A178489 -id:A178489 - cp's OEIS frontend

A255270 Integer part of fourth root of n.

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3

Offset: 0

Bruno Berselli, Feb 20 2015

n appears (n+1)^4 - n^4 times (A005917).

Bruno Berselli, Table of n, a(n) for n = 0..1000

Cf. A005917.
Cf. sequences of the type floor(n^(1/k)): A000196 (k=2), A048766 (k=3), this sequence (k=4), A178487 (k=5), A178489 (k=6).
Cf. A219009.

[IsZero(n) select 0 else Iroot(n, 4): n in [0..100]];

[Floor(n^(1/4)): n in [0..100]]; // Vincenzo Librandi, Feb 20 2015

A255270 := proc(n)
    floor( n^(1/4)) ;
end proc:
seq(A255270(n),n=0..100) ; # R. J. Mathar, May 08 2020

Mathematica
```
Floor[Range[0, 100]^(1/4)]
```
PARI
```
vector(100, n, n--; floor(n^(1/4)))
```

a(n) = sqrtnint(n, 4); \\ Michel Marcus, Dec 22 2016

from sympy import integer_nthroot
def A255270(n): return integer_nthroot(n,4)[0] # Chai Wah Wu, Jun 06 2025

Sage
```
[floor(n^(1/4)) for n in (0..100)]
```

a(n) = floor(n^(1/4)) = floor(sqrt(A000196(n))).
G.f.: Sum_{k>=1} x^(k^4)/(1 - x). - Ilya Gutkovskiy, Dec 22 2016
a(n) = Sum_{i=1..n} A219009(i)*floor(n/i). - Ridouane Oudra, Feb 26 2023

A178487 a(n) = floor(n^(1/5)): integer part of fifth root of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

M. F. Hasler, Oct 09 2010

Each term k appears (k+1)^5 - k^5 times consecutively (A022521). - Bernard Schott, Mar 07 2023

Sequences a(n) = floor(n^(1/k)): A001477 (k=1), A000196 (k=2), A048766 (k=3), A255270 (k=4), this sequence (k= 5), A178489 (k=6), A057427 (k->oo).

Cf. A022521, A253206.

[n eq 0 select 0 else Iroot(n, 5): n in [0..110]]; // Bruno Berselli, Feb 20 2015

seq(floor(n^(1/5)), n=0..100); # Ridouane Oudra, Feb 26 2023

Floor[Range[0,120]^(1/5)] (* Harvey P. Dale, Aug 15 2012 *)

PARI
```
A178487(n)=floor(sqrtn(n+.5,5))
```

a(n) = sqrtnint(n, 5); \\ Michel Marcus, Dec 22 2016

from sympy import integer_nthroot
def A178487(n): return integer_nthroot(n,5)[0] # Chai Wah Wu, Jun 06 2025

G.f.: Sum_{k>=1} x^(k^5)/(1 - x). - Ilya Gutkovskiy, Dec 22 2016

a(n) = Sum_{i=1..n} A253206(i)*floor(n/i). - Ridouane Oudra, Feb 26 2023

A132337 Sum of the integers from 1 to n, excluding the perfect sixth powers.

Original entry on oeis.org

0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430, 1484

Offset: 1

Cino Hilliard, Nov 07 2007

T. D. Noe, Table of n, a(n) for n = 1..10000

Different from A000096.

Cf. A132336, A178489.

A132337 := proc(n) r := floor(n^(1/6)) ; A000217(n)-A000540(r); end proc: seq(A132337(n),n=1..40) ; # R. J. Mathar

Accumulate[Table[If[IntegerQ[Surd[n,6]],0,n],{n,60}]] (* Harvey P. Dale, Jun 01 2022 *)

g6(n)=for(x=1,n,r=floor(x^(1/6));sum6=r^7/7+r^6/2+r^5/2-r^3/6+r/ 42;sn=x* (x+1)/2;print1(sn-sum6","))

A132337(n)=n*(n+1)/2-(1+n=floor(sqrtn(n+.5,6)))*(2*n+1)*((n^3+2*n^2-1)*n*3+1)*n/42 \\ M. F. Hasler, Oct 09 2010

Let r = floor(n^(1/6)). Then a(n) = n(n+1)/2 - (r^7/7 + r^6/2 + r^5/2 - r^3/6 + r/42) = A000217(n) - A000540(r).

