A000579 -id:A000579 - cp's OEIS frontend

A242023 Decimal expansion of Sum_{n >= 1} (-1)^(n + 1)24/(n(n + 1)(n + 2)(n + 3)).

8, 4, 7, 3, 7, 6, 4, 4, 4, 5, 8, 4, 9, 1, 6, 5, 6, 8, 0, 1, 8, 0, 9, 4, 5, 5, 3, 3, 2, 8, 3, 1, 6, 8, 4, 5, 0, 8, 2, 6, 7, 0, 9, 6, 6, 1, 9, 4, 8, 3, 4, 7, 9, 8, 5, 2, 8, 4, 2, 6, 9, 7, 0, 4, 5, 5, 2, 6, 2, 5, 6, 9, 6, 9

Offset: 0

Richard R. Forberg, Aug 11 2014

Sum of terms of the inverse of Binomial(n,4) or A000332, for n>=4, with alternating signs.
In general the sums of Binomial coefficients of this form appear to have the form  m*log(2) - r, where m is an integer and r is rational as below:
For Binomial(n,1):  m = 1, r = 0. See A002162.
For Binomial(n,2):  m = 4, r = 2. See A000217.
For Binomial(n,3):  m = 12 r = 15/2. See A000292.
For Binomial(n,4):  m = 32, r = 64/3. See A000332.
For Binomial(n,5):  m = 80, r = 655/12. See A000389.
For Binomial(n,6):  m = 192, r = 661/5. See A000579.
For Binomial(n,7):  m = 448, r = 9289/30. See A000580.
For Binomial(n,8):  m = 1024, r = 74432/105. See A000581.
This is generalized as follows:
m grows as A001787(k) = k*2^(k-1) for Binomial(n,k).
r * (k-1)! produces the integer sequence: a(k) = 0, 2, 15, 128, 1310, 15864, 222936, 3572736,  where a(k+1)/a(k) approaches 2*k for large k.
Results are precise to 100 digits or more using Mathematica.

			0.8473764445849165680180945...

G. C. Greubel, Table of n, a(n) for n = 0..5000

Cf. A000332, A000217, A000292, A242024, A002162, A001787.

[32*Log(2) - 64/3]; // G. C. Greubel, Nov 23 2017

Sum[N[(-1)^(n + 1)*24/(n*(n + 1)*(n + 2)*(n + 3)), 150], {n, 1, Infinity}]
RealDigits[32*Log[2] - 64/3, 10, 50][[1]] (* G. C. Greubel, Nov 23 2017 *)

32*log(2) - 64/3 \\ Michel Marcus, Aug 13 2014

sumalt(n=1, (-1)^(n + 1)*24/(n*(n + 1)*(n + 2)*(n + 3))) \\ Michel Marcus, Aug 14 2014

Equals 32*log(2) - 64/3.

Equals 32*(A259284-1). - R. J. Mathar, Jun 30 2021

A049017 Expansion of 1/((1-x)^7 - x^7).

Original entry on oeis.org

1, 7, 28, 84, 210, 462, 924, 1717, 3017, 5110, 8568, 14756, 27132, 54264, 116281, 257775, 572264, 1246784, 2641366, 5430530, 10861060, 21242341, 40927033, 78354346, 150402700, 291693136, 574274008, 1148548016, 2326683921, 4749439975, 9714753412, 19818498700, 40199107690

Offset: 0

N. J. A. Sloane

Differs for n >= 7 (1717 vs. 1716) from A000579(n+6) = binomial(n+6,6); see also row 6 of A027555, A059481 and A213808. - M. F. Hasler, Mar 05 2017

Seiichi Manyama, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,2).

Cf. A000579, A000750, A027555, A049018, A059481, A213808.

Sequences of the form 1/((1-x)^m - x^m): A000079 (m=1,2), A024495 (m=3), A000749 (m=4), A049016 (m=5), A192080 (m=6), this sequence (m=7), A290995 (m=8), A306939 (m=9).

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1-x)^7 - x^7) )); // G. C. Greubel, Apr 11 2023

CoefficientList[Series[1/((1-x)^7-x^7),{x,0,30}],x]  (* Harvey P. Dale, Feb 18 2011 *)

Vec(1/((1-x)^7-x^7)+O(x^99)) \\ M. F. Hasler, Mar 05 2017

def A049017_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 1/((1-x)^7 - x^7) ).list()
A049017_list(40) # G. C. Greubel, Apr 11 2023

G.f.: 1/((1-x)^7 - x^7) = 1/((1-2*x)*(1-5*x+11*x^2-13*x^3+9*x^4-3*x^5+x^6)).

