A000389 -id:A000389 - cp's OEIS frontend

A001779 Expansion of 1/((1+x)(1-x)^8).

1, 7, 29, 91, 239, 553, 1163, 2269, 4166, 7274, 12174, 19650, 30738, 46782, 69498, 101046, 144111, 201993, 278707, 379093, 508937, 675103, 885677, 1150123, 1479452, 1886404, 2385644, 2993972

Offset: 0

N. J. A. Sloane

a(n) is the number of positive terms in the expansion of (a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 - z)^n. Also the convolution of A001769 and A000012; A001753 and A001477; A001752 and A000217; A002623 and A000292; A002620 and A000332; A004526 and A000389. - Sergio Falcon (sfalcon(AT)dma.ulpgc.es), Feb 13 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (7,-20,28,-14,-14,28,-20,7,-1).

Cf. A001769 (first differences), A169795 (binomial transf.)

[1/80640*(2*n+9) *(4*n^6 +108*n^5 +1138*n^4 +5904*n^3 +15628*n^2 +19638*n +8925)+(-1)^n/256 : n in [0..30]]; // Vincenzo Librandi, Oct 08 2011

A001779 := proc(n) 1/80640*(2*n+9) *(4*n^6 +108*n^5 +1138*n^4 +5904*n^3 +15628*n^2 +19638*n +8925)+(-1)^n/256 ; end proc:
seq(A001779(n),n=0..50) ; # R. J. Mathar, Mar 22 2011

CoefficientList[Series[1/((1 + x) (1 - x)^8), {x, 0, 50}], x] (* G. C. Greubel, Nov 24 2017 *)
LinearRecurrence[{7,-20,28,-14,-14,28,-20,7,-1},{1,7,29,91,239,553,1163,2269,4166},30] (* Harvey P. Dale, Jan 21 2023 *)

a(n)=(2*n+9)*(4*n^6+108*n^5+1138*n^4+5904*n^3+15628*n^2+19638*n + 8925)/80640 +(-1)^n/256 \\ Charles R Greathouse IV, Apr 17 2012

a(n) = (-1)^{7-n}*Sum_{i=0..n} ((-1)^(7-i)*binomial(7+i,i)). - Sergio Falcon, Feb 13 2007

a(n)+a(n+1) = A000580(n+8). - R. J. Mathar, Jan 06 2021

A039948 A triangle related to A000045 (Fibonacci numbers).

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 18, 12, 3, 1, 120, 72, 24, 4, 1, 960, 600, 180, 40, 5, 1, 9360, 5760, 1800, 360, 60, 6, 1, 105840, 65520, 20160, 4200, 630, 84, 7, 1, 1370880, 846720, 262080, 53760, 8400, 1008, 112, 8, 1, 19958400, 12337920, 3810240, 786240, 120960, 15120, 1512, 144, 9, 1

Offset: 0

Wolfdieter Lang

			Triangle begins :
    1;
    1,   1;
    4,   2,   1;
   18,  12,   3,  1;
  120,  72,  24,  4, 1;
  960, 600, 180, 40, 5, 1;
... - _Philippe Deléham_, Nov 08 2011

Seiichi Manyama, Rows n = 0..139, flattened
P. R. J. Asveld, Fibonacci-like differential equations with a polynomial nonhomogeneous term, Fib. Quart. 27 (1989), 303-309.

Cf. A000045, A000142, A005442, A005443, A110313 (row sums).

Diagonals include: A000027, A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582, A001287, A001288, A010965, A010966.

[(Factorial(n)/Factorial(k))*Fibonacci(n-k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 20 2022

T[n_,k_]:= (n!/k!)*Fibonacci[n-k+1];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 20 2022 *)

def A039948(n, k): return factorial(n-k)*binomial(n,k)*fibonacci(n-k+1)
flatten([[A039948(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Nov 20 2022

