A005914 -id:A005914 - cp's OEIS frontend

A101103 Partial sums of A101104. First differences of A005914.

1, 13, 36, 60, 84, 108, 132, 156, 180, 204, 228, 252, 276, 300, 324, 348, 372, 396, 420, 444, 468, 492, 516, 540, 564, 588, 612, 636, 660, 684, 708, 732, 756, 780, 804, 828, 852, 876, 900, 924, 948, 972, 996, 1020, 1044, 1068, 1092, 1116, 1140, 1164, 1188, 1212, 1236, 1260

Offset: 1

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

For more information, cross-references etc., see A101104.

For n >= 3, a(n) is equal to the number of functions f:{1,2,3,4}->{1,2,...,n} such that Im(f) contains 3 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Mar 08 2007

G. C. Greubel, Table of n, a(n) for n = 1..5000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Claudio de J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq., Vol. 16 (2013) , Article 13.5.7.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A073762.

Concatenation([1,13],List([3..60],n->24*n-36)); # Muniru A Asiru, Dec 02 2018

I:=[36,60]; [1,13] cat [n le 2 select I[n] else 2*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 01 2018

seq(coeff(series(x*(1+x)*(1+10*x+x^2)/(1-x)^2,x,n+1), x, n), n = 1 .. 60); # Muniru A Asiru, Dec 02 2018

MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 4, 4}, {z, 2, 2}, {k, 0, 34}] OR SeriesAtLevelR = Sum[Eulerian[n, i - 1]*Binomial[n + x - i + r, n + r], {i, 1, n}]; Table[SeriesAtLevelR, {n, 4, 4}, {r, -3, -3}, {x, 3, 35}]
Join[{1, 13},LinearRecurrence[{2, -1},{36, 60},33]] (* Ray Chandler, Sep 23 2015 *)

my(x='x+O('x^60)); Vec(x*(1+x)*(1+10*x+x^2)/(1-x)^2) \\ G. C. Greubel, Dec 01 2018

s=(x*(1+x)*(1+10*x+x^2)/(1-x)^2).series(x, 30); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 01 2018

a(n) = 2*a(n-1) - a(n-2), n > 4.
G.f.: x*(1+x)*(1 + 10*x + x^2)/(1-x)^2.
a(n) = 24*n - 36, n >= 3.
a(n) = Sum_{j=0..n} (-1)^j*binomial(3, j)*(n - j)^4. [Indices shifted, Nov 01 2010]
a(n) = Sum_{i=1..4} A008292(4,i)*binomial(n-i+1,1). [Indices shifted, Nov 01 2010]
Sum_{n>=1} (-1)^(n+1)/a(n) = 157/156 - Pi/48. - Amiram Eldar, Jan 26 2022

Removed redundant information already in A101104. Reduced formulas by expansion of constants - R. J. Mathar, Nov 01 2010

A005917 Rhombic dodecahedral numbers: a(n) = n^4 - (n - 1)^4.

Original entry on oeis.org

1, 15, 65, 175, 369, 671, 1105, 1695, 2465, 3439, 4641, 6095, 7825, 9855, 12209, 14911, 17985, 21455, 25345, 29679, 34481, 39775, 45585, 51935, 58849, 66351, 74465, 83215, 92625, 102719, 113521, 125055, 137345, 150415, 164289, 178991

