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 1-5 of 5 results.

A247112 Floor of sums of the cubes of the non-integer square roots of n, as partitioned by the integer roots: floor(Sum_{j=n^2+1..(n+1)^2-1} j^(3/2)).

Original entry on oeis.org

0, 8, 67, 267, 746, 1690, 3333, 5957, 9892, 15516, 23255, 33583, 47022, 64142, 85561, 111945, 144008, 182512, 228267, 282131, 345010, 417858, 501677, 597517, 706476, 829700, 968383, 1123767, 1297142, 1489846, 1703265, 1938833, 2198032, 2482392, 2793491
Offset: 0

Views

Author

Richard R. Forberg, Dec 02 2014

Keywords

Comments

The fractional portion of each sum converges to 1/40 as n --> infinity.
The corresponding sum for square roots of n is given by A014105 or n*(2n+1) with a fractional portion converging to 1/6.
See A248575 for the corresponding sums for the cube roots.
See A248621 for the corresponding sums of squares of the cube roots.
See A248698 for the corresponding sum of the fourth roots.
Conjecture: the corresponding sums for all fractional (rational) powers of n (e.g., 5/2, 7/2, 9/2, ..., 1/3, 2/3, 4/3, ..., 1/4, 3/4, 5/4, ..., 1/5, 2/5, 3/5, ...) will have polynomial integer formulas or recursive integer formulas for their floor, ceiling and/or rounded values, with convergence to a rational fractional portion, with possibly multiple fractional values in a repeating pattern as they converge. This was clear for some additional examples, less so for higher-order examples.

Crossrefs

Programs

  • Magma
    [n eq 0 select 0 else Floor(&+[j^(3/2): j in [n^2+1..(n+1)^2-1]]): n in [0..50]]; // Bruno Berselli, Dec 03 2014
  • Mathematica
    Table[1/2 (n + 5 n^2 + 6 n^3 + 4 n^4), {n, 0, 50}]

Formula

a(n) = floor(Sum_{j=n^2+1..(n+1)^2-1} j^(3/2)).
a(n) = n *(4*n^3 + 6*n^2 + 5*n + 1)/2.
G.f.: x*(8 + 27*x + 12*x^2 + x^3)/(1 - x)^5. - Bruno Berselli, Dec 03 2014
E.g.f.: exp(x)*x*(16 + 51*x + 30*x^2 + 4*x^3)/2. - Stefano Spezia, Jul 13 2024

A248698 Floor of sums of the non-integer fourth roots of n, as partitioned by the integer roots: floor(Sum_{j=n^4+1..(n+1)^4-1} j^(1/4)).

Original entry on oeis.org

0, 23, 166, 621, 1676, 3715, 7218, 12761, 21016, 32751, 48830, 70213, 97956, 133211, 177226, 231345, 297008, 375751, 469206, 579101, 707260, 855603, 1026146, 1221001, 1442376, 1692575, 1973998, 2289141, 2640596, 3031051, 3463290, 3940193, 4464736, 5039991
Offset: 0

Views

Author

Richard R. Forberg, Dec 02 2014

Keywords

Comments

The fractional portion of each sum converges to 3/10.
See A247112 for references to other related sequences and a conjecture.

Crossrefs

Programs

  • Mathematica
    Table[3 n + 8 n^2 + 8 n^3 + 4 n^4, {n, 0, 50}]
  • PARI
    a(n) = floor(sum(j=n^4+1, (n+1)^4-1, j^(1/4))); \\ Michel Marcus, Dec 22 2014
    
  • PARI
    concat(0, Vec(-x*(x^3+21*x^2+51*x+23)/(x-1)^5 + O(x^100))) \\ Colin Barker, Dec 30 2014

Formula

a(n) = floor(Sum_{j=n^4+1..(n+1)^4-1} j^(1/4)).
a(n) = 3*n + 8*n^2 + 8*n^3 + 4*n^4.
G.f.: -x*(x^3+21*x^2+51*x+23) / (x-1)^5. - Colin Barker, Dec 30 2014
E.g.f.: exp(x)*x*(23 + 60*x + 32*x^2 + 4*x^3). - Stefano Spezia, Jul 13 2024

A248621 Floor of sums of the squares of the non-integer cube roots of n, as partitioned by the integer roots: floor(Sum_{j=n^3+1..(n+1)^3-1} j^(2/3)).

