A101094 -id:A101094 - cp's OEIS frontend

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

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

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

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

D. J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al.), National Center for Mathematics Education, University of Gothenburg, Sweden, in press.
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube. [broken link: domain now owned by a domain grabber]
Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883.
Eric Weisstein, Link to section of MathWorld: Eulerian Number.
Eric Weisstein, Link to section of MathWorld: Nexus number.
Eric Weisstein, Link to section of MathWorld: Finite Differences.
Dominika Závacká, Cristina Dalfó, and Miquel Angel Fiol, Integer sequences from k-iterated line digraphs, CEUR: Proc. 24th Conf. Info. Tech. - Appl. and Theory (ITAT 2024) Vol 3792, 156-161. See p. 161, Table 2.
Index entries for linear recurrences with constant coefficients, signature (1).

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.

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 *)

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

I wish the sequence was as interesting as the list of references! - N. J. A. Sloane

New name from Joerg Arndt, Nov 30 2014

A213564 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = (n-1+h)^2, n>=1, h>=1, and ** = convolution.

Original entry on oeis.org

1, 7, 4, 27, 21, 9, 77, 67, 43, 16, 182, 167, 127, 73, 25, 378, 357, 297, 207, 111, 36, 714, 686, 602, 467, 307, 157, 49, 1254, 1218, 1106, 917, 677, 427, 211, 64, 2079, 2034, 1890, 1638, 1302, 927, 567, 273, 81, 3289, 3234, 3054, 2730, 2282, 1757

Offset: 1

Clark Kimberling, Jun 18 2012

Principal diagonal:  A213565
Antidiagonal sums:  A101094
Row 1,  (1,3,6,...)**(1,4,9,...):  A005585
Row 2,  (1,3,6,...)**(4,9,16,...):  (k^5 +25*k^4 + 60*k^3 + 215*k^2 + 59*k)/60
Row 3,  (1,3,6,...)**(9,16,25,...):  (k^5 +35*k^4 + 30*k^3 + 505*k^2 + 149*k)/60
For a guide to related arrays, see A213500.

			Northwest corner (the array is read by falling antidiagonals):
1....7.....27....77....182
4....21....67....167...357
9....43....127...297...602
16...73....207...467...917
25...111...307...677...1302
36...157...427...927...1757

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A213500.

b[n_] := n (n + 1)/2; c[n_] := n^2
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213564 *)
d = Table[t[n, n], {n, 1, 40}] (* A213565 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A101094 *)

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) - T(n,k-6).

G.f. for row n: f(x)/g(x), where f(x) = n^2 - (2*n^2 - 2n - 1)*x + ((n - 1)^2)*x^2 and g(x) = (1 - x)^6.

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

Luciano Ancora, Feb 17 2015

			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)

Luciano Ancora, Table of n, a(n) for n = 1..1000
Luciano Ancora, Partial sums of m-th powers with Faulhaber polynomials.
Luciano Ancora, Pascal's triangle and recurrence relations for partial sums of m-th powers.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A000537, A000578, A003215, A024166, A101094, A101097, A101102, A254469, A254870, A254871, A254872.

[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

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 *)

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

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

A101854 a(n) = n(n+1)(n^2 + 21*n + 50)/24.

Original entry on oeis.org

6, 24, 61, 125, 225, 371, 574, 846, 1200, 1650, 2211, 2899, 3731, 4725, 5900, 7276, 8874, 10716, 12825, 15225, 17941, 20999, 24426, 28250, 32500, 37206, 42399, 48111, 54375, 61225, 68696, 76824, 85646, 95200, 105525, 116661, 128649, 141531, 155350

Offset: 1

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

5th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 3: 1,4,1.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

5th row of the array shown in A101853.

Partial sums of A101853.

Table[25 n/12+(71n^2)/24+(11n^3)/12+n^4/24,{n,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{6,24,61,125,225},40] (* Harvey P. Dale, Nov 05 2011 *)

G.f.: x*(6 - 6*x + x^2)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009; checked and corrected by R. J. Mathar, Sep 16 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), n > 5. - Harvey P. Dale, Nov 05 2011
E.g.f.: exp(x)*x*(144 + 144*x + 28*x^2 + x^3)/24. - Stefano Spezia, Oct 14 2022

Formula moved to be the definition by Eric M. Schmidt, Dec 12 2013

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

Luciano Ancora, Apr 23 2015

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).

			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).

Index entries for linear recurrences with constant coefficients, signature (5, -9, 7, -2).

Cf. A000225, A000670, A050488, A208744, A257449, A257450.

[13*(2^n-1)-3*n^2-9*n: n in [1..30]]; // Vincenzo Librandi, Apr 24 2015

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 *)

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

Edited by Bruno Berselli, Apr 28 2015

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

Ilya Gutkovskiy, Oct 11 2017

nonn

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

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Index entries for sequences related to sums of cubes

Cf. A000537, A000578, A002054, A024166, A101094, A101097, A101102, A265612, A302352.