Offset: 1

N. J. A. Sloane

Final digits of a(n), i.e., a(n) mod 10, are repeated periodically with period of length 5 {1,5,5,5,9}. There is a symmetry in this list since the sum of two numbers equally distant from the ends is equal to 10 = 1 + 9 = 5 + 5 = 2*5. Last two digits of a(n), i.e., a(n) mod 100, are repeated periodically with period of length 50. - Alexander Adamchuk, Aug 11 2006
a(n) = VarScheme(n,2) in the scheme displayed in A128195. - Peter Luschny, Feb 26 2007
If Y is a 3-subset of a 2n-set X then, for n >= 2, a(n-2) is the number of 4-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
The numbers are the constant number found in magic squares of order n, where n is an odd number, see the comment in A006003. A Magic Square of side 1 is 1; 3 is 15; 5 is 65 and so on. - David Quentin Dauthier, Nov 07 2008
Two times the area of the triangle with vertices at (0,0), ((n - 1)^2, n^2), and (n^2, (n - 1)^2). - J. M. Bergot, Jun 25 2013
Bisection of A006003. - Omar E. Pol, Sep 01 2018
Construct an array M with M(0,n) = 2*n^2 + 4*n + 1 = A056220(n+1), M(n,0) = 2*n^2 + 1 = A058331(n) and M(n,n) = 2*n*(n+1) + 1 = A001844(n). Row(n) begins with all the increasing odd numbers from A058331(n) to A001844(n) and column(n) begins with all the decreasing odd numbers from A056220(n+1) to A001844(n). The sum of the terms in row(n) plus those in column(n) minus M(n,n) equals a(n+1). The first five rows of array M are [1, 7, 17, 31, 49, ...]; [3, 5, 15, 29, 47, ...]; [9, 11, 13, 27, 45, ...]; [19, 21, 23, 25, 43, ...]; [33, 35, 37, 39, 41, ...]. - J. M. Bergot, Jul 16 2013 [This contribution was moved here from A047926 by Petros Hadjicostas, Mar 08 2021.]
For n>=2, these are the primitive sides s of squares of type 2 described in A344332. - Bernard Schott, Jun 04 2021
(a(n) + 1) / 2 = A212133(n) is the number of cells in the n-th rhombic-dodecahedral polycube. - George Sicherman, Jan 21 2024

J. H. Conway and R. K. Guy, The Book of Numbers, p. 53.
E. Deza and M. M. Deza, Figurate Numbers, World Scientific Publishing, 2012, pp. 123-124.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Mario Defranco and Paul E. Gunnells, Hypergraph matrix models and generating functions, arXiv:2204.11361 [math.CO], 2022.
Milan Janjic, Two Enumerative Functions
T. P. Martin, Shells of atoms, Phys. Rep., 273 (1996), 199-241, eq. (9).
Andy Nicol, Illustration of Rhombic Dodecahedral Numbers
C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Rhombic Dodecahedral Number.
Eric Weisstein's World of Mathematics, Nexus Number.
D. Zeitlin, A family of Galileo sequences, Amer. Math. Monthly 82 (1975), 819-822.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(2*n^3 - 3*n^2 + n) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A063493, A063494, A063495, A063496.
Column k=3 of A047969.
Cf. A128195, A176850, A005408, A176271, A212133.
Cf. A001844, A000583, A000290.
Cf. A007588, A000292, A000332, A002817, A342354.
Cf. A031215, A008292.
Cf. A016754, A344330, A344332.

a005917 n = a005917_list !! (n-1)
a005917_list = map sum $ f 1 [1, 3 ..] where
   f x ws = us : f (x + 2) vs where (us, vs) = splitAt x ws
-- Reinhard Zumkeller, Nov 13 2014

[n^4 - (n-1)^4: n in [1..50]]; // Vincenzo Librandi, Aug 01 2011

Table[n^4-(n-1)^4,{n,40}]  (* Harvey P. Dale, Apr 01 2011 *)
#[[2]]-#[[1]]&/@Partition[Range[0,40]^4,2,1] (* More efficient than the above Mathematica program because it only has to calculate each 4th power once *) (* Harvey P. Dale, Feb 07 2015 *)
Differences[Range[0,40]^4] (* Harvey P. Dale, Aug 11 2023 *)

a(n)=n^4-(n-1)^4 \\ Charles R Greathouse IV, Jul 31 2011

A005917_list, m = [], [24, -12, 2, 1]
for _ in range(10**2):
    A005917_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

