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A076183
a(n) = the least positive integer k satisfying Omega(k) = Omega(k-1)+...+Omega(k-n) if such k exists; = 0 otherwise. (Omega(n) (A001222) denotes the number of prime factors of n, counting multiplicity.)
1
3, 3, 4, 1440, 18432, 516096, 2621440, 150994944, 4416602112, 91729428480, 253671505920, 184717953466368, 4714705859903488, 74309393851613184, 1215971899390033920
OFFSET
1,1
COMMENTS
1. What is the value of a(7)? For n=7, I have not found a solution k less than 10^6. 2. Is a(n) > 0 for all n, i.e. does a solution k to the "k-th Omega recursion" always exist? If not, what is the first n with a(n) = 0?
a(13) <= 4714705859903488. - David Wasserman, Apr 12 2005
a(16) > 2^63. [From Donovan Johnson, Sep 27 2008]
EXAMPLE
k=3 is the least solution of Omega(k) = Omega(k-1), so a(1) = 3. k=3 is the least solution of Omega(k) = Omega(k-1)+Omega(k-2), so a(2) = 3. k=4 is the least solution of Omega(k) = Omega(k-1)+Omega(k-2)+Omega(k-3), so a(3) = 4. k=1440 is the least solution of Omega(k) = Omega(k-1)+Omega(k-2)+Omega(k-3)+Omega(k-4), so a(4) = 1440.
MATHEMATICA
(*Code to find a(6)*) Omega[n_] := Apply[Plus, Transpose[FactorInteger[n]][[2]]]; ub = 10^6; For[i = 2, i <= ub, i++, a[i] = Omega[i]]; start = 8; For[j = start, j <= ub, j++, If[a[j] == a[j - 1] + a[j - 2] + a[j - 3] + a[j - 4] + a[j - 5] + a[j - 6], Print[j]]]
CROSSREFS
Sequence in context: A306521 A300369 A304296 * A348679 A011445 A197137
KEYWORD
hard,nonn
AUTHOR
Joseph L. Pe, Nov 01 2002
EXTENSIONS
More terms from David Wasserman, Apr 12 2005
a(13)-a(15) from Donovan Johnson, Sep 27 2008
STATUS
approved