A255294 Number of 2's in expansion of F^n mod 3, where F = 1/x+2+x+1/y+y.
0, 1, 5, 1, 8, 20, 5, 28, 64, 1, 8, 28, 8, 49, 101, 20, 109, 241, 5, 28, 80, 28, 149, 289, 64, 305, 437, 1, 8, 28, 8, 49, 109, 28, 149, 305, 8, 49, 149, 49, 272, 524, 101, 532, 1096, 20, 109, 305, 109, 572, 1096, 241, 1160
Offset: 0
Keywords
Examples
The pairs [no. of 1's, no. of 2's] are [1, 0], [4, 1], [8, 5], [4, 1], [17, 8], [29, 20], [8, 5], [37, 28], [49, 64], [4, 1], [17, 8], [37, 28], [17, 8], [76, 49], [128, 101], [29, 20], [136, 109], [196, 241], [8, 5], [37, 28], [89, 80], [37, 28], [176, 149], [292, 289], [49, 64], [260, 305], [584, 437], [4, 1], [17, 8], [37, 28], ...
Programs
-
Maple
# C3 Counts 1's and 2's C3 := proc(f) local c,ix,iy,f2,i,t1,t2,n1,n2; f2:=expand(f) mod 3; n1:=0; n2:=0; if whattype(f2) = `+` then t1:=nops(f2); for i from 1 to t1 do t2:=op(i, f2); ix:=degree(t2, x); iy:=degree(t2, y); c:=coeff(coeff(t2,x,ix),y,iy); if (c mod 3) = 1 then n1:=n1+1; else n2:=n2+1; fi; od: RETURN([n1,n2]); else ix:=degree(f2, x); iy:=degree(f2, y); c:=coeff(coeff(f2,x,ix),y,iy); if (c mod 3) = 1 then n1:=n1+1; else n2:=n2+1; fi; RETURN([n1,n2]); fi; end; F4:=1/x+2+x+1/y+y mod 3; g:=(F,n)->expand(F^n) mod 3; [seq(C3(g(F4,n))[2],n=0..60)];
Comments