A255292 Number of 2's in expansion of F^n mod 3, where F = 1/x+1+x+1/y+y.
0, 0, 9, 0, 0, 37, 9, 45, 44, 0, 0, 45, 0, 0, 177, 37, 185, 156, 9, 45, 72, 45, 225, 228, 44, 220, 573, 0, 0, 45, 0, 0, 185, 45, 225, 220, 0, 0, 225, 0, 0, 877, 177, 885, 716, 37, 185, 256, 185, 925, 788, 156, 780, 2281, 9
Offset: 0
Keywords
Examples
The pairs [no. of 1's, no. of 2's] are [1, 0], [5, 0], [4, 9], [5, 0], [25, 0], [12, 37], [4, 9], [20, 45], [69, 44], [5, 0], [25, 0], [20, 45], [25, 0], [125, 0], [52, 177], [12, 37], [60, 185], [281, 156], [4, 9], [20, 45], [97, 72], [20, 45], [100, 225], [353, 228], [69, 44], [345, 220], [448, 573], [5, 0], [25, 0], [20, 45], ...
Programs
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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; F3:=1/x+1+x+1/y+y mod 3; g:=(F,n)->expand(F^n) mod 3; [seq(C3(g(F3,n))[2],n=0..60)];
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