A048766 -id:A048766 - cp's OEIS frontend

A135663 a(n) = floor(sqrt(n) - n^(1/4)).

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, 2, 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, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7

Offset: 1

Mohammad K. Azarian, Nov 25 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A000196, A048766, A134914, A135660, A135661, A135662.

[Floor(n^(1/2)-n^(1/4)): n in [1..100]]; // Vincenzo Librandi, Feb 18 2013

Table[Floor[n^(1/2) - n^(1/4)], {n, 100}] (* Vincenzo Librandi, Feb 18 2013 *)

More terms from N. J. A. Sloane, Aug 08 2008

Offset changed from 0 to 1 by Vincenzo Librandi, Feb 18 2013

A135664 a(n) = ceiling(sqrt(n) - n^(1/5)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 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, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset: 1

Mohammad K. Azarian, Nov 25 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A000196, A048766, A134914, A135660, A135661, A135662, A135663.

[Ceiling(n^(1/2) - n^(1/5)): n in [1..100]]; // Vincenzo Librandi, Feb 16 2013

Table[Ceiling[n^(1/2) - n^(1/5)], {n, 100}] (* Vincenzo Librandi, Feb 16 2013 *)

Offset corrected by Mohammad K. Azarian, Nov 19 2008

A135665 a(n) = floor(sqrt(n) - n^(1/5)).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 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, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

Mohammad K. Azarian, Nov 25 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A000196, A048766, A134914, A135660, A135661, A135662, A135663, A135664.

[Floor(n^(1/2)-n^(1/5)): n in [1..100]]; // Vincenzo Librandi, Feb 18 2013

Table[Floor[n^(1/2) - n^(1/5)], {n, 100}] (* Vincenzo Librandi, Feb 18 2013 *)

Offset corrected by Mohammad K. Azarian, Nov 19 2008

A211668 Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 3.

Original entry on oeis.org

0, 0, 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, 3, 3, 3, 3, 3, 3

Offset: 1

Hieronymus Fischer, Apr 30 2012

For the general case of "Number of iterations f(f(f(...(n)...))) such that the result is < q, where f(x) = x^(1/p), p > 1, q > 1", the resulting g.f. is g(x) = 1/(1-x)*Sum_{k>=0} x^(q^(p^k))

= (x^q + x^(q^p) + x^(q^(p^2)) + x^(q^(p^3)) + ...)/(1-x).

			a(n) = 1, 2, 3, 4, 5 for n = 3^1, 3^2, 3^4, 3^8, 3^16, i.e., n = 3, 9, 81, 6561, 43946721.

Cf. A001069, A010096, A211662, A211666, A211670.

a[n_] := Length[NestWhileList[Sqrt, n, # >= 3 &]] - 1; Array[a, 100] (* Amiram Eldar, Dec 08 2018 *)

a(n) = {my(nbi = 0); if (n < 3, return (nbi)); r = n; nbi= 1; while ((nr = sqrt(r)) >= 3, nbi++; r = nr); return (nbi);} \\ Michel Marcus, Oct 23 2014

A211668(n, c=0)={while(n>=3, n=sqrtint(n); c++); c} \\ M. F. Hasler, Dec 07 2018

from sympy import integer_log
A048766=lambda n: integer_log(n,3)[0].bit_length() # Natalia L. Skirrow, May 17 2023

a(3^(2^n)) = a(3^(2^(n-1))) + 1, for n >= 1.
G.f.: g(x) = 1/(1-x)*Sum_{k >= 0} x^(3^(2^k))
  = (x^3 + x^9 + x^81 + x^6561 + x^43946721 + ...)/(1 - x).

Edited by Michel Marcus, Oct 23 2014 and M. F. Hasler, Dec 07 2018

A025455 a(n) is the number of partitions of n into 2 positive cubes.

Original entry on oeis.org

0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

David W. Wilson

nonn

In other words, number of solutions to the equation x^3 + y^3 = n with x >= y > 0. - Antti Karttunen, Aug 28 2017

The first term > 1 is a(1729) = 2. - Michel Marcus, Apr 23 2019

Antti Karttunen, Table of n, a(n) for n = 0..100000
Index entries for sequences related to sums of cubes

Cf. A025456, A025468, A003108, A003325, A000578, A048766, A001235 (two or more ways, positions where a(n) > 1).

Cf. also A025426, A216284.

