A013937 -id:A013937 - cp's OEIS frontend

A060431 Number of cubefree numbers <= n.

1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 21, 22, 23, 23, 24, 25, 26, 27, 27, 28, 29, 30, 31, 32, 33, 34, 34, 35, 36, 37, 38, 39, 40, 41, 41, 42, 43, 44, 45, 46, 46, 47, 47, 48, 49, 50, 51, 52, 53, 54, 54, 55, 56, 57, 58, 59, 60, 61, 61

Offset: 1

Vladeta Jovovic, Apr 06 2001

I. M. Vinogradov, Elements of the Theory of Numbers,(in Russian), Moscow, 1981, p. 36.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..500 from Harry J. Smith)

Cf. A013928, A013937, A088453.

a060431 n = a060431_list !! (n-1)
a060431_list = scanl1 (+) a212793_list -- Reinhard Zumkeller, May 27 2012

[&+[MoebiusMu(d)*Floor(n div d^3):d in [1..n]]:n in [1..75]]; // Marius A. Burtea, Oct 02 2019

a(n)=sum(k=1,n,moebius(k)*floor(n/k^3)) \\ Benoit Cloitre, Jun 13 2007

for (n=1, 500, a=sum(k=1, n, moebius(k)*floor(n/k^3)); write("b060431.txt", n, " ", a)) \\ Harry J. Smith, Jul 05 2009

a(n)=my(s); forsquarefree(k=1,sqrtnint(n,3), s+=n\k[1]^3*moebius(k)); s \\ Charles R Greathouse IV, Jan 08 2018

from sympy import mobius, integer_nthroot
def A060431(n): return sum(mobius(k)*(n//k**3) for k in range(1, integer_nthroot(n,3)[0]+1)) # Chai Wah Wu, Aug 06 2024

a(n) = Sum_{d>=1} mu(d)*floor(n/d^3), mu(d) = Moebius function A008683.
a(n) is asymptotic to (1/zeta(3))*n, see A088453. - Benoit Cloitre, Jun 13 2007
a(n) = Sum_{k = 1..n} A212793(k). - Reinhard Zumkeller, May 27 2012

A013938 a(n) = Sum_{k=1..n} floor(n/k^4).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 85, 87

Offset: 1

N. J. A. Sloane, Henri Lifchitz

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Cf. A013936, A013937.

Table[Sum[Floor[n/k^4], {k, 1, n}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 12 2019 *)

a(n) = sum(k=1, n, n\k^4); \\ Michel Marcus, Feb 11 2017

G.f.: (1/(1 - x))*Sum_{k>=1} x^(k^4)/(1 - x^(k^4)). - Ilya Gutkovskiy, Feb 11 2017

a(n) ~ zeta(4)*n = Pi^4*n/90. - Vaclav Kotesovec, Oct 12 2019

A309082 a(n) = n - floor(n/2^3) + floor(n/3^3) - floor(n/4^3) + ...

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 21, 22, 23, 25, 26, 27, 28, 29, 29, 30, 31, 32, 33, 34, 35, 36, 36, 37, 38, 39, 40, 41, 42, 43, 43, 44, 45, 46, 47, 48, 50, 51, 51, 52, 53, 54, 55, 56, 57, 58, 57, 58, 59, 60, 61, 62, 63, 64, 64, 65, 66, 67

Offset: 1

Ilya Gutkovskiy, Jul 11 2019

nonn

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000

Cf. A013937, A059851, A061704, A309081, A309083.

Table[Sum[(-1)^(k + 1) Floor[n/k^3], {k, 1, n}], {n, 1, 75}]
nmax = 75; CoefficientList[Series[1/(1 - x) Sum[(-1)^(k + 1) x^(k^3)/(1 - x^(k^3)), {k, 1, Floor[nmax^(1/3)] + 1}], {x, 0, nmax}], x] // Rest
Table[Sum[Boole[IntegerQ[d^(1/3)] && OddQ[d]], {d, Divisors[n]}] - Sum[Boole[IntegerQ[d^(1/3)] && EvenQ[d]], {d, Divisors[n]}], {n, 1, 75}] // Accumulate

G.f.: (1/(1 - x)) * Sum_{k>=1} (-1)^(k+1) * x^(k^3)/(1 - x^(k^3)).

a(n) ~ 3*zeta(3)*n/4. - Vaclav Kotesovec, Oct 12 2019

A309126 a(n) = n + 2^3 * floor(n/2^3) + 3^3 * floor(n/3^3) + 4^3 * floor(n/4^3) + ...

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 16, 17, 18, 19, 20, 21, 22, 23, 32, 33, 34, 35, 36, 37, 38, 39, 48, 49, 50, 78, 79, 80, 81, 82, 91, 92, 93, 94, 95, 96, 97, 98, 107, 108, 109, 110, 111, 112, 113, 114, 123, 124, 125, 126, 127, 128, 156, 157, 166, 167, 168, 169, 170, 171, 172, 173, 246, 247, 248, 249, 250, 251, 252

Offset: 1

Ilya Gutkovskiy, Jul 13 2019

nonn

Partial sums of A113061.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Cf. A013937, A024916, A113061, A309125, A309127.

