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%I #53 Mar 11 2024 08:28:58
%S 6,12,25,48,94,184,363,719,1430,2851,5691,11371,22728,45443,90870
%N a(n) is the largest k in an n_nacci(k) sequence (Fibonacci(k) for n=2, tribonacci(k) for n=3, etc.) such that n_nacci(k) >= 2^(k-n-1).
%C From _Frank M Jackson_, Jul 02 2023: (Start)
%C Define the n_nacci sequence, basically row n in A092921, with an offset of 0, n_nacci(k) = 0 for 0 <= k <= n-2 and n_nacci(n-1) = 1. Thereafter, n_nacci(k) for k >= n continues as the sum of its previous n terms.
%C This means that n_nacci(k) = 2^(k-n) for n <= k <= 2n-1. In the limit as n tends to infinity the n_nacci sequence after an initial large set of zeros followed by 1 has successive terms of ascending powers of 2.
%C As the n-acci constants, (A001622, A058265, A086088, A103814,...) are smaller than 2, for each n_nacci sequence there is a largest k such that n_nacci(k) >= 2^(k-n-1). (End)
%e For n=3, the tribonacci sequence is 0,0,1,1,2,4,7,...,149,274,504,... and the 13th term is 504 < 512 so a(n)=12 because 274 is greatest term >= 2^(12-3-1) = 256.
%p nAcci := proc(n,k)
%p option remember ;
%p if k <= n-2 then
%p 0;
%p elif k = n-1 then
%p 1;
%p else
%p add( procname(n,i),i=k-n..k-1) ;
%p end if;
%p end proc:
%p A202805 := proc(n)
%p local k ;
%p for k from n do
%p if nAcci(n,k) < 2^(k-n-1) then
%p return k-1;
%p end if;
%p end do:
%p end proc:
%p for n from 2 do
%p print(n,A202805(n)) ;
%p end do: # _R. J. Mathar_, Mar 11 2024
%t fib[n_, m_] := (Block[{nacci}, (Do[nacci[g]=0, {g, 0, m - 2}];
%t nacci[m-1]=1;nacci[p_] := (nacci[p]=Sum[nacci[h], {h, p-m, p-1}]);nacci[n])]);
%t crossover[q_] := (Block[{$RecursionLimit=Infinity}, (k=0;While[fib[k+q+1, q]>=2^k, k++];k+q)]);
%t Table[crossover[j], {j, 2, 12}]
%Y Cf. A000045, A000073, A000078.
%K nonn,more
%O 2,1
%A _Frank M Jackson_, Dec 24 2011
%E Edited by _N. J. A. Sloane_, May 20 2023
%E There seems to be an error in the Comment. See "History" tab. - _N. J. A. Sloane_, Jun 24 2023
%E Removed musing about what might define "complete" sequences. - _R. J. Mathar_, Mar 11 2024
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