This notion of rank is closely related to the Erdős-Selfridge classification of primes.

Crossrefs

For records see A177854.

Programs

  • Mathematica
    rank[1]=0; rank[2]=0; rank[3]=1;
    SetAttributes[rank,Listable];
    rank[p_] := rank[p] = 1+Min[Max@@rank[First/@FactorInteger[p-1]], Max@@rank[First/@FactorInteger[p+1]]]; rank[Prime[Range[100]]]

Formula

a(A000720(A141453(n)))=1 n>1. [From R. J. Mathar, May 28 2010]

A239374 Smallest product of consecutive distinct prime factors of t = prime(n)^2 - 1 in ascending order that provides more than 1/3 factored parts for Brillhart-Lehmer-Selfridge primality test for prime(n).

Original entry on oeis.org

2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 6
Offset: 2

Views

Author

Lei Zhou, Mar 17 2014

Keywords

Comments

The first greater than 2 element of this sequence is a(99).

Examples

			n = 2: prime(2) = 3, 3^2 - 1 = 8 = 2^3, 2^3 > 3, 100% factorization.  So a(2) = 2.
n = 45: prime(45) = 197, 197^2 - 1 = 38808 = 2^3*3^2*7^2*11, 2^3 = 8,  log_197(8) = 0.3936 > 1/3, 39.36% factorization.  So a(45) = 2.
n = 99: prime(99) = 523, 523^2 - 1 = 273528 = 2^3*3^2*29*131, 2^3 = 8, log_523(8) = 0.3322 < 1/3, log_523(2^3*3^2) = 0.6832 > 1/3, 68.32% factorization. So a(99) = 6.
		

Crossrefs

Programs

  • Mathematica
    Table[p = Prime[n]; ck = p^(1/3); sp = p^2 - 1; dp = sp; prod = 1; fp = Union[Transpose[FactorInteger[p + 1]][[1]], Transpose[FactorInteger[p - 1]][[1]]]; i = 0; While[i++; m = fp[[i]]; prod = prod*m; While[Divisible[sp, m], sp = sp/m]; (dp/sp) < ck]; prod, {n, 2, 100}]
Showing 1-2 of 2 results.