A087108 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,4). The p-th row (p>=1) contains a(i,p) for i=1 to 4p-3, where a(i,p) satisfies Sum_{i=1..n} C(i+3,4)^p = 5 C(n+4,5) * Sum_{i=1..4p-3} a(i,p) C(n-1,i-1)/(i+4).

Original entry on oeis.org

1, 1, 4, 6, 4, 1, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 124, 3126, 33124, 191251, 681000, 1596120, 2543520, 2780820, 2058000, 987000, 277200, 34650, 1, 624, 49376, 1350624, 18308751, 146500500, 763418870, 2749648020, 7101675070, 13440210000

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+4,4)^p of degree 4*p in terms of falling factorials: C(x+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+4,4)^p = Sum_{k = 0..4*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,24,176,...,70, so Sum_{i=1..n} C(i+3,4)^3 = 5 * C(n+4,5) * [ a(1,3)/5 + a(2,3)*C(n-1,1)/6 + a(3,3)*C(n-1,2)/7 + ... + a(9,3)*C(n-1,8)/13 ] = 5 * C(n+4,5) * [ 1/5 + 24*C(n-1,1)/6 + 176*C(n-1,2)/7 + ... + 70*C(n-1,8)/13 ]. Cf. A086024 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
  n = 0 | 1
  n = 1 | 1   4    6     4      1
  n = 2 | 1  24  176   624   1251   1500    1070  420  70
  n = 3 | 1 124 3126 33124 191251 681000 1596120 ...
  ...
Row 2: C(i+4,4)^2 = C(i,0) + 24*C(i,1) + 176*C(i,2) + 624*C(i,3) + 1251*C(i,4) + 1500*C(i,5) + 1070*C(i,6) + 420*C(i,7) + 70*C(i,8). Hence, Sum_{i = 0..n-1} C(i+4,4)^2 =  C(n,1) + 24*C(n,2) + 176*C(n,3) + 624*C(n,4) + 1251*C(n,5) + 1500*C(n,6) + 1070*C(n,7) + 420*C(n,8) + 70*C(n,9) .(End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A236463.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+4, 4)^n, i = 0..k), k = 0..4*n), n = 0..6); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 5, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 4, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 4*p - 3}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 5, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 4, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 4*p-3, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+5, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+4, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+4,4)^n. Equivalently, let v_n denote the sequence (1, 5^n, 15^n, 35^n, ...) regarded as an infinite column vector, where 1, 5, 15, 35, ... is the sequence binomial(n+4,4) - see A000332. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = C(k+4,4)*T(n,k) + 4*C(k+3,4)*T(n,k-1) + 6*C(k+2,4)*T(n,k-2) + 4*C(k+1,4)*T(n,k-3) + C(k,4)*T(n,k-4) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 4*n.
n-th row polynomial R(n,x) = (1 + x)^4 o (1 + x)^4 o ... o (1 + x)^4 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n,x) = Sum_{i >= 0} binomial(i+4,4)^n*x^i/(1 + x)^(i+1).
R(n+1,x) = 1/4! * (1 + x)^4 * (d/dx)^4(x^4*R(n,x)).
(1 - x)^(4*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A236463. (End)

Edited by Dean Hickerson, Aug 16 2003

A087109 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,5). The p-th row (p>=1) contains a(i,p) for i=1 to 5p-4, where a(i,p) satisfies Sum_{i=1..n} C(i+4,5)^p = 6 C(n+5,6) * Sum_{i=1..5p-4} a(i,p) C(n-1,i-1)/(i+5).

