A299251 - OEIS

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).

0, 0, 1, 2, 4, 7, 11, 15, 21, 28, 37, 45, 55, 67, 80, 95, 110, 127, 146, 164, 187, 209, 235, 260, 286, 315, 346, 380, 413, 449, 485, 522, 564, 605, 651, 695, 743, 792, 844, 898, 950, 1006, 1064, 1123, 1185, 1250, 1318, 1384, 1451, 1523, 1596, 1670, 1747, 1828

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OFFSET

1,4

COMMENTS

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.)

LINKS

Table of n, a(n) for n=1..54.

Luc Rousseau, Accompanying diagram

FORMULA

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

MATHEMATICA

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

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

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

PROG

(PARI)

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