■(2^n+1)/m型の数(その18)
m=5,n=9の場合も確認してみたい.
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n=9とする.
(9,0)+(9,5)=1+126=127
(2^9+φ^9・2cos9π/5+φ^-9・2cos18π/5)/5=(512+φ^10+φ^-10)/5=(512+34+89)/5=127
(9,1)+(9,6)=9+84=93
(2^9+φ^9・2cos7π/5+φ^-9・2cos14π/5)/5=(512−φ^8−φ^-8)/5=(512−13−34)/5=93
(9,2)+(9,7)=36+36=72
(2^9+φ^9・2cos5π/5+φ^-9・2cos10π/5)/5=(512−2φ^9+2φ^-9)/5=(512−42−110)/5=72
(9,3)+(9,8)=84+9=93
(2^9+φ^9・2cos3π/5+φ^-9・2cos6π/5)/5=(512−φ^8−φ^-8)/5=93
(9,4)+(9,9)=126+1=127
(2^9+φ^9・2cosπ/5+φ^-9・2cos2π/5)/5=(512+φ^10+φ^-10)/5=127
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