A068797 -id:A068797 - cp's OEIS frontend

A067665 The start of a record-setting run of consecutive integers i with distinct A001222(i).

1, 6, 15, 60, 726, 6318, 189375, 755968, 683441871, 33714015615

Offset: 1

G. L. Honaker, Jr., Feb 03 2002

The list of indices of record terms in A068796;
n is in the sequence if A068796(n) is larger than A068796(m) when n is larger than m. For the known terms, f(a(n)) = n+1. Is that true for all n? In other words, is the monotonic subsequence of A068796 identical to A020725?
a(11) > 10^13. - Giovanni Resta, Jan 08 2014

			The values of f(n) for n=1 to 15 are 2,1,2,2,2,3,3,2,1,3,2,3,2,1,4. Records occur at f(1)=2, f(6)=3 and f(15)=4.

J.-M. De Koninck, J. B. Friedlander, and F. Luca, On strings of consecutive integers with a distinct number of prime factors, Proc. Amer. Math. Soc., 137 (2009), 1585-1592.

Cf. A001222, A067650, A068069, A068796, A068797.

bigomega[n_] := Plus@@Last/@FactorInteger[n]; f[n_] := For[k=1; s={bigomega[n]}, True, k++, If[MemberQ[s, z=bigomega[n+k]], Return[k], AppendTo[s, z]]]; For[n=1; max=0, True, n++, If[f[n]>max, Print[n, " ", max=f[n]]]]

a(n,lim=1e12,startAt=1)={
  forstep(i=startAt-1,lim,10^6-n,
    my(v=vectorsmall(min(10^6,lim\1-i),j,bigomega(j+i)));
    for(j=n,#v,if(#vecsort(v[j-n+1..j],,8)==n,return(j+i-n+1)))
  )
}; \\ Charles R Greathouse IV, Jul 03 2013

More terms from Shyam Sunder Gupta, Feb 08 2002
Edited by Robert G. Wilson v, Feb 20 2002
Edited by Dean Hickerson, Mar 05 2002
a(10) from Donovan Johnson, Oct 15 2008

A068796 Maximum k such that k consecutive integers starting at n have distinct numbers of prime factors (counted with multiplicity).

Original entry on oeis.org

2, 1, 2, 2, 2, 3, 3, 2, 1, 3, 2, 3, 2, 1, 4, 3, 2, 2, 3, 2, 1, 3, 3, 2, 1, 2, 1, 2, 2, 4, 3, 2, 1, 1, 3, 3, 2, 1, 4, 3, 2, 2, 2, 1, 4, 3, 4, 3, 2, 2, 4, 4, 3, 2, 2, 2, 1, 3, 2, 5, 4, 3, 3, 4, 3, 2, 3, 2, 4, 3, 2, 4, 3, 2, 1, 2, 5, 5, 4, 4, 3, 3, 3, 2, 1, 1, 3, 2, 4, 3, 2, 2, 1, 1, 4, 3, 2, 1, 3, 3, 2, 3, 4

Offset: 1

Dean Hickerson, Mar 05 2002

nonn

The number of prime factors (counted with multiplicity) of n is bigomega(n) = A001222(n).

			a(6)=3 because 6, 7, 8 and 9 have, respectively, 2, 1, 3 and 2 prime factors; the first 3 of these are distinct.

Zak Seidov, Table of n, a(n) for n = 1..10000

Cf. A001222, A067665, A068797.

bigomega[n_] := Plus@@Last/@FactorInteger[n]; a[n_] := For[k=1; s={bigomega[n]}, True, k++, If[MemberQ[s, z=bigomega[n+k]], Return[k], AppendTo[s, z]]]
ss={}; Do[s={PrimeOmega[n]};k=1;While[FreeQ[s, (b=PrimeOmega[n+k])],s=AppendTo[s,b];k++];ss=AppendTo[ss,k],{n,103}]; (* Zak Seidov, Nov 09 2015 *)

cp's OEIS Frontend

A067665 The start of a record-setting run of consecutive integers i with distinct A001222(i).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Extensions