A000581 -id:A000581 - cp's OEIS frontend

A010968 a(n) = binomial(n,15).

1, 16, 136, 816, 3876, 15504, 54264, 170544, 490314, 1307504, 3268760, 7726160, 17383860, 37442160, 77558760, 155117520, 300540195, 565722720, 1037158320, 1855967520, 3247943160, 5567902560, 9364199760, 15471286560, 25140840660, 40225345056, 63432274896

Offset: 15

N. J. A. Sloane

nonn

There are no primes in this sequence. - Artur Jasinski, Dec 02 2007

T. D. Noe, Table of n, a(n) for n = 15..1000
Milan Janjic, Two Enumerative Functions University of Banja Luka (Bosnia and Herzegovina, 2017).
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Index entries for linear recurrences with constant coefficients, signature (16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1).

Cf. A000581, A001787, A110555, A242091.

[ Binomial(n,15): n in [15..70]]; // Vincenzo Librandi, Mar 26 2011

seq(binomial(n,15),n=15..37); # Zerinvary Lajos, Aug 06 2008

Table[Binomial[n,15],{n,15,50}] (* Vladimir Joseph Stephan Orlovsky, Apr 22 2011 *)

for(n=15, 50, print1(binomial(n,15), ", ")) \\ G. C. Greubel, Aug 31 2017

a(n) = -A110555(n+1,15). - Reinhard Zumkeller, Jul 27 2005
a(n+14) = n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7)(n+8)(n+9)(n+10)(n+11)(n+12)(n+13)(n+14)/15!. - Artur Jasinski, Dec 02 2007; R. J. Mathar, Jul 07 2009
G.f.: x^15/(1-x)^16. - Zerinvary Lajos, Aug 06 2008; R. J. Mathar, Jul 07 2009
a(n) = n/(n-15) * a(n-1), n > 15. - Vincenzo Librandi, Mar 26 2011
From Amiram Eldar, Dec 10 2020: (Start)
Sum_{n>=15} 1/a(n) = 15/14.
Sum_{n>=15} (-1)^(n+1)/a(n) = A001787(15)*log(2) - A242091(15)/14! = 245760*log(2) - 1023103525/6006 = 0.9438350048... (End)

Some formulas adjusted to the offset by R. J. Mathar, Jul 07 2009

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)

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

A306477 Number of ways to write n as C(w+2,2) + C(x+3,4) + C(y+5,6) + C(z+7,8) with w,x,y,z nonnegative integers, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Original entry on oeis.org

1, 3, 4, 4, 3, 3, 5, 6, 5, 5, 8, 8, 6, 4, 6, 10, 10, 8, 6, 6, 6, 10, 9, 6, 6, 7, 7, 6, 8, 10, 10, 7, 4, 7, 7, 9, 13, 12, 9, 6, 5, 6, 11, 12, 12, 13, 10, 9, 8, 9, 11, 15, 12, 8, 8, 10, 14, 11, 7, 8, 12, 9, 8, 9, 10, 11, 13, 8, 5, 9, 10, 13, 14, 12, 8, 7, 6, 12, 14, 14

Offset: 1

Zhi-Wei Sun, Feb 18 2019

nonn

Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as C(w,2) + C(x,4) + C(y,6) + C(z,8), where w,x,y,z are integers greater than one.
I'd like to call this conjecture "the 2-4-6-8 conjecture". I have verified it for all n = 1..3*10^7.
On Feb. 20, 2019, Yaakov Baruch reported on Mathoverflow that he had verified the 2-4-6-8 conjecture for n up to 5*10^8. - Zhi-Wei Sun, Feb 20 2019
On Feb. 24, 2019, Max A. Alekseyev reported on Mathoverflow that he had verified the 2-4-6-8 conjecture for n up to 2*10^11.
I'd like to offer 2468 US dollars as the prize for the first correct proof of my 2-4-6-8 conjecture, or 2468 RMB as the prize for the first explicit counterexample. - Zhi-Wei Sun, Feb 24 2019
Yaakov Baruch reported on March 12, 2019 that he had checked the 2-4-6-8 conjecture for all n = 1..2*10^12 with no counterexample found. - Zhi-Wei Sun, Mar 12 2019

