■完全ベキ乗数列(その6)

  {an}={1,4,8,9,16,25,27,32,36,・・・}

  Σ1/(an−1)=Σ1/(2^k−1)+Σ1/(3^k−1)+・・・

  (k≧2)

のあとは

 Σ1/(5^k−1)+Σ1/(6^k−1)+

 Σ1/(7^k−1)+Σ1/(10^k−1)+

 Σ1/(11^k−1)+Σ1/(12^k−1)+

 Σ1/(13^k−1)+Σ1/(14^k−1)+

 Σ1/(15^k−1)+Σ1/(17^k−1)+

と続く.

 すなわり,排他的数列

  {bn}={2,3,5,6,7,11,12,13,14,17,・・・}

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 直接的な関係はないが

  N=Πn^2/(n^2−1)=Πn/(n−1)・n/(n+1)

=2/1・2/3・3/2・3/4・・・n/(n−1)・n/(n+1)

はうまくキャンセルアウトして

  N=2/1・n/(n+1)→2

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[1]N=Πn^k/(n^k−1)  n=2〜∞

k=2:N=2

k=3:N=3πsech(π√3/2)

k=4:N=4πcosech(π√3/2)

k=6:N=6π^2(sech(π√3/2))^2

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

[2]N=Π(n^k+1)/n^k  n=1〜∞

k=2:N=sinh(π)/π

k=3:N=cosh(π√3/2)/π

 なお,

k=4:N=sin((−1)^1/4π)sin((−1)^3/4π)/π^2

k=6:N=sin((−1)^1/6π)sin((−1)^5/6π)sinh(π)/π^3

と計算される.

−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

[3]

  Π((n^3−1)/(n^3+1)=2/3   n=2〜∞

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