■タクシー数のラマヌジャン解(その26)

(γ-α)ω^n・(11296-136ω^-1)+(α-β)ω^-n・(11296-136ω)

=(11296)・{(ω^-1+1)ω^n-(ω+1)ω^-n}-136{(ω^-1+1)ω^(n-1)-(ω+1)ω^(-n+1)}

=(11296)・{(ω^n-ω^-n)+(ω^(n-1)-ω^(-n+1))}-136{(ω^(n-1)-ω^(-n+1)+(ω^(n-2)-ω^(-n+2))}

g1=ω-ω^-1=δ

g2=ω^2-ω^-2

gn=ω^n-ω^-nが求められれば良いのであるが・・・

ω+ω^-1=83

(γ-α)ω^n・(11296-136ω^-1)+(α-β)ω^-n・(11296-136ω)

=(11296)・{(ω^-1+1)ω^n-(ω+1)ω^-n}-136{(ω^-1+1)ω^(n-1)-(ω+1)ω^(-n+1)}

=(11296)・{(ω^n-ω^-n)+(ω^(n-1)-ω^(-n+1))}-136{(ω^(n-1)-ω^(-n+1)+(ω^(n-2)-ω^(-n+2))}

=(11296)・{gn+gn-1}-136{gn-1+gn-2}

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g(-1)+g(-2)=-δ{84}

g0+g(-1)=-δ

g1+g0=δ

g2+g1=δ{84}

g3+g2=δ{6971}

g4+g3=δ{578509}

g5+g4=δ{48009276}

g6+g5=δ{3984191399}

g7+g6=δ{330639876841}

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an=-{(β-γ)(-1)^n・(-43)+(γ-α)ω^n・(11296-136ω^-1)+(α-β)ω^-n・(11296-136ω)}/(α-β)(β-γ)(γ-α)

=-{δ(-1)^n・(-43)+(11296)・{gn+gn-1}-136{gn-1+gn-2}}/(-85δ)

={δ(-1)^n・(-43)+(11296)・{gn+gn-1}-136{gn-1+gn-2}}/(85δ)

a0={-43-(11296)+136・84}/85=1

a1={+43+(11296)+136}/85=135

a2={-43+(11296)・84-136・1}/85=11161

a3={+43+(11296)・6971-136・84}/85=926271

a4={-43+(11296)・578509-136・6971}/85=76869289

a5={+43+(11296)・48009276-136・578509}/85=6379224759

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