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.

A254681 Fifth partial sums of fourth powers (A000583).

Original entry on oeis.org

1, 21, 176, 936, 3750, 12342, 35112, 89232, 207207, 446875, 906048, 1743248, 3206268, 5670588, 9690000, 16062144, 25912029, 40797009, 62837104, 94875000, 140670530, 205134930, 294610680, 417203280, 583171875, 805386231
Offset: 1

Views

Author

Luciano Ancora, Feb 12 2015

Keywords

Examples

			Fourth differences:  1, 12,  23,  24, (repeat 24)  ...   (A101104)
Third differences:   1, 13,  36,  60,   84,   108, ...   (A101103)
Second differences:  1, 14,  50, 110,  194,   302, ...   (A005914)
First differences:   1, 15,  65, 175,  369,   671, ...   (A005917)
-------------------------------------------------------------------------
The fourth powers:   1, 16,  81, 256,  625,  1296, ...   (A000583)
-------------------------------------------------------------------------
First partial sums:  1, 17,  98, 354,  979,  2275, ...   (A000538)
Second partial sums: 1, 18, 116, 470, 1449,  3724, ...   (A101089)
Third partial sums:  1, 19, 135, 605, 2054,  5778, ...   (A101090)
Fourth partial sums: 1, 20, 155, 760, 2814,  8592, ...   (A101091)
Fifth partial sums:  1, 21, 176, 936, 3750, 12342, ...   (this sequence)
		

Crossrefs

Programs

  • Magma
    [Binomial(n+5,6)*n*(n+5)*(2*n+5)/42: n in [1..30]]; // G. C. Greubel, Dec 01 2018
    
  • Maple
    seq(coeff(series((x+11*x^2+11*x^3+x^4)/(1-x)^10,x,n+1), x, n), n = 1 .. 30); # Muniru A Asiru, Dec 02 2018
  • Mathematica
    Table[n^2(1+n)(2+n)(3+n)(4+n)(5+n)^2(5+2n)/30240, {n,26}] (* or *)
    CoefficientList[Series[(1 + 11 x + 11 x^2 + x^3)/(1-x)^10, {x,0,25}], x]
    CoefficientList[Series[(1/30240)E^x (30240 + 604800 x + 2041200 x^2 + 2368800 x^3 + 1233540 x^4 + 326592 x^5 + 46410 x^6 + 3540 x^7 + 135 x^8 + 2 x^9), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* Stefano Spezia, Dec 02 2018 *)
    Nest[Accumulate[#]&,Range[30]^4,5] (* Harvey P. Dale, Jan 03 2022 *)
  • PARI
    my(x='x+O('x^30)); Vec((x+11*x^2+11*x^3+x^4)/(1-x)^10) \\ G. C. Greubel, Dec 01 2018
    
  • Sage
    [binomial(n+5,6)*n*(n+5)*(2*n+5)/42 for n in (1..30)] # G. C. Greubel, Dec 01 2018

Formula

G.f.:(x + 11*x^2 + 11*x^3 + x^4)/(1 - x)^10.
a(n) = n^2*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)^2*(5 + 2*n)/30240.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^4.
E.g.f.: (1/30240)*exp(x)*(30240 + 604800*x + 2041200*x^2 + 2368800*x^3 + 1233540*x^4 + 326592*x^5 + 46410*x^6 + 3540*x^7 + 135*x^8 + 2*x^9). - Stefano Spezia, Dec 02 2018
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=1} 1/a(n) = 172032*log(2)/125 - 2382233/2500.
Sum_{n>=1} (-1)^(n+1)/a(n) = 42*Pi^2/25 - 43008*Pi/125 + 2663213/2500. (End)

A101104 a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4.

Original entry on oeis.org

1, 12, 23, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24
Offset: 1

Views

Author

Cecilia Rossiter, Dec 15 2004

Keywords

Comments

Original name: The first summation of row 4 of Euler's triangle - a row that will recursively accumulate to the power of 4.

Crossrefs

For other sequences based upon MagicNKZ(n,k,z):
..... | n = 1 | n = 2 | n = 3 | n = 4 | n = 5 | n = 6 | n = 7
---------------------------------------------------------------------------
z = 0 | A000007 | A019590 | .......MagicNKZ(n,k,0) = A008292(n,k+1) .......
z = 1 | A000012 | A040000 | A101101 | thisSeq | A101100 | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | A101095 | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181
Cf. A101095 for an expanded table and more about MagicNKZ.

