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 A299251 a(n) = ((Sum_{k=1..floor((n+1)^2/4)} d(k)) - T(n)) / 2, where d(n) = number of divisors of n (A000005) and T(n) = the n-th triangular number (A000217). 1

%I

%S 0,0,1,2,4,7,11,15,21,28,37,45,55,67,80,95,110,127,146,164,187,209,

%T 235,260,286,315,346,380,413,449,485,522,564,605,651,695,743,792,844,

%U 898,950,1006,1064,1123,1185,1250,1318,1384,1451,1523,1596,1670,1747,1828

%N a(n) = ((Sum_{k=1..floor((n+1)^2/4)} d(k)) - T(n)) / 2, where d(n) = number of divisors of n (A000005) and T(n) = the n-th triangular number (A000217).

%C Twice this sequence is an attempt to find a counterpart to A161664: both compare triangular numbers T(n) and partial sums of numbers of divisors S(n). A161664 computes the excess of T(n) compared to S(n), whereas 2*a(n) computes the excess of S(n') compared to T(n), where n' is chosen equal to floor((n+1)^2/4). This choice appears structurally natural and economical when illustrated in a diagram. (See provided link.)

%H Luc Rousseau, <a href="/A299251/a299251.png">Accompanying diagram</a>

%F a(n) = (A006218(A002620(n + 1)) - A000217(n)) / 2.

%t F[n_] := Floor[(1/4)*n^2]

%t A[n_] := (Sum[DivisorSigma[0, k], {k, 1, F[n + 1]}] - n*(n + 1)/2)/2

%t Table[A[n], {n, 1, 100}]

%o (PARI)

%o f(n)=floor(n^2/4)

%o a(n)=(sum(k=1,f(n+1),numdiv(k))-n*(n+1)/2)/2

%o for(n=1,100,print1(a(n),", "))

%Y Cf. A000005, A000217, A002620, A006218, A161664.

%K nonn

%O 1,4

%A _Luc Rousseau_, Feb 06 2018

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Last modified September 20 06:19 EDT 2019. Contains 327212 sequences. (Running on oeis4.)