			a(1) = 1 with 1 = C(2,2) + C(3,4) + C(5,6) + C(7,8).
a(4655) = 2 with 4655 = C(85,2) + C(14,4) + C(9,6) + C(7,8) = C(94,2) + C(7,4) + C(9,6) + C(11,8).
a(9590) = 2 with 9590 = C(35,2) + C(21,4) + C(7,6) + C(14,8) = C(136,2) + C(7,4) + C(10,6) + C(11,8).
a(24935) = 2 with 24935 = C(49,2) + C(29,4) + C(7,6) + C(8,8) = C(140,2) + C(26,4) + C(10,6) + C(10,8).
a(33845) = 2 with 33845 = C(104,2) + C(8,4) + C(19,6) + C(13,8) = C(148,2) + C(26,4) + C(16,6) + C(9,8).
a(192080) = 2 with 192080 = C(7,2) + C(26,4) + C(25,6) + C(9,8) = C(414,2) + C(39,4) + C(8,6) + C(17,8).
a(23343989) = 1 with 23343989 = C(365,2) + C(76,4) + C(40,6) + C(34,8).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Positive integers written as C(w,2) + C(x,4) + C(y,6) + C(z,8) with w,x,y,z in {2,3,...}, Question 323541 on Mathoverflow, Feb. 19, 2019.

Cf. A000217, A000332, A000579, A000581, A306459, A306460, A306462, A306471.

f[m_,n_]:=f[m,n]=Binomial[m+n-1,m]; TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[r=0;Do[If[f[8,z]>=n,Goto[cc]];Do[If[f[6,y]>=n-f[8,z],Goto[bb]];Do[If[f[4,x]>=n-f[8,z]-f[6,y],Goto[aa]];If[TQ[n-f[8,z]-f[6,y]-f[4,x]],r=r+1],{x,0,n-1-f[8,z]-f[6,y]}];Label[aa],{y,0,n-1-f[8,z]}];Label[bb],{z,0,n-1}];Label[cc];tab=Append[tab,r],{n,1,80}];Print[tab]

A053137 Binomial coefficients C(2*n+8,8).

Original entry on oeis.org

1, 45, 495, 3003, 12870, 43758, 125970, 319770, 735471, 1562275, 3108105, 5852925, 10518300, 18156204, 30260340, 48903492, 76904685, 118030185, 177232627, 260932815, 377348994, 536878650, 752538150, 1040465790, 1420494075, 1916797311, 2558620845, 3381098545

Offset: 0

Wolfdieter Lang

Even-indexed members of ninth column of Pascal's triangle A007318.

Number of standard tableaux of shape (2n+1,1^8). - Emeric Deutsch, May 30 2004

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Milan Janjić, Two Enumerative Functions.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A007318, A053136, A000581, A053130.

[Binomial(2*n+8,8): n in [0..30]]; // Vincenzo Librandi, Oct 07 2011

Table[Binomial[2*n+8, 8], {n, 0, 30}] (* G. C. Greubel, Sep 03 2018 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,45,495,3003,12870,43758,125970,319770,735471},30] (* Harvey P. Dale, Jul 02 2022 *)

a(n)=binomial(2*n+8,8) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = binomial(2*n+8, 8) = A000581(2*n+8).
G.f.: (1+36*x+126*x^2+84*x^3+9*x^4) / (1-x)^9 = (1+3*x) * (3*x^3+27*x^2+33*x+1) / (1-x)^9.
From Amiram Eldar, Nov 03 2022: (Start)
Sum_{n>=0} 1/a(n) = 512*log(2) - 5308/15.
Sum_{n>=0} (-1)^n/a(n) = 16*Pi + 32*log(2) - 1072/15. (End)

A097299 Ninth column (m=8) of (1,6)-Pascal triangle A096956.

Original entry on oeis.org

6, 49, 225, 765, 2145, 5247, 11583, 23595, 45045, 81510, 140998, 234702, 377910, 591090, 901170, 1343034, 1961256, 2812095, 3965775, 5509075, 7548255, 10212345, 13656825, 18067725, 23666175, 30713436, 39516444, 50433900, 63882940

Offset: 0

Wolfdieter Lang, Aug 13 2004

Paolo Xausa, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000581, A096956.

Cf. other columns: A096957 (m = 3), A096958 (m = 4), A096959 (m = 5), A097297 (m = 6), A097298 (m = 7), A097300 (m = 9).

A097299[n_] := (n + 48)*Binomial[n + 7, 7]/8;
Array[A097299, 50, 0] (* Paolo Xausa, May 02 2025 *)

a(n) = A096956(n+8, 8) = 6*b(n) - 5*b(n-1) = (n+48)*binomial(n+7, 7)/8, with b(n) = A000581(n+8) = binomial(n+8, 8).

G.f.: (6-5*x)/(1-x)^9.

A155856 Triangle T(n,k) = binomial(2n-k, k)(n-k)!, read by rows.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 10, 6, 1, 24, 42, 30, 10, 1, 120, 216, 168, 70, 15, 1, 720, 1320, 1080, 504, 140, 21, 1, 5040, 9360, 7920, 3960, 1260, 252, 28, 1, 40320, 75600, 65520, 34320, 11880, 2772, 420, 36, 1, 362880, 685440, 604800, 327600, 120120, 30888, 5544, 660, 45, 1

Offset: 0

Paul Barry, Jan 29 2009

Row sums of B^{-1}*A155856*B^{-1} are A000166 with B=A007318.