Programs

  • Mathematica
    MagicNKZ = Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 1, 1}, {k, 0, 34}]
    Join[{1, 12, 23},LinearRecurrence[{1},{24},56]] (* Ray Chandler, Sep 23 2015 *)

Formula

a(k) = MagicNKZ(4,k,1) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n (cf. A101095). That is, a(k) = Sum_{j=0..k+1} (-1)^j*binomial(4, j)*(k-j+1)^4.
a(1)=1, a(2)=12, a(3)=23, and a(n)=24 for n>=4. - Joerg Arndt, Nov 30 2014
G.f.: x*(1+11*x+11*x^2+x^3)/(1-x). - Colin Barker, Apr 16 2012

Extensions

New name from Joerg Arndt, Nov 30 2014
Original Formula edited and Crossrefs table added by Danny Rorabaugh, Apr 22 2015

A101095 Fourth difference of fifth powers (A000584).

Original entry on oeis.org

1, 28, 121, 240, 360, 480, 600, 720, 840, 960, 1080, 1200, 1320, 1440, 1560, 1680, 1800, 1920, 2040, 2160, 2280, 2400, 2520, 2640, 2760, 2880, 3000, 3120, 3240, 3360, 3480, 3600, 3720, 3840, 3960, 4080, 4200, 4320, 4440, 4560, 4680, 4800, 4920, 5040, 5160, 5280
Offset: 1

Views

Author

Cecilia Rossiter, Dec 15 2004

Keywords

Comments

Original Name: Shells (nexus numbers) of shells of shells of shells of the power of 5.
The (Worpitzky/Euler/Pascal Cube) "MagicNKZ" algorithm is: MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n, with k>=0, n>=1, z>=0. MagicNKZ is used to generate the n-th accumulation sequence of the z-th row of the Euler Triangle (A008292). For example, MagicNKZ(3,k,0) is the 3rd row of the Euler Triangle (followed by zeros) and MagicNKZ(10,k,1) is the partial sums of the 10th row of the Euler Triangle. This sequence is MagicNKZ(5,k-1,2).

Crossrefs

Fourth differences of A000584, third differences of A022521, second differences of A101098, and first differences of A101096.
For other sequences based upon MagicNKZ(n,k,z):
...... | n = 1 | n = 2 | n = 3 | n = 4 | n = 5 | n = 6 | n = 7 | n = 8
--------------------------------------------------------------------------------------
z = 0 | A000007 | A019590 | ....... MagicNKZ(n,k,0) = T(n,k+1) from A008292 .......
z = 1 | A000012 | A040000 | A101101 | A101104 | A101100 | ....... | ....... | .......
z = 2 | A000027 | A005408 | A008458 | A101103 | thisSeq | ....... | ....... | .......
z = 3 | A000217 | A000290 | A003215 | A005914 | A101096 | ....... | ....... | .......
z = 4 | A000292 | A000330 | A000578 | A005917 | A101098 | ....... | ....... | .......
z = 5 | A000332 | A002415 | A000537 | A000583 | A022521 | ....... | A255181 | .......
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

Programs

  • Magma
    I:=[1,28,121,240,360]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, May 07 2015
    
  • Mathematica
    MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 2, 2}, {k, 0, 34}]
    CoefficientList[Series[(1 + 26 x + 66 x^2 + 26 x^3 + x^4)/(1 - x)^2, {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
    Join[{1,28,121,240},Differences[Range[50]^5,4]] (* or *) LinearRecurrence[{2,-1},{1,28,121,240,360},50] (* Harvey P. Dale, Jun 11 2016 *)
  • PARI
    a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015
  • Sage
    [1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015
    

Formula

a(k+1) = Sum_{j=0..k+1} (-1)^j*binomial(n + 1 - z, j)*(k - j + 1)^n; n = 5, z = 2.
For k>3, a(k) = Sum_{j=0..4} (-1)^j*binomial(4, j)*(k - j)^5 = 120*(k - 2).
a(n) = 2*a(n-1) - a(n-2), n>5. G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (1-x)^2. - Colin Barker, Mar 01 2012

Extensions

MagicNKZ material edited, Crossrefs table added, SeriesAtLevelR material removed by Danny Rorabaugh, Apr 23 2015
Name changed and keyword 'uned' removed by Danny Rorabaugh, May 06 2015

A101101 a(1)=1, a(2)=5, and a(n)=6 for n >= 3.