a(n) = (2*n - 1)*(2*n^2 - 2*n + 1).
Sum_{i=1..n} a(i) = n^4 = A000583(n). First differences of A000583.
G.f.: x*(1+x)*(1+10*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
More generally, g.f. for n^m - (n - 1)^m is Euler(m, x)/(1 - x)^m, where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292). E.g.f.: x*(exp(y/(1 - x)) - exp(x*y/(1 - x)))/(exp(x*y/(1 - x))-x*exp(y/(1 - x))). - Vladeta Jovovic, May 08 2002
a(n) = sum of the next (2*n - 1) odd numbers; i.e., group the odd numbers so that the n-th group contains (2*n - 1) elements like this: (1), (3, 5, 7), (9, 11, 13, 15, 17), (19, 21, 23, 25, 27, 29, 31), ... E.g., a(3) = 65 because 9 + 11 + 13 + 15 + 17 = 65. - Xavier Acloque, Oct 11 2003
a(n) = 2*n - 1 + 12*Sum_{i = 1..n} (i - 1)^2. - Xavier Acloque, Oct 16 2003
a(n) = (4*binomial(n,2) + 1)*sqrt(8*binomial(n,2) + 1). - Paul Barry, Mar 14 2004
Binomial transform of [1, 14, 36, 24, 0, 0, 0, ...], if the offset is 0. - Gary W. Adamson, Dec 20 2007
Sum_{i=1..n-1}(a(i) + a(i+1)) = 8*Sum_{i=1..n}(i^3 + i) = 16*A002817(n-1) for n > 1. - Bruno Berselli, Mar 04 2011
a(n+1) = a(n) + 2*(6*n^2 + 1) = a(n) + A005914(n). - Vincenzo Librandi, Mar 16 2011
a(n) = -a(-n+1). a(n) = (1/6)*(A181475(n) - A181475(n-2)). - Bruno Berselli, Sep 26 2011
a(n) = A045975(2*n-1,n) = A204558(2*n-1)/(2*n - 1). - Reinhard Zumkeller, Jan 18 2012
a(n+1) = Sum_{k=0..2*n+1} (A176850(n,k) - A176850(n-1,k))*(2*k + 1), n >= 1. - L. Edson Jeffery, Nov 02 2012
a(n) = A005408(n-1) * A001844(n-1) = (2*(n - 1) + 1) * (2*(n - 1)*n + 1) = A000290(n-1)*12 + 2 + a(n-1). - Bruce J. Nicholson, May 17 2017
a(n) = A007588(n) + A007588(n-1) = A000292(2n-1) + A000292(2n-2) + A000292(2n-3) = A002817(2n-1) - A002817(2n-2). - Bruce J. Nicholson, Oct 22 2017
a(n) = A005898(n-1) + 6*A000330(n-1) (cf. Deza, Deza, 2012, p. 123, Section 2.6.2). - Felix Fröhlich, Oct 01 2018
a(n) = A300758(n-1) + A005408(n-1). - Bruce J. Nicholson, Apr 23 2020
G.f.: polylog(-4, x)*(1-x)/x. See the Simon Plouffe formula above (with expanded numerator), and the g.f. of the rows of A008292 by Vladeta Jovovic, Sep 02 2002. - Wolfdieter Lang, May 10 2021

A206399 a(0) = 1; for n > 0, a(n) = 41*n^2 + 2.

Original entry on oeis.org

1, 43, 166, 371, 658, 1027, 1478, 2011, 2626, 3323, 4102, 4963, 5906, 6931, 8038, 9227, 10498, 11851, 13286, 14803, 16402, 18083, 19846, 21691, 23618, 25627, 27718, 29891, 32146, 34483, 36902, 39403, 41986, 44651, 47398, 50227, 53138, 56131, 59206, 62363, 65602

Offset: 0

Bruno Berselli, Feb 07 2012

Apart from the first term, numbers of the form (r^2 + 2*s^2)*n^2 + 2 = (r*n)^2 + (s*n - 1)^2 + (s*n + 1)^2: in this case is r = 3, s = 4. After 1, all terms are in A000408.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Sequences of the same type: A005893, A005897, A005899, A005901, A005903, A005905, A005914, A005918, A005919, A008527, A010000-A010023.

Cf. A000408.