Original entry on oeis.org

1, 1, 5, 10, 10, 5, 1, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 215, 8830, 148480, 1352615, 7665757, 29224020, 78518790, 152794740, 218270220, 229279512, 175227360, 94864770, 34504470, 7567560, 756756, 1, 1295, 191890

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+5,5)^p of degree 5*p in terms of falling factorials: C(x+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+5,5)^p = Sum_{k = 0..5*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,35,370,...,252, so Sum_{i=1..n} C(i+4,5)^3 = 6 * C(n+5,6) * [ a(1,3)/6 + a(2,3)*C(n-1,1)/7 + a(3,3)*C(n-1,2)/8 + ... + a(11,3)*C(n-1,10)/16 ] = 6 * C(n+5,6) * [ 1/6 + 35*C(n-1,1)/7 + 370*C(n-1,2)/8 + ... + 252*C(n-1,10)/16 ]. Cf. A086026 for more details.
From _Peter Bala_, Mar 11 2018: (Start)
Table begins
1
1  5  10   10    5     1
1 35 370 1920 5835 11253 14240 11830 6230 1890 252
...
Row 2: C(i+5,5)^2 = C(i,0) + 35*C(i,1) + 370*C(i,2) + 1920*C(i,3) + 5835*C(i,4) + 11253*C(i,5) + 14240*C(i,6) + 11830*C(i,7) + 6230*C(i,8) + 1890*C(i,9) + 252*C(i,10). Hence, Sum_{i = 0..n-1} C(i+5,5)^2 = C(n,1) + 35*C(n,2) + 370*C(n,3) + 1920*C(n,4) + 5835*C(n,5) + 11253*C(n,6) + 14240*C(n,7) + 11830*C(n,8) + 6230*C(n,9) + 1890*C(n,10) + 252*C(n,11). (End)

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A000579, A086025, A086026, A087110, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237202.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+5, 5)^n, i = 0..k), k = 0..5*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 6, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 5, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 5*p - 4}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 6, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 5, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 5*p-4, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+6, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+5, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+5,5)^n. Equivalently, let v_n denote the sequence (1, 6^n, 21^n, 56^n, ...) regarded as an infinite column vector, where 1, 6, 21, 56, ... is the sequence binomial(n+5,5) - see A000389. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..5} C(5,i)*C(k+5-i,5)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 5*n.
n-th row polynomial R(n,x) = (1 + x)^5 o (1 + x)^5 o ... o (1 + x)^5 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/5!*(1 + x)^5 * (d/dx)^5(x^5*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+5,5)^n*x^i/(1 + x)^(i+1).
(1 - x)^(5*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237202. (End)

Edited by Dean Hickerson, Aug 16 2003

A087110 This table shows the coefficients of combinatorial formulas needed for generating the sequential sums of p-th powers of binomial coefficients C(n,6). The p-th row (p>=1) contains a(i,p) for i=1 to 6p-5, where a(i,p) satisfies Sum_{i=1..n} C(i+5,6)^p = 7 C(n+6,7) * Sum_{i=1..6p-5} a(i,p) C(n-1,i-1)/(i+6).

Original entry on oeis.org

1, 1, 6, 15, 20, 15, 6, 1, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 342, 21267, 527876, 7020525, 58015362, 324610399, 1297791264, 3839203452, 8595153000, 14760228672, 19560928464, 19987430694

Offset: 1

André F. Labossière, Aug 11 2003

From Peter Bala, Mar 11 2018: (Start)

The table entries T(n,k) are the coefficients when expressing the polynomial C(x+6,6)^p of degree 6*p in terms of falling factorials: C(x+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(x,k). It follows that Sum_{i = 0..n-1} C(i+6,6)^p = Sum_{k = 0..6*p} T(p,k)*C(n,k+1). (End)

			Row 3 contains 1,48,687,...,924, so Sum_{i=1..n} C(i+5,6)^3 = 7 * C(n+6,7) * [ a(1,3)/7 + a(2,3)*C(n-1,1)/8 + a(3,3)*C(n-1,2)/9 + ... + a(13,3)*C(n-1,12)/19 ] = 7 * C(n+6,7) * [ 1/7 + 48*C(n-1,1)/8 + 687*C(n-1,2)/9 + ... + 924*C(n-1,12)/19 ]. Cf. A086028 for more details.

G. C. Greubel, Table of n, a(n) for the first 40 rows, flattened
Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Cf. A000292, A024166, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A087107, A000332, A086020, A086021, A086022, A087108, A000389, A086023, A086024, A087109, A000579, A086025, A086026, A000580, A086027, A086028, A087111, A027555, A086029, A086030.

