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.

Previous Showing 11-14 of 14 results.

A254869 Seventh partial sums of cubes (A000578).

Original entry on oeis.org

1, 15, 111, 561, 2211, 7293, 21021, 54483, 129558, 286858, 598026, 1184118, 2242266, 4083366, 7184166, 12257850, 20348031, 32951985, 52179985, 80958735, 123288165, 184562235, 271965915, 394962165, 565884540, 800652996, 1119632580, 1548656956
Offset: 1

Views

Author

Luciano Ancora, Feb 17 2015

Keywords

Examples

			2nd differences:   0,  6,  12,  18,   24,   30, ... (A008588)
1st differences:   1,  7,  19,  37,   61,   91, ... (A003215)
-------------------------------------------------------------------
The cubes:         1,  8,  27,  64,  125,  216, ... (A000578)
-------------------------------------------------------------------
1st partial sums:  1,  9,  36, 100,  225,  441, ... (A000537)
2nd partial sums:  1, 10,  46, 146,  371,  812, ... (A024166)
3rd partial sums:  1, 11,  57, 203,  574, 1386, ... (A101094)
4th partial sums:  1, 12,  69, 272,  846, 2232, ... (A101097)
5th partial sums:  1, 13,  82, 354, 1200, 3432, ... (A101102)
6th partial sums:  1, 14,  96, 450, 1650, 5082, ... (A254469)
7th partial sums:  1, 15, 111, 561, 2211, 7293, ... (this sequence)
		

Crossrefs

Programs

  • Magma
    [n*(1+n)*(2+n)*(3+n)*(4+n)*(5+n)*(6+n)*(7+n)*(7+7*n+n^2)/604800: n in [1..30]]; // Vincenzo Librandi, Feb 19 2015
  • Mathematica
    Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (7 + 7 n + n^2)/604800, {n, 26}] (* or *)
    CoefficientList[Series[(- 1 - 4 x - x^2)/(- 1 + x)^11, {x, 0, 25}], x]
    Nest[Accumulate,Range[30]^3,7] (* or *) LinearRecurrence[{11,-55,165,-330,462,-462,330,-165,55,-11,1},{1,15,111,561,2211,7293,21021,54483,129558,286858,598026},30] (* Harvey P. Dale, Apr 24 2017 *)
  • PARI
    vector(50, n, n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 7*n + n^2)/604800) \\ Derek Orr, Feb 19 2015
    

Formula

G.f.: x*(1 + 4*x + x^2)/(1 - x)^11.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 7*n + n^2)/604800.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + n^3.
Sum_{n>=1} 1/a(n) = 1920*sqrt(3/7)*Pi*tan(sqrt(21)*Pi/2) - 251488/49. - Amiram Eldar, Jan 26 2022

A257448 a(n) = 13*(2^n - 1) - 3*n^2 - 9*n.

Original entry on oeis.org

1, 9, 37, 111, 283, 657, 1441, 3051, 6319, 12909, 26149, 52695, 105859, 212265, 425161, 851043, 1702903, 3406725, 6814477, 13630095, 27261451, 54524289, 109050097, 218101851, 436205503, 872412957, 1744828021, 3489658311, 6979319059, 13958640729
Offset: 1

Views

Author

Luciano Ancora, Apr 23 2015

Keywords

Comments

These numbers belong to a family of sequences obtained as follows:
. A000225: 1*(2^n-1);
. A050488: 3*(2^n-1) - 2*n;
. a(n): 13*(2^n-1) - 3*n^2 - 9*n;
. A257449: 75*(2^n-1) - 4*n^3 - 18*n^2 - 52*n;
. A257450: 541*(2^n-1) - 5*n^4 - 30*n^3 - 130*n^2 - 375*n,
where the sequence 1, 3, 13, 75, 541, ... is A000670 (after the first term), and A208744 gives the triangle of coefficients:
2;
3, 9;
4, 18, 52;
5, 30, 130, 375;
6, 45, 260, 1125, 3246;
7, 63, 455, 2625, 11361, 32781, etc.
Also, the antidiagonal sums in the array are given by the formula (6*n^2 + 6*k*n + (k-1)*k)*(k+n)!/((k+3)!*(n-1)!) for k = 0, 1, 2, 3, 4, ... (see Example field).

Examples

			By the second comment, the array begins (antidiagonals in A046902):
k=0: 1,  8, 27,  64,  125,  216, ...  A000578
k=1: 1,  9, 36, 100,  225,  441, ...  A000537
k=2: 1, 10, 46, 146,  371,  812, ...  A024166
k=3: 1, 11, 57, 203,  574, 1386, ...  A101094
k=4: 1, 12, 69, 272,  846, 2232, ...  A101097
k=5: 1, 13, 82, 354, 1200, 3432, ...  A101102
k=6: 1, 14, 96, 450, 1650, 5082, ...  A254469
...
See also A254469 (Example field).
		