Table[Sum[k^3 Floor[n/k^3], {k, 1, n}], {n, 1, 70}]
nmax = 70; CoefficientList[Series[1/(1 - x) Sum[k^3 x^(k^3)/(1 - x^(k^3)), {k, 1, Floor[nmax^(1/3)] + 1}], {x, 0, nmax}], x] // Rest

a(n) = sum(k=1, n, k^3*(n\k^3)); \\ Seiichi Manyama, Aug 30 2021

G.f.: (1/(1 - x)) * Sum_{k>=1} k^3 * x^(k^3)/(1 - x^(k^3)).

a(n) ~ zeta(4/3)*n^(4/3)/4 - n/2. - Vaclav Kotesovec, Aug 30 2021

A306533 Square array A(n,k), n >= 1, k >= 0, read by antidiagonals: A(n,k) = Sum_{j=1..n} floor(n/j^k).

Original entry on oeis.org

1, 1, 4, 1, 3, 9, 1, 2, 5, 16, 1, 2, 3, 8, 25, 1, 2, 3, 5, 10, 36, 1, 2, 3, 4, 6, 14, 49, 1, 2, 3, 4, 5, 7, 16, 64, 1, 2, 3, 4, 5, 6, 8, 20, 81, 1, 2, 3, 4, 5, 6, 7, 10, 23, 100, 1, 2, 3, 4, 5, 6, 7, 9, 12, 27, 121, 1, 2, 3, 4, 5, 6, 7, 8, 10, 13, 29, 144, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 14, 35, 169

Offset: 1

Ilya Gutkovskiy, Feb 22 2019

			Square array begins:
   1,   1,  1,  1,  1,  1,  ...
   4,   3,  2,  2,  2,  2,  ...
   9,   5,  3,  3,  3,  3,  ...
  16,   8,  5,  4,  4,  4,  ...
  25,  10,  6,  5,  5,  5,  ...
  36,  14,  7,  6,  6,  6,  ...

Columns k=0..4 give A000290, A006218, A013936, A013937, A013938.

Cf. A306534.

Table[Function[k, Sum[Floor[n/j^k], {j, 1, n}]][i - n], {i, 0, 12}, {n, 1, i}] // Flatten

G.f. of column k (for k > 0): (1/(1 - x)) * Sum_{j>=1} x^(j^k)/(1 - x^(j^k)).

A069470 a(n) = Sum_{k>=1} floor(n/(k*(k+1)/2)).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 9, 10, 11, 13, 15, 16, 19, 20, 21, 24, 25, 26, 29, 30, 32, 35, 36, 37, 40, 41, 42, 44, 46, 47, 52, 53, 54, 56, 57, 58, 62, 63, 64, 66, 68, 69, 73, 74, 75, 79, 80, 81, 84, 85, 87, 89, 90, 91, 94, 96, 98, 100, 101, 102, 107, 108, 109, 112, 113, 114, 118

Offset: 0

Henry Bottomley, Mar 25 2002

nonn

The summation has floor(1/2 + sqrt(2*n)) = A002024(n) nonzero terms. - Enrique Pérez Herrero, Apr 05 2010

			a(11) = floor(11/1) + floor(11/3) + floor(11/6) + floor(11/10) + floor(11/15) + ... = 11 + 3 + 1 + 1 + 0 + ... = 16.

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A000217, A006218, A007862, A013936, A013937, A013938, A013939.

Cf. A002024, A004275. - Enrique Pérez Herrero, Apr 05 2010

[(&+[Floor(n/(k*(k+1)/2)): k in [1..100]]): n in [0..30]]; // G. C. Greubel, May 23 2018

A069470[n_]:=Sum[Floor[(2*n)/(k*(1 + k))], {k, 1, Floor[1/2 + Sqrt[2*n]]}] (* Enrique Pérez Herrero, Apr 05 2010 *)

for(n=0, 30, print1(sum(k=1, 100, floor(n/(k*(k+1)/2))), ", ")) \\ G. C. Greubel, May 23 2018

a(n) = a(n-1) + A007862(n).
It appears that limit((sum(floor((1/2)*n/(k*(k+1))), k=1..n))/n, n=infinity) = 1/2. - Stephen Crowley, Aug 12 2009
From Enrique Pérez Herrero, Apr 05 2010: (Start)
a(n) <= floor((2*n^2)/(1 + n)) = A004275(n).
a(n) <= floor((2*n*floor((1 + 2*sqrt(2*n))/2))/(1+floor((1+2*sqrt(2*n))/2))). (End)
G.f.: (1/(1 - x)) * Sum_{k>=1} x^(k*(k+1)/2)/(1 - x^(k*(k+1)/2)). - Ilya Gutkovskiy, Jul 11 2019

cp's OEIS Frontend

A060431 Number of cubefree numbers <= n.

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A013938 a(n) = Sum_{k=1..n} floor(n/k^4).

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A309082 a(n) = n - floor(n/2^3) + floor(n/3^3) - floor(n/4^3) + ...

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A309126 a(n) = n + 2^3 * floor(n/2^3) + 3^3 * floor(n/3^3) + 4^3 * floor(n/4^3) + ...

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A306533 Square array A(n,k), n >= 1, k >= 0, read by antidiagonals: A(n,k) = Sum_{j=1..n} floor(n/j^k).

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A069470 a(n) = Sum_{k>=1} floor(n/(k*(k+1)/2)).

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