cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A120721 Partial sums of A079645.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 28, 36, 46, 58, 72, 88, 106, 126, 148, 172, 198, 225, 255, 288, 324, 363, 405, 450, 498, 549, 603, 660, 720, 783, 847, 915, 987, 1063, 1143, 1227, 1315, 1407, 1503, 1603, 1707, 1815, 1927, 2043, 2163, 2287, 2412, 2542, 2677, 2817, 2962, 3112, 3267
Offset: 1

Views

Author

N. J. A. Sloane, Oct 29 2006

Keywords

Links

Crossrefs

Cf. A079645, A032378, A112873.

Programs

  • Magma
    A079645:=[n: n in [1..500] | n mod Floor(n^(1/3)) eq 0 ];
[(&+[A079645[k]: k in [1..n]]): n in [1..100]]; // G. C. Greubel, Jul 20 2023
  • Mathematica
    Accumulate[Select[Range[300],Divisible[#,Floor[CubeRoot[#]]]&]] (* Harvey P. Dale, Jun 19 2023 *)
  • SageMath
    A079645=[j for j in (1..500) if j%(floor(j^(1/3)))==0]
def A120721(n): return sum(A079645[k] for k in range(n+1))
[A120721(n) for n in range(101)] # G. C. Greubel, Jul 20 2023

Formula

a(n) = Sum_{j=1..n} A079645(j).
a(n) ~ 2^(5/2)*n^(5/2)/(5*3^(3/2)) - 5*n^2/6. - Vaclav Kotesovec, Oct 13 2024

A032378 Noncubes k that are divisible by floor(k^(1/3)).

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 10, 12, 14, 16, 18, 20, 22, 24, 26, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 222, 228, 234, 240
Offset: 1

Views

Author

N. J. A. Sloane, Dec 22 2001, corrected Oct 29 2006

Keywords

Comments

The Concrete Math Club Casino problem - non-cube winning slots.

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. 2nd Edition. Addison-Wesley, Reading, MA, 1994. Section 3.2, pp. 74-76.

Links

Crossrefs

Cf. A066353, A079645, A112873.

Programs

  • Magma
    [k*j: j in [(k^2+1)..(k^2+3*k+3)], k in [1..6]]; // G. C. Greubel, Jul 20 2023
  • Maple
    t1:=[]; for n from 1 to 500 do t2:=floor(n^(1/3)); if n mod t2 = 0 and t2^3 <> n then t1:=[op(t1),n]; fi; od:
# Alternate:
seq(seq(n,n=k^3+k..(k+1)^3-1,k),k=1..6); # Robert Israel, Mar 24 2020
  • Mathematica
    Select[Range[300],!IntegerQ[Surd[#,3]]&&Divisible[#,Floor[Surd[#,3]]]&] (* Harvey P. Dale, May 13 2020 *)
  • Python
    from itertools import count, islice
from sympy import integer_nthroot
def A032378_gen(): # generator of terms
    return 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)))
A032378_list = list(islice(A032378_gen(),40)) # Chai Wah Wu, Oct 12 2024
  • SageMath
    flatten([[k*j for j in range((k^2+1),(k^2+3*k+3)+1)] for k in range(1,7)]) # G. C. Greubel, Jul 20 2023

A113768 a(1) = 1, a(n+1) = a(n) + floor(a(n)^(1/3)).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 133, 138, 143, 148, 153, 158, 163, 168, 173, 178, 183, 188, 193, 198, 203, 208, 213, 218, 224, 230, 236, 242, 248, 254, 260
Offset: 1

Views

Author

Jonathan Vos Post, Jan 19 2006

Keywords

Comments

First 17 terms identical to A079645 (Integer part of the cube root of n divides n). Replacing cube root by square root gives A033638.

Links

Crossrefs

Cf. A033638, A048766, A079645.

Programs

  • Magma
    [n le 1 select 1 else Self(n-1)+Floor(Self(n-1)^(1/3)): n in [1..75]]; // Vincenzo Librandi, Jul 29 2019
  • Maple
    A[1]:= 1:
for n from 1 to 100 do A[n+1] := A[n] + floor(A[n]^(1/3)) od:
seq(A[i],i=1..100); # Robert Israel, Jul 28 2019
  • Mathematica
    NestList[#+Floor[Surd[#,+3]]&,1,70] (* Harvey P. Dale, Jan 21 2013 *)

Formula

Conjecture: a(n) ~ (2/3)*n*sqrt((2/3)*n). - José María Grau Ribas, Feb 13 2024

Extensions

Corrected and extended by Harvey P. Dale, Jan 21 2013

A204315 Numbers j such that floor(j^(1/4)) divides j.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126
Offset: 1

Views

Author

Benoit Cloitre, Jan 14 2012

Keywords

Examples

			26 is a term as floor(26^(1/4)) = 2 divides 26. - _David A. Corneth_, Oct 04 2023

Links

Crossrefs

Cf. A079645, A228318.

Programs

  • Maple
    isA204315 := proc(n)
    if modp(n,floor(root[4](n))) = 0 then
        true ;
    else
        false ;
    fi ;
end proc:
for n from 1 to 130 do
    if isA204315(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Sep 10 2017
  • Mathematica
    Select[Range[150],Mod[#,Floor[Surd[#,4]]]==0&] (* Harvey P. Dale, Oct 04 2023 *)
  • PARI
    a(n) = {my(k = 0, t = 0); while(t < n, k++; t = 4*k^3/3 + 5*k^2 + 26*k/3); (k+1)^4 - 1 - k * (t - n)} \\ David A. Corneth, Oct 06 2023
  • PARI
    first(n) = {my(res = vector(n), t = 0); for(i = 1, oo, forstep(j = i^4, (i + 1)^4 - 1, i, t++; if(t > n, return(res)); res[t] = j))} \\ David A. Corneth, Oct 06 2023

Formula

Let f(x) = 4*x^3/3 + 5*x^2 + 26*x/3 and let k be the smallest integer x such that f(x) >= n. Then a(n) = (k+1)^4 - 1 - k * (f(k) - n). - David A. Corneth, Oct 06 2023
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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