

A276703


Let A_n be the sequence defined in the same way as A159559 but with initial term prime(n), n>=2; a(n) = max(A_n(m)  A159559(m)), m>=2.


5



0, 4, 14, 14, 14, 70, 70, 70, 90, 90, 90, 90, 90, 90, 90, 90, 90, 121, 121, 121, 121, 121, 121, 126, 126, 126, 126, 126, 172, 172, 172, 172, 172, 172, 174, 174, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260, 2260
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OFFSET

2,2


COMMENTS



LINKS



EXAMPLE

A_3(2)=5 and, by the definition of A159559 we have A_3(3)=7, A_3(4)=8, A_3(5)=11, A_3(6)=12, A_3(7)=13, A_3(8)=14, A_3(9)=15, A_3(10)=16, A_3(11)=17. Since A229019(3)=11, then comparing with the first 11 terms of A159559, we conclude that a(3)=A_3(5)A_2(5)=4.


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



