OFFSET
1,3
COMMENTS
Consider a version of Pascal's Triangle: a triangular array with a single 1 on row 0, with numbers below equal to the sum of the two numbers above it if and only if that sum appears in the Fibonacci sequence A000045. If the sum does not appear in A000045, a 1 is put in its place.
So the first few rows would be as follows:
row 0: 1
row 1: 1 1
row 2: 1 2 1
row 3: 1 3 3 1
row 4: 1 1 1 1 1
row 5: 1 2 2 2 2 1
row 6: 1 3 1 1 1 3 1
row 7: 1 1 1 2 2 1 1 1
row 8: 1 2 2 3 1 3 2 2 1
row 9: 1 3 1 5 1 1 5 1 3 1
...
a(n) is the row number in which the n-th Fibonacci number first appears in this triangular array.
a(16) > 2.2*10^5. - David A. Corneth, Mar 25 2019
a(16) > 3.2*10^6. - Daniel Suteu, Mar 26 2019
a(16) > 1.5*10^7. - Bert Dobbelaere, Apr 02 2019
LINKS
David A. Corneth, PARI program
MATHEMATICA
Block[{s = Array[Fibonacci, 20], t}, t = Nest[Append[#1, (PadLeft[#1[[-1]], #2] + PadRight[#1[[-1]], #2]) /. k_Integer /; FreeQ[s, k] -> 1] & @@ {#, Length@ # + 1} &, {{1}}, 10^4]; -1 + TakeWhile[Map[FirstPosition[t, #][[1]] &, s], IntegerQ]] (* Michael De Vlieger, Mar 24 2019 *)
PROG
(PARI) isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
lista(nn) = {print1(0, ", ", 0, ", "); v = [1, 1]; nextf = 3; for (n=2, nn, w = vector(n+1); w[1] = v[1]; for (j=2, n, w[j] = v[j-1]+ v[j]; if (!isfib(w[j]), w[j] = 1)); w[n+1] = v[n]; sw = vecsort(w, , 8); if (vecsearch(sw, fibonacci(nextf)), print1(n, ", "); nextf++); v = w; ); } \\ Michel Marcus, Mar 22 2019
(PARI) See Corneth link \\ David A. Corneth, Mar 25 2019
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Elliott Line, Mar 22 2019
EXTENSIONS
a(14)-a(15) from Michel Marcus, Mar 22 2019
STATUS
approved