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Partial sums of A147810(n) = tau(n^2 + 1)/2.
3

%I #14 Dec 01 2023 15:52:33

%S 1,2,4,5,7,8,11,13,15,16,18,20,24,25,27,28,32,35,37,38,42,44,48,49,51,

%T 52,56,58,60,62,66,69,73,75,77,78,82,85,87,88,91,93,99,101,103,105,

%U 113,115,117,119,121,123,127,128,132,133,141,143,145,147,149,151,155,157

%N Partial sums of A147810(n) = tau(n^2 + 1)/2.

%C Also, number of inequivalent (i.e., q < r) integer solutions to 1/pqr = 1/p - 1/q - 1/r with p <= n; cf. A147811.

%H Amiram Eldar, <a href="/A147807/b147807.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{p = 1..n} tau(1 + p^2)/2 = n + A147806(n) > n.

%F a(n) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - _Amiram Eldar_, Dec 01 2023

%t Accumulate[DivisorSigma[0, Range[64]^2 + 1]/2] (* _Amiram Eldar_, Oct 25 2019 *)

%o (PARI) s=0;A147807=vector(99,n,s+=numdiv(n^2+1))/2

%Y Cf. A093582, A147806, A147809, A147810, A147811.

%K easy,nonn

%O 1,2

%A _M. F. Hasler_, Dec 13 2008