T(n, m) = n!*Fibonacci(n-m+1)/m!, n >= m >= 0.
T(n, 0) = A005442(n).
T(n, 1) = A005443(n).
E.g.f. for column m: x^m/(m!*(1-x-x^2)), m >= 0.
From G. C. Greubel, Nov 20 2022: (Start)
T(n, n-1) = A000027(n).
T(n, n-2) = 4*A000217(n-1), n >= 2.
T(n, n-3) = 18*A000292(n-2), n >= 3.
T(n, n-4) = 5! * A000332(n), n >= 4.
T(n, n-5) = 8 * 5! * A000389(n), n >= 5.
T(n, n-6) = 13 * 6! * A000579(n), n >= 6.
T(n, n-7) = 21 * 7! * A000580(n), n >= 7.
T(n, n-8) = 34 * 8! * A000581(n), n >= 8.
T(n, n-9) = 55 * 9! * A000582(n), n >= 9.
T(n, n-10) = 89 * 10! * A001287(n), n >= 10.
T(n, n-11) = 12 * 12! * A001288(n), n >= 11.
T(n, n-12) = 233 * 12! * A010965(n), n >= 12.
T(n, n-13) = 89 * 13! * A010966(n), n >= 13.
Sum_{k=0..n} T(n, k) = A110313(n). (End)

A096959 Sixth column (m=5) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 31, 96, 231, 476, 882, 1512, 2442, 3762, 5577, 8008, 11193, 15288, 20468, 26928, 34884, 44574, 56259, 70224, 86779, 106260, 129030, 155480, 186030, 221130, 261261, 306936, 358701, 417136, 482856, 556512, 638792, 730422, 832167, 944832

Offset: 0

Wolfdieter Lang, Aug 13 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A096958 (fifth column), A097297 (seventh column).

[(n+30)*Binomial(n+4, 4)/5: n in [0..30]]; // G. C. Greubel, Nov 24 2017

Table[(n + 30)*Binomial[n + 4, 4]/5, {n, 0, 50}] (* G. C. Greubel, Nov 24 2017 *)

for(n=0,30, print1((n+30)*binomial(n+4, 4)/5, ", ")) \\ G. C. Greubel, Nov 24 2017

a(n) = A096956(n+5, 5).
a(n) = 6*b(n) - 5*b(n-1), with b(n) = A000389(n+5) = binomial(n+5, 5).
a(n) = (n+30)*binomial(n+4, 4)/5.
G.f.: (6-5*x)/(1-x)^6.
E.g.f.: x*(720 + 1140*x + 420*x^2 + 45*x^3 + x^4)*exp(x)/120. - G. C. Greubel, Nov 24 2017

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

A015650 Number of ordered 5-tuples of integers from [ 1..n ] with no global factor.

Original entry on oeis.org

1, 5, 19, 49, 118, 225, 434, 729, 1209, 1850, 2850, 4059, 5878, 8044, 11020, 14566, 19410, 24789, 32103, 40213, 50615, 62260, 77209, 93099, 113504, 135431, 162341, 191396, 227355, 264463, 310838, 359322, 417212, 478408, 551944, 626971, 718360, 812311, 922407, 1036667

Offset: 1

Olivier Gérard

nonn

Chai Wah Wu, Table of n, a(n) for n = 1..10000

Column k=5 of A177976.
Partial sums of A117109.
Cf. A000389, A002088, A013663, A015631, A015634.

a[n_] := Sum[DivisorSum[k, MoebiusMu[k/#]*Binomial[# + 3, 4] &], {k, 1, n}]; Array[a, 40] (* Amiram Eldar, Jun 07 2025 *)

a(n) = sum(k=1, n, sumdiv(k, d, moebius(k/d)*binomial(d+3, 4))); \\ Seiichi Manyama, Jun 12 2021

a(n) = binomial(n+4, 5)-sum(k=2, n, a(n\k)); \\ Seiichi Manyama, Jun 12 2021

my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, moebius(k)*x^k/(1-x^k)^5)/(1-x)) \\ Seiichi Manyama, Jun 12 2021

from functools import lru_cache
@lru_cache(maxsize=None)
def A015650(n):
    if n == 0:
        return 0
    c, j = n+1, 2
    k1 = n//j
    while k1 > 1:
        j2 = n//k1 + 1
        c += (j2-j)*A015650(k1)
        j, k1 = j2, n//j2
    return n*(n+1)*(n+2)*(n+3)*(n+4)//120-c+j # Chai Wah Wu, Apr 18 2021

G.f.: (1/(1 - x)) * Sum_{k>=1} mu(k) * x^k / (1 - x^k)^5. - Ilya Gutkovskiy, Feb 14 2020
a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)/120 - Sum_{j=2..n} a(floor(n/j)) = A000389(n+4) - Sum_{j=2..n} a(floor(n/j)). - Chai Wah Wu, Apr 18 2021
a(n) ~ n^5 / (120*zeta(5)). - Amiram Eldar, Jun 08 2025

A027660 a(n) = C(n+2, 2) + C(n+2, 3) + C(n+2, 4) + C(n+2, 5).

