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%I #21 Dec 17 2020 05:26:34
%S 2,3,5,7,11,42,168,520,1312,2861,5607,10133,17185,27692,42786,63822,
%T 92398,130375,179897,243411,323687,423838,547340,698052,880236,
%U 1098577,1358203,1664705,2024157,2443136,2928742,3488618,4130970
%N a(n) = (1/6)*n^5 - (19/8)*n^4 + (51/4)*n^3 - (253/8)*n^2 + (445/12)*n - 14.
%C Deliberately contrived to begin with first five primes; illustrates absurdity of many "guess the next term" puzzles.
%C The 4th-order formula a(n) = (1/8)*n^4 - (17/12)*n^3 + (47/8)*n^2 - (103/12)*n + 6 is sufficient to yield a(n) = prime(n) for n = 1..5. - _Jon E. Schoenfield_, Sep 15 2018
%H Harry J. Smith, <a href="/A059999/b059999.txt">Table of n, a(n) for n = 1..1000</a>
%H de.rec.denksport, <a href="http://janko.at/Denksport/FAQ.txt">FAQ</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%t Table[(n^5)/6 - (19/8)n^4 + (51/4)n^3 - (253/8)n^2 + (445/12)n - 14, {n, 40}] (* _Alonso del Arte_, Jan 25 2017 *)
%o (PARI) for (n=1, 1000, write("b059999.txt", n, " ", (4*n^5 - 57*n^4 + 306*n^3 - 759*n^2 + 890*n)/24 - 14); ) \\ _Harry J. Smith_, Jul 01 2009
%K dumb,easy,nonn
%O 1,1
%A _Rainer Rosenthal_, Mar 10 2001
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