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

Author

Joseph L. Pe, Nov 04 2002

Keywords

• nonn
• more
• changed

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

Examples

			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.
		

Crossrefs

Cf. A001221, A006049, A076251, A076252.

Programs

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]]]

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) \\ Lambert Klasen (lambert.klasen(AT)gmx.net), Nov 05 2005

Extensions

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

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