cp's OEIS Frontend

A066885 a(n) = (prime(n)^2 + 1)/2.

%I A066885 #47 Oct 03 2024 13:01:07
%S A066885 5,13,25,61,85,145,181,265,421,481,685,841,925,1105,1405,1741,1861,
%T A066885 2245,2521,2665,3121,3445,3961,4705,5101,5305,5725,5941,6385,8065,
%U A066885 8581,9385,9661,11101,11401,12325,13285,13945,14965,16021,16381,18241,18625
%N A066885 a(n) = (prime(n)^2 + 1)/2.
%C A066885 a(n) is the average of the numbers from 1 to prime(n)^2. It's also the average of the primes in a prime(n) X prime(n) example of Haga's conjecture (see link below).
%C A066885 If a(n) is a square c^2, then prime(n) is an NSW prime (A088165) and a prime RMS number (A140480). - _Ctibor O. Zizka_, Aug 26 2008
%C A066885 The sequence starts with a(2) = (3^2 + 1)/2 = 5 since a(1) would be (2^2 + 1)/2 = 5/2. - _Michael B. Porter_, Dec 14 2009
%H A066885 Harry J. Smith, <a href="/A066885/b066885.txt">Table of n, a(n) for n = 2..1000</a>
%H A066885 Carlos Rivera, <a href="http://www.primepuzzles.net/conjectures/conj_026.htm">Conjecture 26. The Calendar-like square Conjecture</a>, The Prime Puzzles and Problems Connection.
%F A066885 a(n) = 1 + A084921(n). - _R. J. Mathar_, Sep 30 2011
%F A066885 a(n) mod 4 = 1. - _Altug Alkan_, Apr 08 2016
%F A066885 Product_{n>=2} (1 - 1/a(n)) = 2/3. - _Amiram Eldar_, Jun 03 2022
%p A066885 A066885:=n->(ithprime(n)^2+1)/2: seq(A066885(n), n=2..50); # _Wesley Ivan Hurt_, Jun 23 2015
%t A066885 a[n_] := (Prime[n]^2+1)/2; Table[a[n], {n, 2, 50}]
%o A066885 (PARI) A066885(n) = (prime(n)^2+1)/2 \\ _Michael B. Porter_, Dec 14 2009
%o A066885 (PARI) { for (n=2, 1000, write("b066885.txt", n, " ", (prime(n)^2 + 1)/2) ) } \\ _Harry J. Smith_, Apr 04 2010
%o A066885 (Magma) [(NthPrime(n)^2+1)/2 : n in [2..50]]; // _Wesley Ivan Hurt_, Jun 23 2015
%Y A066885 Cf. A066883, A066886.
%Y A066885 Cf. A088165, A140480.
%Y A066885 Cf. A084921.
%Y A066885 Partial sums of A124434.
%K A066885 easy,nonn
%O A066885 2,1
%A A066885 _Enoch Haga_, Jan 22 2002
%E A066885 Edited by _Dean Hickerson_, Jun 08 2002