A338994 Table read by antidiagonals: if x(n+1) = A001414(x(n-1)) + A001414(x(n)) with x(0) = i and x(1) = j, then T(i,j) is the first k such that (x(k), x(k+1)) is a fixed point or a member of a cycle. If there is no such k, then T(i,j) = -1.
2, 20, 21, 19, 19, 20, 15, 18, 10, 16, 10, 18, 18, 10, 11, 10, 9, 14, 9, 16, 11, 8, 9, 17, 14, 15, 16, 9, 14, 15, 17, 17, 14, 15, 15, 11, 14, 14, 8, 17, 9, 14, 9, 15, 11, 8, 14, 14, 13, 9, 9, 13, 15, 15, 9, 13, 15, 14, 13, 8, 9, 9, 14, 15, 15, 14, 8, 12, 8, 13, 16, 8, 9, 13, 14, 9, 7, 9, 9, 15, 8
Offset: 1
Examples
Table begins 2, 20, 19, 15, 10, 10, 8, 14, 14, 8, 13, 8, ... 21, 19, 18, 18, 9, 9, 15, 14, 14, 15, 12, 15, ... 20, 10, 18, 14, 17, 17, 8, 14, 14, 8, 8, 8, ... 16, 10, 9, 14, 17, 17, 13, 13, 13, 13, 12, 13, ... 11, 16, 15, 14, 9, 9, 8, 16, 16, 8, 12, 8, ... 11, 16, 15, 14, 9, 9, 8, 16, 16, 8, 12, 8, ... 9, 15, 9, 13, 9, 9, 7, 14, 14, 7, 12, 7, ... 11, 15, 15, 14, 13, 13, 12, 13, 13, 12, 15, 12, ... 11, 15, 15, 14, 13, 13, 12, 13, 13, 12, 15, 12, ... 9, 15, 9, 13, 9, 9, 7, 14, 14, 7, 12, 7, ... 14, 7, 6, 6, 23, 23, 4, 16, 16, 4, 12, 4, ... 9, 15, 9, 13, 9, 9, 7, 14, 14, 7, 12, 7, ... T(1,7) = 8 because starting at x(0)=1, x(1)=7 we have x(2)=7, x(3)=14, x(4)=16, x(5)=17, x(6)=25, x(7)=27, x(8)=19, x(9)=28, and (19,28) is in the cycle (19, 28, 30, 21, 20).
Links
- Robert Israel, Table of n, a(n) for n = 1..10011 (first 141 antidiagonals, flattened)
Programs
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Maple
spf:= proc(n) local t; add(t[1]*t[2],t=ifactors(n)[2]) end proc: Cyc:= {[0, 0], [16, 16], [32, 33], [33, 24], [24, 23], [23, 32], [28, 30], [30, 21], [21, 20], [20, 19], [19, 28], [34, 42], [42, 31], [31, 43], [43, 74], [74, 82], [82, 82], [82, 86], [86, 88], [88, 62], [62, 50], [50, 45], [45, 23], [23, 34]}: f:= proc(t) local count,x; count:= 0; x:= t; while count < 1000 do if member(x,Cyc) then return count fi; x:= [x[2],spf(x[1])+spf(x[2])]; count:= count+1; od; FAIL end proc: seq(seq(f([i,k-i]),i=1..k-1),k=2..14);
Comments