a(n) = A000217(n) - A000540(A178489(n)). - M. F. Hasler, Oct 09 2010

Incorrect formula deleted by Jon E. Schoenfield, Jun 12 2010
Incorrect program replaced by R. J. Mathar, Oct 08 2010
Edited by the Assoc. Editors of the OEIS, Oct 12 2010. Thanks to Daniel Mondot for pointing out that the sequence needed editing.
Incorrect linear recurrence removed by Georg Fischer, Apr 11 2019

A381042 Alternating sum of floor(n^(1/k)), with k >= 2.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7

Offset: 0

Friedjof Tellkamp, Apr 14 2025

nonn

			n:       0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
k=2 (+): 0, 1, 1, 1, 2, 2, 2, 2, 2, 3, ... (A000196)
k=3 (-): 0, 1, 1, 1, 1, 1, 1, 1, 2, 2, ... (A048766)
...
Sum:     0, 0, 0, 0, 1, 1, 1, 1, 0, 1 ... (= this sequence).

Friedjof Tellkamp, Table of n, a(n) for n = 0..10000

Cf. A000196 (k=2), A048766 (k=3), A255270 (k=4), A178487 (k=5), A178489 (k=6).

Cf. A089361 (nonalternating), A382691, A382692.

z = 100; Table[Sum[(-1)^k Floor[n^(1/k)], {k, 2, 2 Floor@Log[2, z/2] - 1}], {n, 0, z}]

a(n) = A000196(n) - A048766(n) + A255270(n) - A178487(n) + ... .
a(n) = Sum_{k>=2} (-1)^k * floor(n^(1/k)) = Sum_{k>=1} (floor(n^(1/(2*k))) - floor(n^(1/(2*k+1)))).
a(n) = Sum_{i=1..n} A382691(i).
a(n) ~ A382692(n).
G.f.: Sum_{j>=1, k>=2} (-1)^k * x^(j^k)/(1-x).

A365335 The number of exponentially odd coreful divisors of the square root of the largest square dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Amiram Eldar, Sep 01 2023

First differs from A160338 at n = 64, and from A178489 at n = 65.
The number of divisors of the square root of the largest square dividing n is A046951(n).
The number of exponentially odd divisors of the square root of the largest square dividing n is A365549(n) and their sum is A365336(n). [corrected, Sep 08 2023]

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000188, A046951, A325837, A355265, A365336, A365549.

Cf. A160338, A178489.

f[p_, e_] := Max[1, Floor[(e+2)/4]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

a(n) = vecprod(apply(x -> max(1, (x+2)\4), factor(n)[, 2]));

a(n) = A325837(A000188(n)).
a(n) > 1 if and only if n is a bicubeful number (A355265).
Multiplicative with a(p^e) = floor((e+2)/4).
Dirichlet g.f.: zeta(s) * zeta(4*s) * Product_{p prime} (1 - 1/p^(4*s) + 1/p^(6*s)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = zeta(4) * Product_{p prime} (1 - 1/p^4 + 1/p^6) = 1.0181534831085... .

Name corrected by Amiram Eldar, Sep 08 2023

A382692 a(n) = floor of alternating sum of k-th roots of n, with k >= 2.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6

Offset: 0

Friedjof Tellkamp, Apr 05 2025

nonn

			a(15) = floor(15^(1/2) - 15^(1/3) + 15^(1/4) - ...) = floor(1.9701...) = 1.

Cf. A000196 (k=2), A048766 (k=3), A255270 (k=4), A178487 (k=5), A178489 (k=6).

Cf. A381042, A382691.

Floor@Table[NSum[n^(1/(2 k)) - n^(1/(2 k + 1)), {k, 1, Infinity}, WorkingPrecision -> 30], {n, 1, 100}]

a(n) = floor(Sum_{k>=2} (-1)^k * n^(1/k)) = floor(Sum_{k>=1} (n^(1/(2*k)) - n^(1/(2*k + 1)))).
a(n) ~ A381042(n).
Series expansion of Sum_{k>=2} (-1)^k * x^(1/k) at x=1: Sum_{i>=0} A(i) * (x-1)^i/i!, where A(i) = KroneckerDelta(i, 1) - Sum_{j=1..i} eta(j) * StirlingS1(i, j), with eta as the Dirichlet eta function.

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A255270 Integer part of fourth root of n.

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A178487 a(n) = floor(n^(1/5)): integer part of fifth root of n.

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A132337 Sum of the integers from 1 to n, excluding the perfect sixth powers.

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A381042 Alternating sum of floor(n^(1/k)), with k >= 2.

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A365335 The number of exponentially odd coreful divisors of the square root of the largest square dividing n.

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A382692 a(n) = floor of alternating sum of k-th roots of n, with k >= 2.

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