1,1

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 probabilties 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}.

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).

Cf. A076815, A076814, A131686.

Sequence in context: A089686 A141964 A142285 this_sequence A141914 A142932 A082077

Adjacent sequences: A076812 A076813 A076814 this_sequence A076816 A076817 A076818

nonn

Zak Seidov (zakseidov(AT)yahoo.com), Oct 17 2002

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