Cf. A087127, A237252.

seq(seq(add( (-1)^(k-i)*binomial(k, i)*binomial(i+6, 6)^n, i = 0..k), k = 0..6*n), n = 0..5); # Peter Bala, Mar 11 2018

a[i_, p_] := Sum[Binomial[i - 1, 2*k - 2]*Binomial[i - 2*k + 7, i - 2*k + 1]^(p - 1) - Binomial[i - 1, 2*k - 1]*Binomial[i - 2*k + 6, i - 2*k]^(p - 1), {k, 1, (2*i + 1 + (-1)^(i - 1))/4}]; Table[If[p == 1, 1, a[i, p]], {p, 1, 10}, {i, 1, 6*p - 5}]//Flatten (* G. C. Greubel, Nov 23 2017 *)

{a(i, p) = sum(k=1, (2*i + 1 + (-1)^(i - 1))/4, binomial(i - 1, 2*k - 2)*binomial(i - 2*k + 7, i - 2*k + 1)^(p - 1) - binomial(i - 1, 2*k - 1)*binomial(i - 2*k + 6, i - 2*k)^(p - 1))}; for(p=1,8, for(i=1, 6*p-5, print1(if(p==1,1,a(i,p)), ", "))) \\ G. C. Greubel, Nov 23 2017

a(i, p) = Sum_{k=1..[2*i+1+(-1)^(i-1)]/4} [ C(i-1, 2*k-2)*C(i-2*k+7, i-2*k+1)^(p-1) -C(i-1, 2*k-1)*C(i-2*k+6, i-2*k)^(p-1) ]
From Peter Bala, Mar 11 2018: (Start)
The following remarks assume the row and column indices start at 0.
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i) * binomial(i+6,6)^n. Equivalently, let v_n denote the sequence (1, 7^n, 28^n, 84^n, ...) regarded as an infinite column vector, where 1, 7, 28, 84, ... is the sequence binomial(n+6,6) - see A000579. Then the n-th row of this table is determined by the matrix product P^(-1)*v_n, where P denotes Pascal's triangle A007318.
Recurrence: T(n+1,k) = Sum_{i = 0..6} C(6,i)*C(k+6-i,6)*T(n,k-i) with boundary conditions T(n,0) = 1 for all n and T(n,k) = 0 for k > 6*n.
n-th row polynomial R(n,x) = (1 + x)^6 o (1 + x)^6 o ... o (1 + x)^6 (n factors), where o denotes the black diamond product of power series defined in Dukes and White.
R(n+1,x) = 1/6!*(1 + x)^6 * (d/dx)^6(x^6*R(n,x)).
R(n,x) = Sum_{i >= 0} binomial(i+6,6)^n*x^i/(1 + x)^(i+1).
(1 - x)^(6*n)*R(n,x/(1 - x)) appears to equal the n-th row polynomial of A237252. (End)

Edited by Dean Hickerson, Aug 16 2003

A221857 Number A(n,k) of shapes of balanced k-ary trees with n nodes, where a tree is balanced if the total number of nodes in subtrees corresponding to the branches of any node differ by at most one; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 1, 1, 0, 1, 1, 4, 3, 4, 1, 0, 1, 1, 5, 6, 1, 4, 1, 0, 1, 1, 6, 10, 4, 9, 4, 1, 0, 1, 1, 7, 15, 10, 1, 27, 1, 1, 0, 1, 1, 8, 21, 20, 5, 16, 27, 8, 1, 0, 1, 1, 9, 28, 35, 15, 1, 96, 81, 16, 1, 0, 1, 1, 10, 36, 56, 35, 6, 25, 256, 81, 32, 1, 0

Offset: 0

Alois P. Heinz, Apr 10 2013

			: A(2,2) = 2  : A(2,3) = 3      : A(3,3) = 3          :
:   o     o   :   o    o    o   :   o      o      o   :
:  / \   / \  :  /|\  /|\  /|\  :  /|\    /|\    /|\  :
: o         o : o      o      o : o o    o   o    o o :
:.............:.................:.....................:
: A(3,4) = 6                                          :
:    o        o        o        o       o        o    :
:  /( )\    /( )\    /( )\    /( )\   /( )\    /( )\  :
: o o      o   o    o     o    o o     o   o      o o :
Square array A(n,k) begins:
  1, 1, 1,  1,   1,   1,  1,  1,  1,   1,   1, ...
  1, 1, 1,  1,   1,   1,  1,  1,  1,   1,   1, ...
  0, 1, 2,  3,   4,   5,  6,  7,  8,   9,  10, ...
  0, 1, 1,  3,   6,  10, 15, 21, 28,  36,  45, ...
  0, 1, 4,  1,   4,  10, 20, 35, 56,  84, 120, ...
  0, 1, 4,  9,   1,   5, 15, 35, 70, 126, 210, ...
  0, 1, 4, 27,  16,   1,  6, 21, 56, 126, 252, ...
  0, 1, 1, 27,  96,  25,  1,  7, 28,  84, 210, ...
  0, 1, 8, 81, 256, 250, 36,  1,  8,  36, 120, ...

