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

79, 258, 397, 428, 429, 502, 503, 609, 787, 788, 925, 926, 927, 1026, 1027, 1028, 1105, 1312, 1334, 1335, 1343, 1348, 1349, 1378, 1422, 1524, 1572, 1601, 1602, 1790, 1791, 1813, 2015, 2081, 2082, 2125, 2126, 2131, 2141, 2142, 2147, 2292, 2448, 2765, 2766

Offset: 1

Author

Zak Seidov, Oct 17 2002

Keywords

• nonn

Comments

Or, numbers n such that sum of 5 consecutive primes squared, starting with p(n), ends with 5.

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.

Cases of sums ending with 5 are much less numerous than cases with 1, 3, 7 and 9.

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.

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}.

Examples

			sum(prime(i)^2,i=79..83)/5=(401^2+409^2+419^2+421^2+431^2)/5=866645/5=173329=A076814(1),
sum(prime(i)^2,i=258..262)/5=(1627^2+1637^2+1657^2+1663^2+1667^2)/5=13617005/5=2723401=A076814(2).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A076815, A076814, A131686.

Programs

Mathematica

PrimePi[Sqrt[#]]&/@Select[Partition[Prime[Range[3000]]^2,5,1],IntegerQ[ Mean[ #]]&][[All,1]] (* Harvey P. Dale, Jul 15 2017 *)

Extensions

Edited and merged with A131359 by Zak Seidov, May 18 2008 at the suggestion of R. J. Mathar