A112873 Partial sums of A032378.

2, 5, 9, 14, 20, 27, 37, 49, 63, 79, 97, 117, 139, 163, 189, 219, 252, 288, 327, 369, 414, 462, 513, 567, 624, 684, 747, 815, 887, 963, 1043, 1127, 1215, 1307, 1403, 1503, 1607, 1715, 1827, 1943, 2063, 2187, 2317, 2452, 2592, 2737, 2887, 3042, 3202, 3367, 3537

Offset: 1

Author

N. J. A. Sloane, Oct 29 2006

Keywords

• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A032378, A066353, A120721.

Programs

Magma

A032378:=[k*j: j in [(k^2+1)..(k^2+3*k+3)], k in [1..15]];
[(&+[A032378[j]: j in [1..n]]): n in [1..100]]; // G. C. Greubel, Jul 20 2023

Mathematica

Accumulate[Select[Range[300],!IntegerQ[Surd[#,3]]&&Divisible[#,Floor[ Surd[ #,3]]]&]] (* Harvey P. Dale, May 13 2020 *)

Python

from itertools import count, islice, accumulate
from sympy import integer_nthroot
def A112873_gen(): # generator of terms
    return accumulate(filter(lambda x: not x%integer_nthroot(x,3)[0],(n+(k:=integer_nthroot(n, 3)[0])+int(n>=(k+1)**3-k) for n in count(1))))
A112873_list = list(islice(A112873_gen(),40)) # Chai Wah Wu, Oct 12 2024

SageMath

A032378=flatten([[k*j for j in range((k^2+1),(k^2+3*k+3)+1)] for k in range(1,15)])
def A112873(n): return sum(A032378[j] for j in range(n+1))
[A112873(n) for n in range(101)] # G. C. Greubel, Jul 20 2023

Formula

From Vaclav Kotesovec, Oct 13 2024: (Start)

a(3*k*(k+3)/2) = 3*k*(k+1)*(k+2)*(8*k^2+21*k+31)/40.

a(n) ~ 2^(5/2)*n^(5/2)/(5*3^(3/2)) - n^2/2 + 13*n^(3/2)/(2^(3/2)*sqrt(3)). (End)