Original entry on oeis.org

1, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6
Offset: 1

Views

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004

Keywords

Comments

Previous name was: The first summation of row 3 of Euler's triangle - a row that will recursively accumulate to the power of 3.
Decimal expansion of 47/30. - Elmo R. Oliveira, Aug 09 2024

Crossrefs

Within the "cube" of related sequences with construction based upon MaginNKZ formula, with n downward, k rightward and z backward:
Before: this_sequence, A008458, A003215, A000578, A000537, A024166 or A024166, A101094, A101097, A101102.
Above: this_sequence, below: A101104, A101100.
Within the "cube" of related sequences with construction based upon SeriesAtLevelR formula, with n downward, x rightward and r backward:
Above: this_sequence, below: A101103, A101096.

Programs

  • Mathematica
    MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 3, 3}, {z, 1, 1}, {k, 0, 34}] (* OR *)
    SeriesAtLevelR = Sum[Eulerian[n, i - 1]*Binomial[n + x - i + r, n + r], {i, 1, n}]; Table[SeriesAtLevelR, {n, 3, 3}, {r, -3, -3}, {x, 4, 35}]
    Join[{1, 5},LinearRecurrence[{1},{6},78]] (* Ray Chandler, Sep 23 2015 *)

Formula

G.f.: x*(1+4*x+x^2)/(1-x). - L. Edson Jeffery, Jan 29 2012

Extensions

I wish the sequence was as interesting as the list of references! - N. J. A. Sloane
New name from Joerg Arndt, Nov 30 2014

A101860 a(n) = (3+n)*(2 + 33*n + n^2)/6.

Original entry on oeis.org

1, 24, 60, 110, 175, 256, 354, 470, 605, 760, 936, 1134, 1355, 1600, 1870, 2166, 2489, 2840, 3220, 3630, 4071, 4544, 5050, 5590, 6165, 6776, 7424, 8110, 8835, 9600, 10406, 11254, 12145, 13080, 14060, 15086, 16159, 17280, 18450, 19670, 20941
Offset: 0

Views

Author

Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 18 2004

Keywords

Comments

The 4th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 4: 1,11,11,1.
The partial sums of A101859 (plus a leading 1). 4th row in the array shown in the examples. The 2nd column is A101104, the 3rd column is A101103, the 4th column is A005914.

Examples

			Array with first column equal to the 4th row of A008292, and column k defined by partial sums of the preceding column k-1:
1  1 1 1  1  1  1  1  1  1  1
11 12 13 14 15 16 17 18 19 20 21
11 23 36 50 65 81 98 116 135 155 176
1  24 60 110 175 256 354 470 605 760 936   A101860
0  24 84 194 369 625 979 1449 2054 2814 3750   A101861
0  24 108 302 671 1296 2275 3724 5778 8592 12342  A101862
0  24 132 434 1105 2401 4676 8400 14178 22770 35112
0  24 156 590 1695 4096 8772 17172 31350 54120 89232
0  24 180 770 2465 6561 15333 32505 63855 117975 207207
... ... ... ... ... ... ... ... ... ...
		

Programs

  • Magma
    I:=[1, 24, 60, 110]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..45]]; // Vincenzo Librandi, Jun 26 2012
    
  • Mathematica
    LinearRecurrence[{4,-6,4,-1},{1,24,60,110},50] (* or *) CoefficientList[Series[(1+20*x-30*x^2+10*x^3)/(x-1)^4,{x,0,50}],x] (* Vincenzo Librandi, Jun 26 2012 *)
  • PARI
    a(n) = (n+3)*(n^2+33*n+2)/6; \\ Altug Alkan, Sep 23 2018

Formula

G.f.: ( 1 + 20*x - 30*x^2 + 10*x^3 ) / (x-1)^4 . - R. J. Mathar, Dec 06 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 26 2012
Showing 1-5 of 5 results.