Original entry on oeis.org

1, 4, 11, 26, 56, 112, 210, 372, 627, 1012, 1573, 2366, 3458, 4928, 6868, 9384, 12597, 16644, 21679, 27874, 35420, 44528, 55430, 68380, 83655, 101556, 122409, 146566, 174406, 206336, 242792

Offset: 0

N. J. A. Sloane

Also, number of 135246-avoiding permutations of n+2 with exactly 1 descent. E.g., there are 57 permutations of 6 with exactly 1 descent. Of these, only the permutation 135246 contains the pattern 135246 so a(4)=56. - Mike Zabrocki, Nov 29 2004

If Y is a 2-subset of an n-set X then, for n>=5, a(n-5) is the number of 5-subsets of X which do not have exactly one element in common with Y. - Milan Janjic, Dec 28 2007

G. C. Greubel, Table of n, a(n) for n = 0..1000
Alexsandar Petojevic, The Function vM_m(s; a; z) and Some Well-Known Sequences, Journal of Integer Sequences, Vol. 5 (2002), Article 02.1.7.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000295, A000332, A000389, A051601.

[(n^2-n+20)*Binomial(n+3,3)/20: n in [0..60]]; // G. C. Greubel, Aug 01 2022

a:= n-> (n+3)*(n+2)*(n+1)*(n^2-n+20)/120;
seq(a(n), n = 0..60);

Sum[Binomial[3+Range[0,60], 2*j+1], {j,2}] (* G. C. Greubel, Aug 01 2022 *)

[binomial(n+3,5) +binomial(n+3,3) for n in range(0, 60)] # Zerinvary Lajos, May 17 2009

a(n) = (n+3)*(n+2)*(n+1)*(n^2 - n + 20)/120.
G.f.: (1 - 2*x + 2*x^2)/(1-x)^6. - Mike Zabrocki, Nov 29 2004
a(n) = binomial(n+3,5) + binomial(n+3,3). - Zerinvary Lajos, Jul 24 2006, corrected Oct 01 2021
a(n) = A000389(n+5) - 2*A000332(n+3). - R. J. Mathar, Oct 01 2021
From G. C. Greubel, Aug 01 2022: (Start)
a(n) = Sum_{j=0..3} binomial(n+2, j+2).
E.g.f.: (1/120)*(120 +360*x +240*x^2 +80*x^3 +15*x^4 +x^5)*exp(x). (End)

A062136 Twelfth column of Losanitsch's triangle A034851 (formatted as lower triangular matrix).

Original entry on oeis.org

1, 6, 42, 182, 693, 2184, 6216, 15912, 37854, 83980, 176484, 352716, 676270, 1248072, 2229096, 3863080, 6519591, 10737090, 17299646, 27313650, 42337659, 64512240, 96770544, 143048880, 208616044, 300402648, 427500360, 601661144, 838033836, 1155900720, 1579738736

Offset: 0

Wolfdieter Lang, Jun 19 2001

Also seventh column (m=6) of triangle A062135.

Number of homeomorphically irreducible (or series-reduced) trees (no vertices of degree 2) with n+9 leaves which become tree P(7) (path on 7 nodes (vertices) or 6 edges (links) when all leaves are omitted. A leave is an edge together with a node of degree 1 at one end). Proof by Polya enumeration. See illustration for A034851.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for sequences related to trees

Cf. A018213.

[(1/(2*Factorial(11)))*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)*(n + 7)*(n + 8)*(n + 9)*(n + 10)*(n + 11) + (1/15)*(1/2^9)*(n + 2)*(n + 4)*(n + 6)*(n + 8)*(n + 10)*(1/2)*(1 + (-1)^n): n in [0..30]]; // G. C. Greubel, Nov 24 2017

Table[(1/(2*11!))*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)*(n + 7)*(n + 8)*(n + 9)*(n + 10)*(n + 11) + (1/15)*(1/2^9)*(n + 2)*(n + 4)*(n + 6)*(n + 8)*(n + 10)*(1/2)*(1 + (-1)^n), {n, 0, 50}] (* G. C. Greubel, Nov 24 2017 *)

for(n=0,50, print1((1/(2*11!))*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)*(n + 6)*(n + 7)*(n + 8)*(n + 9)*(n + 10)*(n + 11) + (1/15)*(1/2^9)*(n + 2)*(n + 4)*(n + 6)*(n + 8)*(n + 10)*(1/2)*(1 + (-1)^n), ", ")) \\ G. C. Greubel, Nov 24 2017

