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A076815 Initial indices of five successive primes squared with integer average.

%I A076815 #13 Jul 15 2017 19:05:33
%S A076815 79,258,397,428,429,502,503,609,787,788,925,926,927,1026,1027,1028,
%T A076815 1105,1312,1334,1335,1343,1348,1349,1378,1422,1524,1572,1601,1602,
%U A076815 1790,1791,1813,2015,2081,2082,2125,2126,2131,2141,2142,2147,2292,2448,2765,2766
%N A076815 Initial indices of five successive primes squared with integer average.
%C A076815 Or, numbers n such that sum of 5 consecutive primes squared, starting with p(n), ends with 5.
%C A076815 Unlike the average of two, three, four and six successive primes squared (with initial indices > 1,2,1,2, respectively), the average of five successive primes squared is rarely an integer.
%C A076815 Cases of sums ending with 5 are much less numerous than cases with 1, 3, 7 and 9.
%C A076815 E.g. for the first 20000, sums with final digits 1, 3, 5, 7 and 9 are 7238, 2380, 466, 2529 and 7386 (and 1 case with final 8, 208=A131686(1)). And for first 200000 sums the corresponding numbers are 71166, 25820, 5956, 26075, 70982.
%C A076815 The explanation of this "deficiency of final 5's" is simple: assuming that final digits {1,3,7,9} of primes are equally often, we get that probabilities for final digits {1,3,5,7,9} of sum of squares of five primes are {10/32,5/32,2/32,5/32,10/32}.
%H A076815 Harvey P. Dale, <a href="/A076815/b076815.txt">Table of n, a(n) for n = 1..1000</a>
%e A076815 sum(prime(i)^2,i=79..83)/5=(401^2+409^2+419^2+421^2+431^2)/5=866645/5=173329=A076814(1),
%e A076815 sum(prime(i)^2,i=258..262)/5=(1627^2+1637^2+1657^2+1663^2+1667^2)/5=13617005/5=2723401=A076814(2).
%t A076815 PrimePi[Sqrt[#]]&/@Select[Partition[Prime[Range[3000]]^2,5,1],IntegerQ[ Mean[ #]]&][[All,1]] (* _Harvey P. Dale_, Jul 15 2017 *)
%Y A076815 Cf. A076815, A076814, A131686.
%K A076815 nonn
%O A076815 1,1
%A A076815 _Zak Seidov_, Oct 17 2002
%E A076815 Edited and merged with A131359 by Zak Seidov, May 18 2008 at the suggestion of R. J. Mathar