Downward diagonals T(n+j, n) = j!*binomial(n+j, n) = j!*seq(j), where seq(j) are sequences A010965, A010967, ..., A011101, A017714, A017716, ..., A017764, for 6 <= j <= 50, respectively. - G. C. Greubel, Jun 04 2021

			Triangle begins:
     1;
     1,    1;
     2,    3,    1;
     6,   10,    6,    1;
    24,   42,   30,   10,    1;
   120,  216,  168,   70,   15,   1;
   720, 1320, 1080,  504,  140,  21,  1;
  5040, 9360, 7920, 3960, 1260, 252, 28, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Veronica Bitonti, Bishal Deb, and Alan D. Sokal, Thron-type continued fractions (T-fractions) for some classes of increasing trees, arXiv:2412.10214 [math.CO], 2024. See p. 58.

Cf. A155857 (row sums), A155858 (diagonal sums).

Table[Binomial[2n-k,k](n-k)!,{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Mar 24 2017 *)

flatten([[factorial(n-k)*binomial(2*n-k, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 04 2021

T(n,k) = binomial(2*n-k, k)*(n-k)!.
Sum_{k=0..n} T(n, k) = A155857(n)
Sum_{k=0..floor(n/2)} T(n-k, k) = A155858(n) (diagonal sums).
G.f.: 1/(1-xy-x/(1-xy-x/(1-xy-2x/(1-xy-2x/(1-xy-3x/(1-.... (continued fraction).
From G. C. Greubel, Jun 04 2021: (Start)
  T(n, 0) = A000142(n).        T(n+1, n) = A000217(n+1).
T(n+1, 1) = A007680(n).        T(n+2, n) = A034827(n+4).
T(n+2, 2) = A175925(n).        T(n+3, n) = A253946(n).
T(2*n, n) = A064352(n)         T(n+4, n) = 4!*A000581(n).
T(n+1, n) = A000217(n+1).      T(n+5, n) = 5!*A001287(n).  (End)

A304366 Numbers with additive persistence = 1.

Original entry on oeis.org

10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 30, 31, 32, 33, 34, 35, 36, 40, 41, 42, 43, 44, 45, 50, 51, 52, 53, 54, 60, 61, 62, 63, 70, 71, 72, 80, 81, 90, 100, 101, 102, 103, 104, 105, 106, 107, 108, 110, 111, 112, 113, 114, 115, 116

Offset: 1

Jaroslav Krizek, May 11 2018

For d >= 2, there are A000581(d+8) terms with d digits. - Robert Israel, Dec 28 2023

			Adding the digits of 10 gives 1, a single-digit number, so 10 is a member. Adding the digits of 39 gives 12, which is a 2-digit number, so 39 is not a member. - _Michael B. Porter_, May 16 2018

Robert Israel, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Additive Persistence.

Cf. A000581, A031286.

Cf. Numbers with additive persistence k: A304367 (k=2), A304368 (k=3), A304373 (k=4).

select(t -> convert(convert(t,base,10),`+`) < 10, [$10 .. 200]); # Robert Israel, Dec 28 2023

Select[Range@ 120, Length@ FixedPointList[Total@ IntegerDigits@ # &, #] == 3 &] (* Michael De Vlieger, May 14 2018 *)

nb(n) = {my(nba = 0); while (n > 9, n = sumdigits(n); nba++); nba;}
isok(n) = nb(n) == 1; \\ Michel Marcus, May 13 2018

A031286(a(n)) = 1.

A056001 a(n) = (n+1)*binomial(n+7, 7).

Original entry on oeis.org

1, 16, 108, 480, 1650, 4752, 12012, 27456, 57915, 114400, 213928, 381888, 655044, 1085280, 1744200, 2728704, 4167669, 6229872, 9133300, 13156000, 18648630, 26048880, 35897940, 48859200, 65739375, 87512256, 115345296, 150629248, 195011080, 250430400, 319159632

Offset: 0

Barry E. Williams, Jun 18 2000

Original name: A second-order recursive sequence.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

Partial sums of A052226.
Cf. A093565 ((8, 1) Pascal, column m=8).
Cf. A007318, A000142, A245334.