G.f.: Pe(6, x^2)/((1-x)^(2*6)*(1+x)^6), with Pe(6, x^2) := Sum_{m=0..3} A034839(6, m)*x^(2*m) = 1+15*x^2+15*x^4+x^6.
a(n) = A034851(n+11,11).
a(2n+1) = A001288(2n+12)/2; a(2n) = (A001288(2n+11)+A000389(n+5))/2. - Gary W. Adamson, Dec 15 2010
a(n) = (1/(2*11!))*(n+1)*(n+2)*(n+3)*(n+4)*(n+5)*(n+6)*(n+7)*(n+8)*(n+9)*(n+10)*(n+11) + (1/15)*(1/2^9)*(n+2)*(n+4)*(n+6)*(n+8)*(n+10)*(1/2)*(1+(-1)^n). - Yosu Yurramendi, Jun 24 2013

A185508 Third accumulation array, T, of the natural number array A000027, by antidiagonals.

Original entry on oeis.org

1, 5, 6, 16, 29, 21, 41, 89, 99, 56, 91, 219, 295, 259, 126, 182, 469, 705, 755, 574, 252, 336, 910, 1470, 1765, 1645, 1134, 462, 582, 1638, 2786, 3605, 3780, 3206, 2058, 792, 957, 2778, 4914, 6706, 7595, 7266, 5754, 3498, 1287, 1507, 4488, 8190, 11634, 13916, 14406, 12894, 9690, 5643, 2002, 2288, 6963, 13035, 19110, 23814, 26068, 25284, 21510, 15510, 8723, 3003, 3367, 10439, 19965, 30030, 38640, 44100

Offset: 1

Clark Kimberling, Jan 29 2011

See A144112 (and A185506) for the definition of accumulation array (aa).

Sequence is aa(aa(aa(A000027))).

			Northwest corner:
   1    5   16   41   91  182
   6   29   89  219  469  910
  21   99  295  705 1470 2786
  56  259  755 1765 3605 6706

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A000027, A185506, A185507, A185509.

Cf. A000389 (column 1), A257199 (row 1).

h[n_,k_]:=k(k+1)(k+2)n(n+1)(n+2)*(4n^2+(5k+23)n+4k^2+3k+41)/2880;
TableForm[Table[h[n,k],{n,1,10},{k,1,15}]]
Table[h[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten

{h(n,k) = k*(k+1)*(k+2)*n*(n+1)*(n+2)*(4*n^2+(5*k+23)*n +4*k^2 +3*k + 41)/2880}; for(n=1,10, for(k=1,n, print1(h(k, n-k+1), ", "))) \\ G. C. Greubel, Nov 23 2017

T(n,k) = F*(4n^2 + (5k+23)n + 4k^2 + 3k+41), where F = k(k+1)(k+2)n(n+1)(n+2)/2880.

A210569 a(n) = (n-3)(n-2)(n-1)n(n+1)/30.

Original entry on oeis.org

0, 0, 0, 0, 4, 24, 84, 224, 504, 1008, 1848, 3168, 5148, 8008, 12012, 17472, 24752, 34272, 46512, 62016, 81396, 105336, 134596, 170016, 212520, 263120, 322920, 393120, 475020, 570024, 679644, 805504, 949344, 1113024, 1298528, 1507968, 1743588, 2007768, 2303028

Offset: 0

Bruno Berselli, Mar 23 2012

The following sequences are provided by the formula n*binomial(n,k) - binomial(n,k+1) = k*binomial(n+1,k+1):
. A000217(n)   for k=1,
. A007290(n+1) for k=2,
. A050534(n)   for k=3,
. a(n)         for k=4,
. A000910(n+1) for k=5.
Sum of reciprocals of a(n), for n>3: 5/16.
From a(2), convolution of oblong numbers (A002378) with themselves. - Bruno Berselli, Oct 24 2016

Bruno Berselli, Table of n, a(n) for n = 0..1000
C. P. Neuman and D. I. Schonbach, Evaluation of sums of convolved powers using Bernoulli numbers, SIAM Rev. 19 (1977), no. 1, 90--99. MR0428678 (55 #1698). See Table 3. - _N. J. A. Sloane_, Mar 23 2014
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000217, A000389, A000910, A002378, A007290, A050534, A177747.

