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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
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
1,4
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.)
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)
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
CROSSREFS
KEYWORD
nonn
AUTHOR
Luc Rousseau, Feb 06 2018
STATUS
approved