

A122550


Floor of the slanted side of a right trapezoid formed by 3 consecutive primes.


0



4, 6, 9, 12, 14, 18, 19, 25, 30, 32, 38, 41, 43, 48, 54, 59, 61, 67, 71, 73, 79, 83, 90, 97, 101, 103, 107, 109, 114, 128, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 212, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277
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OFFSET

1,1


COMMENTS

It is easy to prove that for any 3 consecutive primes p1,p2,p3, a(n) always lies between p2 and p3.


LINKS



EXAMPLE

For the first 3 prime numbers, go up 2, go right 3 and go down 5. Connecting the figure to form a right trapezoid we have the slanted side = sqrt(18). The integer part of this is 4, the first term of the sequence.


PROG

(PARI) g(n) = { for(x=1, n, p1=prime(x); p2=prime(x+1); p3=prime(x+2); y=p3p1; print1(floor(sqrt(p2^2+y^2)), ", ")) )


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



STATUS

approved



