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Let p_n be the polynomial of degree n-1 that interpolates the first n primes (i.e., p_n(i) = prime(i) for 1 <= i <= n.) Then a(n) = p_n(n+1)/2.
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%I #9 Jul 21 2017 13:24:04

%S 1,2,4,4,11,-3,36,-46,133,-213,419,-586,716,-199,-1807,7570,-20637,

%T 47563,-97849,185438,-326192,531721,-785058,980926,-780084,-700944,

%U 5511613,-18000159,46704269,-107137804,225187101,-439627178,799622938,-1347732434,2069035230

%N Let p_n be the polynomial of degree n-1 that interpolates the first n primes (i.e., p_n(i) = prime(i) for 1 <= i <= n.) Then a(n) = p_n(n+1)/2.

%C As n approaches infinity, |a(n)|^(1/n) converges to 2, but a(n+1)/a(n) does not appear to converge.

%F a(n) = Sum_{j=1..n} (-1)^(j+n)*prime(j)*binomial(n,j-1)/2.

%e p_3(x) = (x^2-x+4)/2. p_3(1) = 2, p_3(2) = 3, p_3(3) = 5, so

%e a(3) = p_3(4)/2 = 4.

%t Table[ Sum[(-1)^(j + r)Prime[j] Binomial[r, j - 1]/2, {j, r}], {r, 50}]

%K easy,sign

%O 1,2

%A Joseph Van Name (prism720(AT)yahoo.com), Aug 08 2006

%E Edited and extended by _David Wasserman_, Aug 16 2006

%E Corrected by _N. J. A. Sloane_, Oct 29 2006