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]!

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).

A125166 Triangle R(n,k), companion to A125165, left column n^3.

Original entry on oeis.org

1, 8, 1, 27, 9, 1, 64, 36, 10, 1, 125, 100, 46, 11, 1, 216, 225, 146, 57, 12, 1, 343, 441, 371, 203, 69, 13, 1, 512, 784, 812, 574, 272, 82, 14, 1, 729, 1296, 1596, 1386, 846, 354, 96, 15, 1, 1000, 2025, 2892, 2982, 2232, 1200, 450, 111, 16, 1

Offset: 0

Gary W. Adamson, Nov 22 2006

Riordan array ((1 + 4*x + x^2)/(1 - x)^4, x/(1 - x)). - Philippe Deléham, Dec 09 2013

			With other offset (k >= 1) from first formula: R(5,3) = 146 = R(4,3) + R(4,2) = 46 + 100.
The Riordan triangle R begins:
  n\k|     0    1    2    3    4    5   6   7   8  9
  --------------------------------------------------
   0 |     1
   1 |     8    1
   2 |    27    9    1
   3 |    64   36   10    1
   4 |   125  100   46   11    1
   5 |   216  225  146   57   12    1
   6 |   343  441  371  203   69   13   1
   7 |   512  784  812  574  272   82  14   1
   8 |   729 1296 1596 1386  846  354  96  15   1
   9 |  1000 2025 2892 2982 2232 1200 450 111  16  1
... refomatted and extended by _Wolfdieter Lang_, Mar 25 2025.

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).

Cf. A000578 (column 0), A000537 (column 1), A024166 (column 2), A101094 (column 3).
Cf. A257448 (row sums).
Cf. A008292, A028246, A097805, A125165, A050488, A382057.

A125166[n_, k_] := A125166[n, k] = Switch[k, 0, (n + 1)^3, n, 1, _, A125166[n - 1, k - 1] + A125166[n - 1, k]];
Table[A125166[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Apr 08 2025 *)

y = polygen(QQ, 'y')
x = y.parent()[['x']].gen()
gf = (1 + 4*x + x^2)/((x - 1)^3*(y*x + x - 1))
[list(u) for u in list(gf.O(11))]  # Peter Luschny, Apr 02 2025

Binomial transform of an infinite matrix M with diagonal D, subdiagonal (D-1)..., etc; as follows: (D) = (1,1,1...); (D-1) = (7,7,7...); (D-2) = (12,12,12...); (D-3) = (6,6,6...). Alternatively, given left border n^3: (1, 8, 27, 64...); for k>1, R(n,k) = R(n-1,k) + R(n-1,k-1).
From Wolfdieter Lang, Mar 27 2025: (Start)
Riordan triangle (see a comment above):
R(n, 0) = (n+1)^3, R(n, k) = R(n-1, k-1) + R(n-1, k), for k >= 1. (from the (finite) A-sequence {1, 1} with offset 0),
R(n, 0) = Sum_{k=0..n-1} Z(j)*R(n-1, k), for n >= 1, and R(0, 0) = 1, with the Riordan Z-sequence A382057. For Riordan A- and Z-sequences see the first W. Lang link in A006232.
R(n, k) = Sum{j=0..n} (j+1)^3*A097805(n-j, k) (convolution property).
R(n, k) = Sum{j=0..3} A028246(4, j+1)*binomial(n, k+j). (Proof for k=0 with a standard (n+1)^3  formula, and for k >= 1 using the Pascal type recurrence for the triangle.)
O.g.f. column k (with leading 0s): ((1 + 4*x + x^2)/(1 - x)^4)*(x/(1-x))^k. (Numerator polynomial from row 3 of the triangle A008292.)
O.g.f. row polynomials P(n, y) = Sum_{k=0..n} R(n, k)*y^k:
   G(y, x) = (1 + 4*x + x^2)/((1 - x)^3*(1 - (1+y)*x)). (End)

a(27) corrected by Georg Fischer, Feb 18 2020

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

Luciano Ancora, Jul 08 2015

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

			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

Cf. A000537, A024166, A101094, A101097, A101102, A254469.

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}]

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

cp's OEIS Frontend

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

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A213564 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = (n-1+h)^2, n>=1, h>=1, and ** = convolution.

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A254869 Seventh partial sums of cubes (A000578).

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A101854 a(n) = n*(n+1)*(n^2 + 21*n + 50)/24.

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

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

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A125166 Triangle R(n,k), companion to A125165, left column n^3.

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SageMath

A101854 a(n) = n(n+1)(n^2 + 21*n + 50)/24.

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

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