Crossrefs

Programs

  • Magma
    [13*(2^n-1)-3*n^2-9*n: n in [1..30]]; // Vincenzo Librandi, Apr 24 2015
  • Mathematica
    Table[13 (2^n - 1) - 3 n^2 - 9n, {n, 30}]
    CoefficientList[Series[x (1 + 4 x + x^2)/((1 - x)^3*(1 - 2 x)), {x, 0, 30}], x] (* Michael De Vlieger, Nov 14 2016 *)

Formula

G.f.: x*(1+4*x+x^2)/((1-x)^3*(1-2*x)).
a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4) for n>4. - Ray Chandler, Jul 25 2015

Extensions

Edited by Bruno Berselli, Apr 28 2015

A293550 a(n) = Sum_{k=0..n} k^3*binomial(2*n-k,n).

Original entry on oeis.org

0, 1, 11, 69, 354, 1650, 7293, 31213, 130832, 540702, 2212550, 8989090, 36327810, 146228940, 586823265, 2349424125, 9389012160, 37467344310, 149345215290, 594753416790, 2366845396500, 9413555798556, 37423053793026, 148719333293394, 590842248405024, 2346813893147500
Offset: 0

Views

Author

Ilya Gutkovskiy, Oct 11 2017

Keywords

Comments

Main diagonal of iterated partial sums array of cubes (starting with the first partial sums). For nonnegative integers see A002054, for squares see A265612.

Crossrefs

Programs

  • Mathematica
    Table[Sum[k^3 Binomial[2 n - k, n], {k, 0, n}], {n, 0, 25}]
    Table[SeriesCoefficient[x (1 + 4 x + x^2)/(1 - x)^(n + 5), {x, 0, n}], {n, 0, 25}]
    Table[2^(2 n + 1) n^2 (13 n + 7) Gamma[n + 3/2]/(Sqrt[Pi] Gamma[n + 5]), {n, 0, 25}]
    CoefficientList[Series[(6 - 6 Sqrt[1 - 4 x] - 36 x + 24 Sqrt[1 - 4 x] x + 55 x^2 - 19 Sqrt[1 - 4 x] x^2 - 15 x^3 + Sqrt[1 - 4 x] x^3)/(2 Sqrt[1 - 4 x] x^4), {x, 0, 25}], x]
    CoefficientList[Series[(E^(2 x) (36 - 24 x + 13 x^2) BesselI[0, 2 x])/x^2 + (E^(2 x) (-36 + 24 x - 31 x^2 + 13 x^3) BesselI[1, 2 x])/x^3, {x, 0, 25}], x]* Range[0, 25]!

Formula

a(n) = [x^n] x*(1 + 4*x + x^2)/(1 - x)^(n+5).
a(n) = 2^(2*n+1)*n^2*(13*n + 7)*Gamma(n+3/2)/(sqrt(Pi)*Gamma(n+5)).
a(n) ~ 26*4^n/sqrt(Pi*n).

A259914 Staircase path through the array P(n,k) of the k-th partial sums of cubes (A000578).

Original entry on oeis.org

1, 9, 10, 46, 57, 203, 272, 846, 1200, 3432, 5082, 13728, 21021, 54483, 85696, 215254, 346086, 848198, 1388900, 3337236, 5549786, 13119614, 22108704, 51557260, 87885070, 202588830, 348817770, 796117860, 1382941125, 3129153795
Offset: 1

Views

Author

Luciano Ancora, Jul 08 2015

Keywords

Comments

The term "stepped path" in the name field is the same used in A001405 and A259775.

Examples

			The array begins:
[1], [9],  36,   100,   225,    441,  ...  A000537
1,  [10], [46],  146,   371,    812,  ...  A024166
1,   11,  [57], [203],  574,   1386,  ...  A101094
1,   12,   69,  [272], [846],  2232,  ...  A101097
1,   13,   82,   354, [1200], [3432], ...  A101102
1,   14,   96,   450,  1650,  [5082], ...  A254469
		

Crossrefs

Programs

  • Mathematica
    Table[DifferenceRoot[Function[{a, n},
             {(-650880 - 1496112*n - 1426512*n^2 - 722164*n^3 - 204716*n^4 - 30812*n^5 - 1924*n^6)*a[n] + (-56736 - 140412*n - 132006*n^2 - 58114*n^3 - 12090*n^4 - 962*n^5)*a[1 + n] + (78624 + 229884*n + 273800*n^2 + 167579*n^3 + 54567*n^4 + 8665*n^5 + 481*n^6)*a[2 + n] == 0, a[1] == 1, a[2] == 9}]][n], {n, 30}]

Formula

Conjecture: 2*(n+7)*(145672*n^2-236343*n+123525)*a(n) +(-78613*n^3-794662*n^2+327391*n+20220)*a(n-1) +2*(-582688*n^3-1889455*n^2-2148719*n-832650)*a(n-2) +4*(n-1)*(78613*n^2+133361*n+64050)*a(n-3) = 0. - R. J. Mathar, Jul 16 2015
Previous Showing 11-14 of 14 results.