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.

A051699 Distance from n to closest prime.

Original entry on oeis.org

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

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Examples

			Closest primes to 0,1,2,3,4 are 2,2,2,3,3.
		

Crossrefs

Programs

  • Maple
    A051699 := proc(n) if isprime(n) then 0; elif n<= 2 then 2-n ; else min(nextprime(n)-n, n-prevprime(n)) ; end if ; end proc; # R. J. Mathar, Nov 01 2009
  • Mathematica
    FormatSequence[ Table[Min[Abs[n-If[n<2, 2, Prime[{#, #+1}&[PrimePi[n]]]]]], {n, 0, 101}], 51699, 0, Name->"Distance to closest prime." ]
    (* From version 6 on: *) a[?PrimeQ] = 0; a[n] := Min[NextPrime[n]-n, n-NextPrime[n, -1]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Apr 05 2012 *)
  • PARI
    a(n)=if(n<1,2*(n==0),vecmin(vector(n,k,abs(n-prime(k)))))
    
  • PARI
    a(n)=if(n<1,2*(n==0),min(nextprime(n)-n,n-precprime(n)))
    
  • Python
    from sympy import prevprime, nextprime, isprime
    def A051699(n): return nextprime(n) - n if n <= 1 else 0 if isprime(n) else min(n-prevprime(n), nextprime(n)-n) # Ya-Ping Lu, Mar 22 2025

Formula

Conjecture: S(n) = Sum_{k=1..n} a(k) is asymptotic to C*n*log(n) with C=0.29...... - Benoit Cloitre, Aug 11 2002
C = lim_{n->oo} S(n)/(n*log(n)) = 0.44 approximately. - Ya-Ping Lu, Apr 06 2025
Comment from Giorgio Balzarotti, Sep 18 2005: by means of the Prime Number Theorem is possible to derive the following inequality: c1*n*log(n) < S(n) < c2*n*log(n), where c1 = 1/4 and c2 = 3/8 (for n > 130). For a more accurate estimation of the values for c1 and c2, it necessary to know the number of twin primes with respect to the total number of primes.
abs(a(n)-a(n+1)) = 1 if n != 2; a((p+q)/2 +- k) = (q-p)/2 - k, where p < q are two consecutive primes and k = 0, 1, 2, ..., (q-p)/2. - Ya-Ping Lu, Mar 22 2025

Extensions

More terms from James Sellers