Original entry on oeis.org

0, 16, 120, 456, 1240, 2760, 5376, 9520, 15696, 24480, 36520, 52536, 73320, 99736, 132720, 173280, 222496, 281520, 351576, 433960, 530040, 641256, 769120, 915216, 1081200, 1268800, 1479816, 1716120, 1979656, 2272440, 2596560, 2954176, 3347520, 3778896, 4250680
Offset: 0

Views

Author

Richard R. Forberg, Dec 02 2014

Keywords

Comments

The fractional portion of each sum converges to 1/10.
See A248575 for the corresponding sums of the cube root.
See A247112 for the cube of the square roots, other references and a conjecture.

Crossrefs

Programs

  • Mathematica
    Table[2 n + 5 n^2 + 6 n^3 + 3 n^4, {n, 0, 50}]
  • PARI
    a(n) = floor(sum(j=n^3+1, (n+1)^3-1, j^(2/3))); \\ Michel Marcus, Dec 22 2014
    
  • PARI
    concat(0, Vec(-8*x*(x+2)*(2*x+1)/(x-1)^5 + O(x^100))) \\ Colin Barker, Dec 30 2014

Formula

a(n) = floor(Sum_{j=n^3+1..(n+1)^3-1} j^(2/3)).
a(n) = 2*n + 5*n^2 + 6*n^3 + 3*n^4.
G.f.: -8*x*(x+2)*(2*x+1) / (x-1)^5. - Colin Barker, Dec 30 2014
E.g.f.: exp(x)*x*(16 + 44*x + 24*x^2 + 3*x^3). - Stefano Spezia, Jul 13 2024

A374384 a(n) = floor(Sum_{k=n^3..(n+1)^3} k^(1/3)).

Original entry on oeis.org

1, 12, 51, 134, 281, 508, 835, 1278, 1857, 2588, 3491, 4582, 5881, 7404, 9171, 11198, 13505, 16108, 19027, 22278, 25881, 29852, 34211, 38974, 44161, 49788, 55875, 62438, 69497, 77068, 85171, 93822, 103041, 112844, 123251, 134278, 145945, 158268, 171267, 184958, 199361
Offset: 0

Views

Author

Amrit Awasthi, Jul 07 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Floor[Sum[(n^3+k)^(1/3),{k,0,3n^2+3n+1}]],{n,0,40}] (* Stefano Spezia, Jul 07 2024 *)
  • PARI
    a(n) = 3*n^3+9*n^2\2+4*n+1; \\ Michel Marcus, Jul 09 2024

Formula

a(n) = floor(3*n^3+9*n^2/2+4*n+1).
a(2*n) = 24*n^3 + 18*n^2 + 8*n + 1.
a(2*n-1) = 24*n^3-18*n^2+8*n-2 for n > 0.
a(2*n) = A248575(2*n) + 4*n + 1.
a(2*n-1) = A248575(2*n-1) + 4*n - 2.
From Stefano Spezia, Jul 09 2024: (Start)
G.f.: (1 + 9*x + 17*x^2 + 7*x^3 + 2*x^3)/((1 - x)^4*(1 + x)).
E.g.f.: exp(x)*(1 + 11*x + 14*x^2 + 3*x^3). (End)

A374489 a(n) = floor(Sum_{k=n^4..(n+1)^4} k^(1/4)).

Original entry on oeis.org

1, 26, 171, 628, 1685, 3726, 7231, 12776, 21033, 32770, 48851, 70236, 97981, 133238, 177255, 231376, 297041, 375786, 469243, 579140, 707301, 855646, 1026191, 1221048, 1442425, 1692626, 1974051, 2289196, 2640653, 3031110, 3463351, 3940256, 4464801, 5040058, 5669195
Offset: 0

Views

Author

Amrit Awasthi, Jul 09 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{1,26,171,628,1685},40] (* Harvey P. Dale, Nov 07 2024 *)

Formula

a(n) = 4*n^4+8*n^3+8*n^2+5*n+1.
From Stefano Spezia, Jul 09 2024: (Start)
G.f.: (1 + 21*x + 51*x^2 + 23*x^3)/(1 - x)^5.
E.g.f.: exp(x)*(1 + 25*x + 60*x^2 + 32*x^3 + 4*x^4). (End)
Showing 1-5 of 5 results.