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A202805
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).
1
6, 12, 25, 48, 94, 184, 363, 719, 1430, 2851, 5691, 11371, 22728, 45443, 90870
OFFSET
2,1
COMMENTS
From Frank M Jackson, Jul 02 2023: (Start)
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.
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.
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)
EXAMPLE
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.
MAPLE
nAcci := proc(n, k)
option remember ;
if k <= n-2 then
0;
elif k = n-1 then
1;
else
add( procname(n, i), i=k-n..k-1) ;
end if;
end proc:
A202805 := proc(n)
local k ;
for k from n do
if nAcci(n, k) < 2^(k-n-1) then
return k-1;
end if;
end do:
end proc:
for n from 2 do
print(n, A202805(n)) ;
end do: # R. J. Mathar, Mar 11 2024
MATHEMATICA
fib[n_, m_] := (Block[{nacci}, (Do[nacci[g]=0, {g, 0, m - 2}];
nacci[m-1]=1; nacci[p_] := (nacci[p]=Sum[nacci[h], {h, p-m, p-1}]); nacci[n])]);
crossover[q_] := (Block[{$RecursionLimit=Infinity}, (k=0; While[fib[k+q+1, q]>=2^k, k++]; k+q)]);
Table[crossover[j], {j, 2, 12}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Frank M Jackson, Dec 24 2011
EXTENSIONS
Edited by N. J. A. Sloane, May 20 2023
There seems to be an error in the Comment. See "History" tab. - N. J. A. Sloane, Jun 24 2023
Removed musing about what might define "complete" sequences. - R. J. Mathar, Mar 11 2024
STATUS
approved