Alois P. Heinz, Antidiagonals n = 0..140, flattened
Jeffrey Barnett, Counting Balanced Tree Shapes, 2007
Samuele Giraudo, Intervals of balanced binary trees in the Tamari lattice, arXiv:1107.3472 [math.CO], Apr 2012

Columns k=1-10 give: A000012, A110316, A131889, A131890, A131891, A131892, A131893, A229393, A229394, A229395.
Rows n=0+1, 2-3, give: A000012, A001477, A179865.
Diagonal and upper diagonals give: A028310, A000217, A000292, A000332, A000389, A000579, A000580, A000581, A000582, A001287, A001288.
Lower diagonals give: A000012, A000290, A092364(n) for n>1.

A:= proc(n, k) option remember; local m, r; if n<2 or k=1 then 1
      elif k=0 then 0 else r:= iquo(n-1, k, 'm');
      binomial(k, m)*A(r+1, k)^m*A(r, k)^(k-m) fi
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

a[n_, k_] := a[n, k] = Module[{m, r}, If[n < 2 || k == 1, 1, If[k == 0, 0, {r, m} = QuotientRemainder[n-1, k]; Binomial[k, m]*a[r+1, k]^m*a[r, k]^(k-m)]]]; Table[a[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Apr 17 2013, translated from Maple *)

A231303 Recurrence a(n) = a(n-2) + n^M for M=4, starting with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 16, 82, 272, 707, 1568, 3108, 5664, 9669, 15664, 24310, 36400, 52871, 74816, 103496, 140352, 187017, 245328, 317338, 405328, 511819, 639584, 791660, 971360, 1182285, 1428336, 1713726, 2042992, 2421007, 2852992, 3344528, 3901568, 4530449, 5237904

Offset: 0

Stanislav Sykora, Nov 07 2013

In physics, a(n)/2^(M-1) is the trace of the spin operator |S_z|^M for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of their 4th powers is 2*82/16 = a(3)/8 (analogously for other values of M).

Partial sums of A062392. - Bruce J. Nicholson, Jun 29 2019

			a(4) = 4^4 + 2^4 = 272; a(5) = 5^4 + 3^4 + 1^4 = 707.

Stanislav Sykora, Table of n, a(n) for n = 0..9999
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A001477 (M=1), A000292 (M=2), A105636 (M=3), A231304 (M=5), A231305 (M=6), A231306 (M=7), A231307 (M=8), A231308 (M=9), A231309 (M=10).

Cf. A000389, A000579, A000332, A000583.

List([0..40], n-> n*(3*n^4+15*n^3+20*n^2-8)/30); # G. C. Greubel, Jul 01 2019

[1/30*n*(3*n^4+15*n^3+20*n^2-8): n in [0..40]]; // Vincenzo Librandi, Dec 23 2015

Table[SeriesCoefficient[x*(1+10*x+x^2)/(1-x)^6, {x, 0, n}], {n, 0, 40}] (* Michael De Vlieger, Dec 22 2015 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 16, 82, 272, 707}, 40] (* Vincenzo Librandi, Dec 23 2015 *)

nmax=40;a = vector(nmax);a[2]=1;for(i=3,#a,a[i]=a[i-2]+(i-1)^4); print(a);

concat(0, Vec(x*(1+10*x+x^2)/(1-x)^6 + O(x^40))) \\ Colin Barker, Dec 22 2015

[n*(3*n^4+15*n^3+20*n^2-8)/30 for n in (0..40)] # G. C. Greubel, Jul 01 2019

a(n) = Sum_{k=0..floor(n/2)} (n - 2*k)^4.
From Colin Barker, Dec 22 2015: (Start)
a(n) = (1/30)*n*(3*n^4 + 15*n^3 + 20*n^2 - 8).
G.f.: x*(1 + 10*x + x^2) / (1-x)^6.
(End)
E.g.f.: x*(30 + 210*x + 185*x^2 + 45*x^3 + 3*x^4)*exp(x)/30. - G. C. Greubel, Apr 24 2016
From Bruce J. Nicholson, Jun 29 2019: (Start)
a(n) = 12*A000389(n+3) + A000292(n);
a(n) = (12*A000579(n+4)+A000332(n+3)) - (12*A000579(n+3)+A000332(n+2));
a(n) - a(n-2) = A000583(n). (End)

A258993 Triangle read by rows: T(n,k) = binomial(n+k,n-k), k = 0..n-1.