List([0..30], n-> (n+1)*Binomial(n+7,7)); # G. C. Greubel, Aug 29 2019

a056001 n = (n + 1) * a007318' (n + 7) 7
-- Reinhard Zumkeller, Aug 31 2014

[(n+1)*Binomial(n+7,7): n in [0..30]]; // G. C. Greubel, Aug 29 2019

seq((n+1)*binomial(n+7,7), n=0..30); # G. C. Greubel, Aug 29 2019

Table[(n+1)Binomial[n+7, 7], {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011; corrected by Bruno Berselli, Jan 23 2015 *)

vector(30, n, n*binomial(n+6,7)) \\ G. C. Greubel, Aug 29 2019

[(n+1)*binomial(n+7,7) for n in (0..30)] # G. C. Greubel, Aug 29 2019

G.f.: (1+7*x)/(1-x)^9.
a(n) = A245334(n+7,7)/A000142(7). - Reinhard Zumkeller, Aug 31 2014
a(n) = A000581(n+8)+7*A000581(n+7). - R. J. Mathar, Oct 24 2014
E.g.f.: (5040 +75600*x +194040*x^2 +170520*x^3 +66150*x^4 +12642*x^5 + 1225*x^6 +57*x^7 +x^8)*exp(x)/5040. - G. C. Greubel, Aug 29 2019
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=0} 1/a(n) = 7*Pi^2/6 - 37583/3600.
Sum_{n>=0} (-1)^n/a(n) = 7*Pi^2/12 - 2912*log(2)/15 + 155701/1200. (End)

A117411 Skew triangle associated to the Euler numbers.

Original entry on oeis.org

1, 0, 1, 0, -4, 1, 0, 0, -12, 1, 0, 0, 16, -24, 1, 0, 0, 0, 80, -40, 1, 0, 0, 0, -64, 240, -60, 1, 0, 0, 0, 0, -448, 560, -84, 1, 0, 0, 0, 0, 256, -1792, 1120, -112, 1, 0, 0, 0, 0, 0, 2304, -5376, 2016, -144, 1, 0, 0, 0, 0, 0, -1024, 11520, -13440, 3360, -180, 1, 0, 0, 0, 0, 0, 0, -11264, 42240, -29568, 5280, -220, 1

Offset: 0

Paul Barry, Mar 13 2006

Inverse is A117414. Row sums of the inverse are the Euler numbers A000364.

Triangle, read by rows, given by [0,-4,4,0,0,0,0,0,0,0,...] DELTA [1,0,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 01 2009

			Triangle begins
  1;
  0,  1;
  0, -4,   1;
  0,  0, -12,   1;
  0,  0,  16, -24,    1;
  0,  0,   0,  80,  -40,     1;
  0,  0,   0, -64,  240,   -60,      1;
  0,  0,   0,   0, -448,   560,    -84,      1;
  0,  0,   0,   0,  256, -1792,   1120,   -112,      1;
  0,  0,   0,   0,    0,  2304,  -5376,   2016,   -144,      1;
  0,  0,   0,   0,    0, -1024,  11520, -13440,   3360,   -180,    1;
  0,  0,   0,   0,    0,     0, -11264,  42240, -29568,   5280, -220,    1;
  0,  0,   0,   0,    0,     0,   4096, -67584, 126720, -59136, 7920, -264, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000364, A006495 (row sums), A098158, A117413, A117414.

Cf. A000217, A000332, A000579, A000581, A262710.

A117411:= func< n,k | (-4)^(n-k)*(&+[Binomial(n,k-j)*Binomial(j,n-k): j in [0..n-k]]) >;
[A117411(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Sep 07 2022

T[n_,k_]:= T[n,k]= (-4)^(n-k)*Sum[Binomial[n, k-j]*Binomial[j, n-k], {j,0,n-k}];
Table[T[n,k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 07 2022 *)

def A117411(n,k): return (-4)^(n-k)*sum(binomial(n,k-j)*binomial(j,n-k) for j in (0..n-k))
flatten([[A117411(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Sep 07 2022

Sum_{k=0..n} T(n, k) = A006495(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A117413(n).
T(n, k) = (-4)^(n-k)*Sum_{j=0..n-k} C(n,k-j)*C(j,n-k).
G.f.: (1-x*y)/(1-2x*y+x^2*y(y+4)). - Paul Barry, Mar 14 2006
T(n, k) = (-4)^(n-k)*A098158(n,k). - Philippe Deléham, Nov 01 2009
T(n, k) = 2*T(n-1,k-1) - 4*T(n-2,k-1) - T(n-2,k-2), T(0,0) = T(1,1) = 1, T(1,0) = 0, T(n,k) = 0 if k > n or if k < 0. - Philippe Deléham, Oct 31 2013
From G. C. Greubel, Sep 07 2022: (Start)
T(n, n) = 1.
T(n, n-1) = -4*A000217(n-1), n >= 1.
T(n, n-2) = (-4)^2 * A000332(n), n >= 2.
T(n, n-3) = (-4)^3 * A000579(n), n >= 3.
T(n, n-4) = (-4)^4 * A000581(n), n >= 4.
T(2*n, n) = A262710(n). (End)

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