[n eq 0 select 1 else 41*n^2+2: n in [0..39]];

I:=[1,43,166,371]; [n le 4 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..41]]; // Vincenzo Librandi, Aug 18 2013

Join[{1}, 41 Range[39]^2 + 2]
CoefficientList[Series[(1 + x) (1 + 39 x + x^2) / (1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2013 *)

a(n)=if(n,41*n^2+2,1) \\ Charles R Greathouse IV, Sep 24 2015

O.g.f.: (1 + x)*(1 + 39*x + x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 4. - Wesley Ivan Hurt, Dec 18 2020
E.g.f.: exp(x)*(41*x^2 + 41*x + 2) - 1. - Elmo R. Oliveira, Nov 29 2024

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

Luciano Ancora, Feb 12 2015

			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)

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 .
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000538, A000583, A005914, A005917, A101089, A101090, A101091, A101103, A101104, A254682, A254683, A254684.

[Binomial(n+5,6)*n*(n+5)*(2*n+5)/42: n in [1..30]]; // G. C. Greubel, Dec 01 2018

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

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 *)

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

[binomial(n+5,6)*n*(n+5)*(2*n+5)/42 for n in (1..30)] # G. C. Greubel, Dec 01 2018

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

Cecilia Rossiter, Dec 15 2004

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

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 [Dead link]
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
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
Index entries for linear recurrences with constant coefficients, signature (1).

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
z = 6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177
z = 7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523
z = 8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015
z = 9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541
Cf. A101095 for an expanded table and more about MagicNKZ.

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 *)

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

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

Cecilia Rossiter, Dec 15 2004

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).

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
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.
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Archive Machine link]
Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
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
Index entries for linear recurrences with constant coefficients, signature (2,-1).

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 =  6 | A000389 | A005585 | A024166 | A000538 | A000584 | A022522 | A255177 | A255182
z =  7 | A000579 | A040977 | A101094 | A101089 | A000539 | A001014 | A022523 | A255178
z =  8 | A000580 | A050486 | A101097 | A101090 | A101092 | A000540 | A001015 | A022524
z =  9 | A000581 | A053347 | A101102 | A101091 | A101099 | A101093 | A000541 | A001016
z = 10 | A000582 | A054333 | A254469 | A254681 | A254644 | A254640 | A250212 | A000542
z = 11 | A001287 | A054334 | A254869 | A254470 | A254682 | A254645 | A254641 | A253636
z = 12 | A001288 | A057788 | ....... | A254870 | A254471 | A254683 | A254646 | A254642
z = 13 | A010965 | ....... | ....... | ....... | A254871 | A254472 | A254684 | A254647
z = 14 | A010966 | ....... | ....... | ....... | ....... | A254872 | ....... | .......
--------------------------------------------------------------------------------------
Cf. A047969.

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

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 *)

a(n)=if(n>3, 120*n-240, 33*n^2-72*n+40) \\ Charles R Greathouse IV, Oct 11 2015

[1,28,121]+[120*(k-2) for k in range(4,36)] # Danny Rorabaugh, Apr 23 2015

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

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

A101860 a(n) = (3+n)(2 + 33n + 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

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

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.

			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
... ... ... ... ... ... ... ... ... ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

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

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 *)

a(n) = (n+3)*(n^2+33*n+2)/6; \\ Altug Alkan, Sep 23 2018

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

A245019 Number of ordered n-tuples of positive integers, whose minimum is 0 and maximum is 4.

Original entry on oeis.org

0, 2, 24, 194, 1320, 8162, 47544, 266114, 1448520, 7727522, 40616664, 211117634, 1088079720, 5571427682, 28384443384, 144041002754, 728708854920, 3677645732642, 18524892775704, 93171895169474, 468051525534120, 2349032799986402

Offset: 1

Ovidiu Bagdasar, Sep 17 2014

For given k and n positive integers, let T(k,n) represent the number of n-tuples of positive integers, whose minimum is zero and maximum is k. In this notation, the sequence corresponds to a(n) = T(4,n).

			For n=2 the a(2)=2 solutions are (0,4) and (4,0).

O. Bagdasar, On Some Functions Involving the lcm and gcd of Integer Tuples, Scientific publications of the state university of Novi Pazar, Ser. A: Appl. Maths. Inform. and Mech., Vol. 6, 2 (2014), 91-100.
Index entries for linear recurrences with constant coefficients, signature (12,-47,60).