Original entry on oeis.org

1, 1, 3, 1, 6, 5, 1, 10, 15, 7, 1, 15, 35, 28, 9, 1, 21, 70, 84, 45, 11, 1, 28, 126, 210, 165, 66, 13, 1, 36, 210, 462, 495, 286, 91, 15, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1, 66, 715, 3003, 6435, 8008, 6188, 3060, 969, 190, 21

Offset: 1

Reinhard Zumkeller, Jun 22 2015

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

			.  n\k |  0  1    2    3     4     5     6     7    8    9  10 11
. -----+-----------------------------------------------------------
.   1  |  1
.   2  |  1  3
.   3  |  1  6    5
.   4  |  1 10   15    7
.   5  |  1 15   35   28     9
.   6  |  1 21   70   84    45    11
.   7  |  1 28  126  210   165    66    13
.   8  |  1 36  210  462   495   286    91    15
.   9  |  1 45  330  924  1287  1001   455   120   17
.  10  |  1 55  495 1716  3003  3003  1820   680  153   19
.  11  |  1 66  715 3003  6435  8008  6188  3060  969  190  21
.  12  |  1 78 1001 5005 12870 19448 18564 11628 4845 1330 231 23  .

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

If a diagonal of 1's is added on the right, this becomes A085478.
Essentially the same as A143858.
Cf. A007318, A004736, A094727.
Cf. A027941 (row sums), A117671 (central terms), A143858, A000967, A258708.
T(n,k): A000217 (k=1), A000332 (k=2), A000579 (k=3), A000581 (k=4), A001287 (k=5), A010965 (k=6), A010967 (k=7), A010969 (k=8), A010971 (k=9), A010973 (k=10), A010975 (k=11), A010977 (k=12), A010979 (k=13), A010981 (k=14), A010983 (k=15), A010985 (k=16), A010987 (k=17), A010989 (k=18), A010991 (k=19), A010993 (k=20), A010995 (k=21), A010997 (k=22), A010999 (k=23), A011001 (k=24), A017714 (k=25), A017716 (k=26), A017718 (k=27), A017720 (k=28), A017722 (k=29), A017724 (k=30), A017726 (k=31), A017728 (k=32), A017730 (k=33), A017732 (k=34), A017734 (k=35), A017736 (k=36), A017738 (k=37), A017740 (k=38), A017742 (k=39), A017744 (k=40), A017746 (k=41), A017748 (k=42), A017750 (k=43), A017752 (k=44), A017754 (k=45), A017756 (k=46), A017758 (k=47), A017760 (k=48), A017762 (k=49), A017764 (k=50).
T(n+k,n): A005408 (k=1), A000384 (k=2), A000447 (k=3), A053134 (k=4), A002299 (k=5), A053135 (k=6), A053136 (k=7), A053137 (k=8), A053138 (k=9), A196789 (k=10).
Cf. A165253.

Flat(List([1..12], n-> List([0..n-1], k-> Binomial(n+k,n-k) ))); # G. C. Greubel, Aug 01 2019

a258993 n k = a258993_tabl !! (n-1) !! k
a258993_row n = a258993_tabl !! (n-1)
a258993_tabl = zipWith (zipWith a007318) a094727_tabl a004736_tabl

[Binomial(n+k,n-k): k in [0..n-1], n in [1..12]]; // G. C. Greubel, Aug 01 2019

Table[Binomial[n+k,n-k], {n,1,12}, {k,0,n-1}]//Flatten (* G. C. Greubel, Aug 01 2019 *)

T(n,k) = binomial(n+k,n-k);
for(n=1, 12, for(k=0,n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019

[[binomial(n+k,n-k) for k in (0..n-1)] for n in (1..12)] # G. C. Greubel, Aug 01 2019

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

A369809 Expansion of 1/(1 - x^6/(1-x)^7).

