login
Fibonacci regression array: For n>=2 and 1<=k<=n-1, T(n,k) is the last term before the first nonpositive term in the sequence n, k, n-k, 2k-n, 2n-3k, 5k-3n, ...
3

%I #14 Jan 20 2018 22:30:22

%S 1,2,1,3,2,2,4,3,1,3,5,4,3,2,4,6,5,4,2,3,5,7,6,5,4,1,4,6,8,7,6,5,3,3,

%T 5,7,9,8,7,6,5,2,4,6,8,10,9,8,7,6,4,2,5,7,9,11,10,9,8,7,6,3,4,6,8,10,

%U 12,11,10,9,8,7,5,1,5,7,9,11,13,12,11,10,9,8,7,4,3,6,8,10,12,14,13,12,11

%N Fibonacci regression array: For n>=2 and 1<=k<=n-1, T(n,k) is the last term before the first nonpositive term in the sequence n, k, n-k, 2k-n, 2n-3k, 5k-3n, ...

%H G. C. Greubel, <a href="/A097367/b097367.txt">Table of n, a(n) for n = 1..1000</a>

%F For n > k >= 1, define d(1)=n, d(2)=k, d(j) = d(j-2) - d(j-1) for j >= 3. Then d(j) = F(j-2)*n - F(j-1)*k for odd j>=1 and d(j) = F(j-1)*k - F(j-2)*n for even j>=2, where F(h)=A000045(h) = h-th Fibonacci number. The sequence d is the aforementioned sequence n, k, n-k, 2k-n, 2n-3k, 5k-3n, ...

%F T(m*n,m*k) = m*T(n,k). - _Robert Israel_, Jan 20 2018

%e Rows 2,3,4,5,6:

%e 1

%e 2 1

%e 3 2 2

%e 4 3 1 3

%e 5 4 3 2 4

%e T(8,5)=1, last term before 0 in 8,5,3,2,1,1,0,1,-1,...

%e T(8,6)=4, last term before -2 in 8,6,2,4,-2,6,-8,14,...

%t f[n_] := Fibonacci[n]; d[n_, k_, 1] := n; d[n_, k_, 2] := k;

%t d[n_, k_, j_] := ((-1)^j) (k*f[j - 1] - n*f[j - 2]);

%t s[n_, k_] := Select[Range[100], d[n, k, # + 1] <= 0 &, 1];

%t t = Table[d[n, k, s[n, k]], {n, 2, 20}, {k, 1, n - 1}]; (* A097367 array *)

%t Flatten[t] (* A097367 sequence *)

%t Table[Min[Flatten[Table[d[n, k, s[n, k]], {k, 1, n - 1}]]], {n, 2, 100}] (* A097368 *)

%t (* _Clark Kimberling_, Oct 14 2016 *)

%Y Cf. A000045, A097368, A097369.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Aug 09 2004