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

Offset: 1

Author

Luc Rousseau, Feb 06 2018

Keywords

• nonn

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

• Luc Rousseau, Accompanying diagram

Crossrefs

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

Programs

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}]

PARI

f(n)=floor(n^2/4)
a(n)=(sum(k=1,f(n+1),numdiv(k))-n*(n+1)/2)/2
for(n=1,100,print1(a(n),", "))

Python

from math import isqrt
def A299251(n): return (-(s:=isqrt(m:=(n+1)**2>>2))**2-(n*(n+1)>>1)>>1)+sum(m//k for k in range(1,s+1)) # Chai Wah Wu, Oct 23 2023

Formula

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