

A168010


a(n) = Sum of all numbers of divisors of all numbers k such that n^2 <= k < (n+1)^2.


4



5, 15, 25, 39, 47, 67, 75, 95, 105, 129, 129, 163, 167, 191, 205, 229, 231, 269, 267, 299, 313, 337, 341, 379, 387, 409, 427, 459, 445, 505, 497, 529, 553, 573, 571, 627, 625, 657, 661, 711, 687, 757, 743, 783, 805, 821, 831, 885, 875, 913, 929, 961, 961, 1011
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OFFSET

1,1


COMMENTS

A straightforward approach to calculate a(n) would require computing tau (A000005) for the 2n+1 integers between n^2 and (n+1)^2. Since Sum_{i=1..n} tau(i) can be computed by summing sqrt(n) terms, we can compute a(n) via the summation of n terms of the form 2*(floor(n*(n+2)/i)floor((n1)*(n+1)/i)) without the need to compute tau. Similarly for the sequence A168012.  Chai Wah Wu, Oct 24 2023


LINKS



EXAMPLE

a(2) = 15 because the numbers k are 4, 5, 6, 7 and 8 (since 2^2 <= k < 3^2) and d(4) + d(5) + d(6) + d(7) + d(8) = 3 + 2 + 4 + 2 + 4 = 15, where d(n) is the number of divisors of n (see A000005).


MATHEMATICA

Table[Total[DivisorSigma[0, Range[n^2, (n+1)^21]]], {n, 60}] (* Harvey P. Dale, Aug 17 2015 *)


PROG

(Python)
a, b = n*(n+2), (n1)*(n+1)
return (sum(a//kb//k for k in range(1, n))<<1)+5 # Chai Wah Wu, Oct 23 2023


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



