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%I #36 Aug 16 2019 16:18:45
%S 3,5,7,7,9,8,10,9,11,9,12,10,12,10,13,11,13,10,14,12,14,10,14,13,15,
%T 11,14,12,16,12,16,12,14,12,17,15,15,10,16,14,18,12,16,14,16,12,16,15,
%U 19,13,16,12,16,14,20,16,16,10,18,16,18,12,17,17,19,14,16,12,18,14,22,16,18,12,16,16,18,14,20
%N a(1) = 3, for n>1, a(n) = A000005(n-1) + A000005(n) + A000005(n+1).
%C (Tau) divisibility of n's 1-area. This is the first step to examine the divisibility of n's k-area. n's k-area is the set of m for which |n-m| is less than or equal to k (n, k, m are natural numbers). 1's 1-area is {1,2}, 5's 1-area {4,5,6}, 3's 2-area {1,2,3,4,5}. We could call this natural area, and still talk about nonnegative or integer area etc.
%e a(10) = 9 as 9 has 3 divisors, 10 has 4 divisors and 11 has 2 divisors. - _David A. Corneth_, Mar 26 2019
%t {3}~Join~Map[Total, Partition[DivisorSigma[0, Range@ 80], 3, 1]] (* _Michael De Vlieger_, Jun 06 2019 *)
%o (PARI) a(n) = if(n == 3, return(1)); sum(i = n-1, n+1, numdiv(i)) \\ _David A. Corneth_, Mar 26 2019
%Y Cf. A000005, A307118, A307119.
%K nonn
%O 1,1
%A _Todor Szimeonov_, Mar 25 2019
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