Assume that p, q, and k are integers for which the conditions 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ and 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ are satisfied. The initial values 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ, 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ, together with the recursive rule 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ define the non-negative integer sequence 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ. In this paper, we solve completely the diophantine equation 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓin the positive integers 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ unconditionally for 1≤p,q≤102≤k≤10G0=0G1=1Gm=kGm-1+Gm-2{Gm}m=0∞G1p+2G2p+⋯+ℓGℓp=Gnqk,p,q,ℓ,nℓ and n. The method works, at least in theory for arbitrary positive integers p, q, and k.