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
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 1..10000
Programs
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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
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Mathematica
Accumulate[Select[Range[300],!IntegerQ[Surd[#,3]]&&Divisible[#,Floor[ Surd[ #,3]]]&]] (* Harvey P. Dale, May 13 2020 *)
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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
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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)
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