

A324795


a(n) = 2*p(n)*p(n+2)p(n+1)^2 where p(k) = kth prime.


2



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OFFSET

1,1


COMMENTS

Theorem: a(n)>0. Proof: Use p(n+1) <= 2 p(n)^2 for n>4. (See Mitrinovic). QED


REFERENCES

D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section VII.18(b).


LINKS



CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