Table[Count[IntegerPartitions[n,{2}],?(AllTrue[Surd[#,3],IntegerQ]&)],{n,0,110}] (* _Harvey P. Dale, Nov 23 2022 *)

(define (A025455 n) (let loop ((x (A048766 n)) (s 0)) (let* ((x3 (A000578 x)) (y3 (- n x3))) (if (< x3 y3) s (loop (- x 1) (+ s (if (and (> y3 0) (= (A000578 (A048766 y3)) y3)) 1 0))))))) ;; Antti Karttunen, Aug 28 2017

If a(n) > 0 then A025456(n + k^3) > 0 for k>0; a(A113958(n)) > 0; a(A003325(n)) > 0. - Reinhard Zumkeller, Jun 03 2006
a(n) >= A025468(n). - Antti Karttunen, Aug 28 2017
a(n) = [x^n y^2] Product_{k>=1} 1/(1 - y*x^(k^3)). - Ilya Gutkovskiy, Apr 23 2019

Secondary offset added by Antti Karttunen, Aug 28 2017

Secondary offset corrected by Michel Marcus, Apr 23 2019

A255270 Integer part of fourth root of n.

Original entry on oeis.org

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

A025468 a(n) is the number of partitions of n into 2 distinct positive cubes.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

David W. Wilson

nonn

In other words, number of solutions to the equation n = x^3 + y^3 with x > y > 0. The first value > 1 is a(1729) = 2. - Antti Karttunen, Aug 29 2017

Antti Karttunen, Table of n, a(n) for n = 0..100000
Index entries for sequences related to sums of cubes

Cf. A000578, A048766, A025455, A025465, A025469.

(define (A025468 n) (let loop ((x (A048766 n)) (s 0)) (let* ((x3 (A000578 x)) (y3 (- n x3))) (if (<= x3 y3) s (loop (- x 1) (+ s (if (and (> y3 0) (= (A000578 (A048766 y3)) y3)) 1 0))))))) ;; Antti Karttunen, Aug 28 2017

From Antti Karttunen, Aug 28-29 2017: (Start)
a(n) = A025465(n) - A025469(n).
a(n) <= A025455(n).
(End)

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

A178489 a(n) = floor(n^(1/6)): integer part of sixth 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, 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, 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

Cf. A000196, A048766, A178487.

Floor[Power[Range[0,110], (6)^-1]] (* Harvey P. Dale, Jul 18 2011 *)

PARI
```
A178489(n)=floor(sqrtn(n+.5,6))
```

a(n) = sqrtnint(n, 6); \\ Michel Marcus, Dec 22 2016

G.f.: Sum_{k>=1} x^(k^6)/(1 - x). - Ilya Gutkovskiy, Dec 22 2016

A105209 Nearest integer to the cube root of n.

Original entry on oeis.org

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, 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, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 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

Offset: 1

Pahikkala Jussi, Apr 13 2005

			a(16) = 3 because 16^(1/3) = 2.519842...

Erwin Voellmy, Fünfstellige Logarithmen und Zahlentafeln, Orell Füssli Verlag, Zürich (1962).

Seung-Jin Bang, Problem 10212, The American Mathematical Monthly, Vol. 99, No. 4 (1992), p. 361, Nearest Integer Zeta Functions, solution to Problem 10212, ibid., Vol. 101, No. 6 (1994), pp. 579-580.

Cf. A000194, A048766.

for n from 1 to 200 do printf(`%d,`,round(n^(1/3))) od: # James Sellers, Apr 21 2005

Round[Surd[Range[110],3]] (* Harvey P. Dale, Feb 28 2015 *)

a(n) = round(n^(1/3)); \\ Michel Marcus, Aug 19 2016

from sympy import integer_nthroot
def A105209(n): return (m:=integer_nthroot(n,3)[0])+((n<<3)>=((m<<1)+1)**3) # Chai Wah Wu, Jun 06 2025

Sum_{n>=1} 1/a(n)^s = 3*zeta(s-2) + zeta(s)/4^s, for s > 3 (Seung-Jin Bang, 1992). - Amiram Eldar, Oct 31 2020

More terms from James Sellers, Apr 21 2005

cp's OEIS Frontend

A135663 a(n) = floor(sqrt(n) - n^(1/4)).

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A135664 a(n) = ceiling(sqrt(n) - n^(1/5)).

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A135665 a(n) = floor(sqrt(n) - n^(1/5)).

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A211668 Number of iterations sqrt(sqrt(sqrt(...(n)...))) such that the result is < 3.

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A025455 a(n) is the number of partitions of n into 2 positive cubes.

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A255270 Integer part of fourth root of n.

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A025468 a(n) is the number of partitions of n into 2 distinct positive cubes.

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