T(1,n) gives A000918, T(2,n-1) gives A028243, T(n,3) gives A008588, T(n,4) gives A005914.

concat(0, Vec(-2*x^2/((3*x-1)*(4*x-1)*(5*x-1)) + O(x^100))) \\ Colin Barker, Sep 17 2014

a(n) = 5^n-2*4^n+3^n.
a(n) = 12*a(n-1)-47*a(n-2)+60*a(n-3). G.f.: -2*x^2 / ((3*x-1)*(4*x-1)*(5*x-1)). - Colin Barker, Sep 17 2014
a(n) = 2*A016753(n-2) for n>1. - Colin Barker, Sep 17 2014

A276189 Triangle read by rows: T(n,k) = 2(6k^2 + 1)*(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.

Original entry on oeis.org

1, 2, 14, 3, 28, 50, 4, 42, 100, 110, 5, 56, 150, 220, 194, 6, 70, 200, 330, 388, 302, 7, 84, 250, 440, 582, 604, 434, 8, 98, 300, 550, 776, 906, 868, 590, 9, 112, 350, 660, 970, 1208, 1302, 1180, 770, 10, 126, 400, 770, 1164, 1510, 1736, 1770, 1540, 974, 11, 140, 450, 880, 1358, 1812, 2170, 2360, 2310, 1948, 1202

Offset: 0

Stefano Maruelli, Aug 24 2016

The row sums of the triangle provide the positive terms of A000583.

Similar triangles can be generated by the formula P(n,k,m) = (Q(k+1,m)-Q(k,m))*(n+1-k), where Q(i,r) = i^r-(i-1)^r, 0 < k <= n, and P(n,0,m) = n+1. T(n,k) is the case m=4, that is T(n,k) = P(n,k,4).

			Triangle starts:
----------------------------------------------
n \ k |  0   1    2    3    4    5    6    7
----------------------------------------------
0     |  1;
1     |  2, 14;
2     |  3, 28,  50;
3     |  4, 42, 100, 110;
4     |  5, 56, 150, 220, 194;
5     |  6, 70, 200, 330, 388, 302;
6     |  7, 84, 250, 440, 582, 604, 434;
7     |  8, 98, 300, 550, 776, 906, 868, 590;
...

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)

Cf. A000583, A005914, A276158.

[IsZero(k) select n+1 else 2*(6*k^2+1)*(n+1-k): k in [0..n], n in [0..10]];

/* As triangle (see the second comment): */ m:=4; Q:=func; P:=func; [[P(n,k,m): k in [0..n]]: n in [0..10]];

Table[If[k == 0, n + 1, 2 (6 k^2 + 1) (n + 1 - k)], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 29 2016 *)

T(n,n-h) = (h+1)*A005914(n-h) for 0 <= h <= n. Therefore, the main diagonal of the triangle is A005914.

Sum_{k=0..n} T(n,k) = T(n,0)^4 = A000583(n+1).

Corrected, rewritten and extended by Bruno Berselli, Aug 31 2016

a(40) ff. corrected by Georg Fischer, Nov 08 2021

A383736 Cluster series for percolation on polyominoid cells.

Original entry on oeis.org

1, 12, 92, 604, 3732, 22766, 136564

Offset: 0

Pontus von Brömssen, May 10 2025

The coordination sequence for polyominoid cells (row 18 of A366768) is A005914, except that the term for distance 1 should be 12.

Index entries for sequences related to cluster series.

Row 18 of A383735.

Cf. A005914, A075678, A075679, A366768, A383737.

cp's OEIS Frontend

A101103 Partial sums of A101104. First differences of A005914.

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A005917 Rhombic dodecahedral numbers: a(n) = n^4 - (n - 1)^4.

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A206399 a(0) = 1; for n > 0, a(n) = 41*n^2 + 2.

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A254681 Fifth partial sums of fourth powers (A000583).

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

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A101095 Fourth difference of fifth powers (A000584).

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

A276189 Triangle read by rows: T(n,k) = 2(6k^2 + 1)*(n + 1 - k) for 0 < k <= n; for k = 0, T(n,0) = n + 1.