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 1, 7, 28, 84, 210, 462, 925, 1730, 3108, 5565, 10388, 20944, 45697, 104673, 242481, 553455, 1229305, 2650221, 5565127, 11465758, 23397041, 47757235, 98317135, 205108561, 433747259, 926655972, 1989584722, 4271185538, 9133958765, 19421679515

Offset: 0

Seiichi Manyama, Feb 01 2024

Number of compositions of 7*n-6 into parts 6 and 7.

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-6,1).

Cf. A099253, A369805, A369806, A369807, A369808.
Cf. A088305, A095263, A290998, A368475, A369794.
Cf. A000579, A017847.

my(N=40, x='x+O('x^N)); Vec(1/(1-x^6/(1-x)^7))

a(n) = sum(k=0, n\6, binomial(n-1+k, n-6*k));

G.f. (1-x)^7/((1-x)^7-x^6).
a(n) = A017847(7*n-6) = Sum_{k=0..floor((7*n-6)/6)} binomial(k,7*n-6-6*k) for n > 0.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 6*a(n-6) + a(n-7) for n > 7.
a(n) = Sum_{k=0..floor(n/6)} binomial(n-1+k,n-6*k).
a(n) = A373912(n)-A373912(n-1). - R. J. Mathar, Jun 24 2024

A053625 Product of 6 consecutive integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 720, 5040, 20160, 60480, 151200, 332640, 665280, 1235520, 2162160, 3603600, 5765760, 8910720, 13366080, 19535040, 27907200, 39070080, 53721360, 72681840, 96909120, 127512000, 165765600, 213127200, 271252800, 342014400, 427518000, 530122320

Offset: 0

Henry Bottomley, Mar 20 2000

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A002378, A007531, A045619, A052762, A052787.

Cf. A000579, A173333.

F:=Factorial;; Concatenation([0,0,0,0,0,0], List([6..30], n-> F(n)/F(n-5) )); # G. C. Greubel, Aug 27 2019

I:=[0,0,0,0,0,0,720]; [n le 7 select I[n] else 7*Self(n-1) -21*Self(n-2)+35*Self(n-3)-35*Self(n-4)+21*Self(n-5)-7*Self(n-6) +Self(n-7): n in [1..30]]; // Vincenzo Librandi, Apr 28 2012

seq(combinat[numbperm](n, 6), n=0..31); # Zerinvary Lajos, Apr 26 2007

CoefficientList[Series[720*x^6/(1-x)^7,{x,0,30}],x] (* Vincenzo Librandi, Apr 28 2012 *)
Times@@@Partition[Range[-5,30],6,1] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,0,0,0,0,0,720},30] (* Harvey P. Dale, Nov 13 2015 *)
Pochhammer[Range[30]-6, 6] (* G. C. Greubel, Aug 27 2019 *)

a(n)=factorback([n-5..n]) \\ Charles R Greathouse IV, Oct 07 2015

[rising_factorial(n-5,6) for n in (0..30)] # G. C. Greubel, Aug 27 2019

a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5) = n!/(n-6)! = A052787(n)*(n-6) = a(n-1)*n/(n-6).
E.g.f.: x^6*exp(x).
a(n) = 720 * A000579(n). - Zerinvary Lajos, Apr 26 2007
For n > 5: a(n) = A173333(n, n-6). - Reinhard Zumkeller, Feb 19 2010
G.f.: 720*x^6/(1-x)^7. - Colin Barker, Mar 27 2012
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). - Vincenzo Librandi, Apr 28 2012
From Amiram Eldar, Mar 08 2022: (Start)
Sum_{n>=6} 1/a(n) = 1/600.
Sum_{n>=6} (-1)^n/a(n) = 4*log(2)/15 - 661/3600. (End)

cp's OEIS Frontend

A242023 Decimal expansion of Sum_{n >= 1} (-1)^(n + 1)*24/(n*(n + 1)*(n + 2)*(n + 3)).

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A049017 Expansion of 1/((1-x)^7 - x^7).

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A221857 Number A(n,k) of shapes of balanced k-ary trees with n nodes, where a tree is balanced if the total number of nodes in subtrees corresponding to the branches of any node differ by at most one; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A231303 Recurrence a(n) = a(n-2) + n^M for M=4, starting with a(0)=0, a(1)=1.

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A242023 Decimal expansion of Sum_{n >= 1} (-1)^(n + 1)24/(n(n + 1)(n + 2)(n + 3)).