editing

approved

f[n_] := Count[Table[Mod[k, 6], {k, Prime[n], Prime[n + 1] - 1}], 3]; Table[f[n], {n, 120}] (* _Ray Chandler_, Oct 17 2006 *)

approved

editing

_Giovanni Teofilatto (g.teofilatto(AT)tiscalinet.it), _, May 02 2005

Corrected and extended by _Ray Chandler (rayjchandler(AT)sbcglobal.net), _, Oct 17 2006

easy,nonn,new

Corrected and extended by Ray Chandler (RayChandlerrayjchandler(AT)alumni.tcusbcglobal.edunet), Oct 17 2006

Number of numbers of the form 6k+3=m with prime(n) <= m 6k+3 < prime(n+1).

0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 2, 1, 1, 0, 2, 0, 1, 1, 1, 1, 1, 0, 2, 0, 1, 0, 2, 2, 1, 0, 1, 1, 0, 2, 1, 1, 1, 0, 1, 1, 0, 2, 2, 1, 0, 1, 2, 1, 2, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 0, 2, 0, 1, 1, 1, 1, 1, 0, 1, 2, 1, 1, 1, 1, 1, 2, 0, 3, 1, 2, 1, 1, 0, 1

f[n_] := Count[Table[Mod[k, 6], {k, Prime[n], Prime[n + 1] - 1}], 3]; Table[f[n], {n, 120}] (*Chandler*)

easy,nonn,new

Corrected and extended by Ray Chandler (RayChandler(AT)alumni.tcu.edu), Oct 17 2006

Number of numbers of the form 6k+3=m with prime(n)<= m < prime(n+1).

0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 2, 1, 1, 0, 2, 0, 1, 1, 1, 1, 1, 2, 1, 0, 1, 0, 1, 2, 1, 0, 1, 1, 0, 2, 1, 1, 1, 1, 1, 0, 2, 2, 1, 0, 1

1,30

a(4)=1 because between prime(4)=7 and prime(5)=11 there is one number of the form 6k+3: 9.

easy,nonn

Giovanni Teofilatto (g.teofilatto(AT)tiscalinet.it), May 02 2005

approved