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A076815
Initial indices of five successive primes squared with integer average.
3
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
1,1
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}.
LINKS
EXAMPLE
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).
MATHEMATICA
PrimePi[Sqrt[#]]&/@Select[Partition[Prime[Range[3000]]^2, 5, 1], IntegerQ[ Mean[ #]]&][[All, 1]] (* Harvey P. Dale, Jul 15 2017 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Zak Seidov, Oct 17 2002
EXTENSIONS
Edited and merged with A131359 by Zak Seidov, May 18 2008 at the suggestion of R. J. Mathar
STATUS
approved