A076253 a(n) = the least positive integer solution of the "n-th omega recurrence" omega(k) = omega(k-1) + ... + omega(k-n), if such k exists; = 0 otherwise. (omega(n) denotes the number of distinct prime factors of n.)

3, 3, 2310, 746130, 601380780, 89419589469210, 489423552293946270

Offset

1,1

Comments

Question: Is a(n) > 0 for all n, i.e. can the n-th omega recurrence be solved for all n?

Note that 601380780 is not squarefree. Using primorials, I easily found candidates up to a(8) - Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

Links

Table of n, a(n) for n=1..7.

Example

k=3 is the smallest solution of omega(k)=omega(k-1), so a(1)=3. k=3 is the smallest solution of omega(k)=omega(k-1)+omega(k-2), so a(2)=3. k=2310 is the smallest solution of omega(k)=omega(k-1)+omega(k-2)+omega(k-3), so a(3)=2310.

Mathematica

(*code to find a(4)*) omega[n_] := Length[FactorInteger[n]]; ub = 2*10^6; For[i = 2, i <= ub, i++, a[i] = omega[i]]; start = 5; For[j = start, j <= ub, j++, If[a[j] == a[j - 1] + a[j - 2] + a[j - 3] + a[j - 4], Print[j]]]

Prog

(PARI) /* find a(5) */ v=[0, 0, 0, 0, 0]; s=0; for(i=1, 5, v[i]=omega(i); s+=v[i]) for(i=6, 10^10, o=omega(i); if(o==s, print(i); break); s-=v[i%5+1]; s+=o; v[i%5+1]=o) (Klasen)

Crossrefs

Cf. A001221.

Sequence in context: A009011 A291095 A009498 * A081174 A292832 A264932

Adjacent sequences: A076250 A076251 A076252 * A076254 A076255 A076256

Keyword

nonn

Author

Joseph L. Pe, Nov 04 2002

Extensions

a(5) from Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

a(6)-a(7) from Donovan Johnson, Feb 07 2009

Status

approved