First differences are in A033488.

Magma
```
[4*Binomial(n+1,5): n in [0..38]];
```

f:=n->(n^5-5*n^3+4*n)/30;
[seq(f(n),n=0..50)]; # N. J. A. Sloane, Mar 23 2014

LinearRecurrence[{6,-15,20,-15,6,-1}, {0,0,0,0,4,24}, 39]
CoefficientList[Series[4x^4/(1-x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 24 2014 *)
Times@@@Partition[Range[-3,40],5,1]/30 (* Harvey P. Dale, Sep 19 2020 *)

makelist(coeff(taylor(4*x^4/(1-x)^6, x, 0, n), x, n), n, 0, 38);

a(n)=(n-3)*(n-2)*(n-1)*n*(n+1)/30 \\ Charles R Greathouse IV, Oct 07 2015

[4*binomial(n+1,5) for n in (0..40)] # G. C. Greubel, May 23 2022

G.f.: 4*x^4/(1-x)^6.
a(n) = n*binomial(n,4)-binomial(n,5) = 4*binomial(n+1,5) = 4*A000389(n+1).
a(n) = 2*A177747(2*n-7), with A177747(-7) = A177747(-5) = A177747(-3) = A177747(-1) = 0.
(n-4)*a(n) = (n+1)*a(n-1).
E.g.f.: (1/30)*x^4*(5+x)*exp(x). - G. C. Greubel, May 23 2022
Sum_{n>=4} (-1)^n/a(n) = 20*log(2) - 655/48. - Amiram Eldar, Jun 02 2022

A256859 a(n) = n(n + 1)(n + 2)*(n^2 - n + 4)/24.

Original entry on oeis.org

1, 6, 25, 80, 210, 476, 966, 1800, 3135, 5170, 8151, 12376, 18200, 26040, 36380, 49776, 66861, 88350, 115045, 147840, 187726, 235796, 293250, 361400, 441675, 535626, 644931, 771400, 916980, 1083760, 1273976, 1490016, 1734425, 2009910, 2319345, 2665776, 3052426

Offset: 1

Luciano Ancora, Apr 14 2015

This is the case k = n of b(n,k) = n*(n+1)*(n+2)*(k*(n-1)+4)/24, where b(n,k) is the n-th hypersolid number in 4 dimensions generated from an arithmetical progression with the first term 1 and common difference k. Therefore, the sequence is the main diagonal of the Table 3 in Sardelis et al. paper (see Links field).

G. C. Greubel, Table of n, a(n) for n = 1..1000
D. A. Sardelis and T. M. Valahas, On Multidimensional Pythagorean Numbers, arXiv:0805.4070v1 [math.GM], 2008.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000389, A261721.

Cf. similar sequences of the form binomial(n+k-2,k-1)+n*binomial(n+k-2,k): A006000 (k=2); A257055 (k=3); this sequence (k=4); A256860 (k=5); A256861 (k=6).

[n*(n + 1)*(n + 2)*(n^2 - n + 4)/24: n in [1..30]]; // G. C. Greubel, Nov 23 2017

Table[n (n + 1) (n + 2) (n^2 - n + 4)/24, {n, 40}]
LinearRecurrence[{6,-15,20,-15,6,-1},{1,6,25,80,210,476},40] (* Harvey P. Dale, Mar 19 2022 *)

vector(40, n, n*(n+1)*(n+2)*(n^2-n+4)/24) \\ Bruno Berselli, Apr 15 2015

G.f.: x*(1 + 4*x^2)/(1 - x)^6.
a(n) = 4*A000389(n+2) + A000389(n+4). - Bruno Berselli, Apr 15 2015
E.g.f.: (24*x + 48*x^2 + 40*x^3 + 12*x^4 + x^5)*exp(x)/24. - G. C. Greubel, Nov 23 2017
a(n) = A261721(n,n-1). - Alois P. Heinz, Apr 15 2020

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A210569 a(n) = (n-3)(n-2)(n-1)n(n+1)/30.

A256859 a(n) = n(n + 1)(n + 2)*(n^2 - n + 4)/24.