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

1,1

Links

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

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)

Mathematica

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

Prog

(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
(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
(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

Crossrefs

Cf. A032378, A066353, A120721.

Sequence in context: A000096 A134189 A109470 * A048093 A024669 A006482

Adjacent sequences: A112870 A112871 A112872 * A112874 A112875 A112876

Keyword

nonn

Author

N. J. A. Sloane, Oct 29 2006

Status

approved