cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A136212 Triple factorial array, read by antidiagonals, where row n+1 is generated from row n by first removing terms in row n at positions {[m*(m+5)/6], m >= 0} and then taking partial sums, starting with all 1's in row 0.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 28, 10, 3, 1, 280, 80, 18, 4, 1, 3640, 880, 162, 28, 5, 1, 58240, 12320, 1944, 280, 39, 6, 1, 1106560, 209440, 29160, 3640, 418, 52, 7, 1, 24344320, 4188800, 524880, 58240, 5714, 600, 66, 8, 1, 608608000, 96342400, 11022480, 1106560, 95064
Offset: 0

Views

Author

Paul D. Hanna, Dec 22 2007

Keywords

Comments

This is the triple factorial variant of Moessner's factorial array described by A125714 and also of the double factorial array A135876. Another very interesting variant is A136217.

Examples

			Square array begins:
(1),(1),(1),1,(1),1,(1),1,(1),1,1,(1),1,1,(1),1,1,(1),1,1,1,(1),1,1,1,...;
(1),(2),(3),4,(5),6,(7),8,(9),10,11,(12),13,14,(15),16,17,(18),19,20,21,..;
(4),(10),(18),28,(39),52,(66),82,(99),118,138,(159),182,206,(231),258,286,..;
(28),(80),(162),280,(418),600,(806),1064,(1350),1696,2074,(2485),2966,3484,..;
(280),(880),(1944),3640,(5714),8680,(12164),16840,(22194),29080,36824,(45474),.;
(3640),(12320),(29160),58240,(95064),151200,(219108),315440,(428652),581680,...;
(58240),(209440),(524880),1106560,(1864456),3082240,...;
where terms in parenthesis are at positions {[m*(m+5)/6], m>=0}
and are removed before taking partial sums to obtain the next row.
To generate the array, start with all 1's in row 0; from then on,
obtain row n+1 from row n by first removing terms in row n at
positions {[m*(m+5)/6], m>=0} and then taking partial sums.
For example, to generate row 2 from row 1:
[(1),(2),(3),4,(5),6,(7),8,(9),10,11,(12),13,14,(15),16,17,(18),...],
remove terms at positions [0,1,2,4,6,8,11,14,17,...] to get:
[4, 6, 8, 10,11, 13,14, 16,17, 19,20,21, 23,24,25, 27,28,29, ...]
then take partial sums to obtain row 2:
[4, 10, 18, 28,39, 52,66, 82,99, 118,138,159, 182,206,231, ...].
Continuing in this way will generate all the rows of this array.
		

Crossrefs

Programs

  • Mathematica
    t[n_, k_] := t[n, k] = Module[{a = 0, m = 0, c = 0, d = 0}, If[n == 0, a = 1, While[d <= k, If[c == Quotient[(m*(m + 5)), 6], m += 1, a += t[n - 1, c]; d += 1]; c += 1]]; a]; Table[t[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 06 2013, translated from Pari *)
  • PARI
    {T(n, k)=local(A=0, m=0, c=0, d=0); if(n==0, A=1, until(d>k, if(c==(m*(m+5))\6, m+=1, A+=T(n-1, c); d+=1); c+=1)); A}

Formula

Columns 0, 1 and 2 form the triple factorials A007559, A008544 and A032031, respectively. Column 4 equals A024216; column 6 equals A024395.

A225117 Triangle read by rows, coefficients of the generalized Eulerian polynomials A_{n, 3}(x) in descending order.

Original entry on oeis.org

1, 2, 1, 4, 13, 1, 8, 93, 60, 1, 16, 545, 1131, 251, 1, 32, 2933, 14498, 10678, 1018, 1, 64, 15177, 154113, 262438, 88998, 4089, 1, 128, 77101, 1475736, 4890287, 3870352, 692499, 16376, 1, 256, 388321, 13270807, 77404933, 117758659, 50476003, 5175013, 65527, 1
Offset: 0

Views

Author

Peter Luschny, May 02 2013

Keywords

Comments

The row sums equal the triple factorial numbers A032031 and the alternating row sums, i.e., Sum_{k=0..n}(-1)^k*T(n,k), are up to a sign A000810. - Johannes W. Meijer, May 04 2013

Examples

			[0]  1
[1]  2*x   +   1
[2]  4*x^2 +  13*x   +    1
[3]  8*x^3 +  93*x^2 +   60*x   +   1
[4] 16*x^4 + 545*x^3 + 1131*x^2 + 251*x + 1
...
The triangle T(n, k) begins:
n \ k 0      1        2        3       4      5     6 7 ...
0:    1
1:    2      1
2:    4     13        1
3:    8     93       60        1
5:   16    545     1131      251       1
6:   32   2933    14498    10678    1018      1
7:   64  15177   154113   262438   88998   4089     1
8:  128  77101  1475736  4890287 3870352 692499 16376 1
...  - _Wolfdieter Lang_, Apr 08 2017
Three term recurrence: T(2,1) = (3*(2-1)+1)*2 + (3*1+2)*1 = 13. - _Wolfdieter Lang_, Apr 10 2017
		

Crossrefs

Coefficients of A_{n,1}(x) = A008292, coefficients of A_{n,2}(x) = A060187, coefficients of A_{n,4}(x) = A225118.

Programs

  • Maple
    gf := proc(n, k) local f; f := (x,t) -> x*exp(t*x/k)/(1-x*exp(t*x));
    series(f(x,t), t, n+2); ((1-x)/x)^(n+1)*k^n*n!*coeff(%, t, n):
    collect(simplify(%), x) end:
    seq(print(seq(coeff(gf(n, 3), x, n-k), k=0..n)), n=0..6);
    # Recurrence
    P := proc(n,x) option remember; if n = 0 then 1 else
      (n*x+(1/3)*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x);
      expand(%) fi end:
    A225117 := (n,k) -> 3^n*coeff(P(n,x),x,n-k):
    seq(print(seq(A225117(n,k), k=0..n)), n=0..5);  # Peter Luschny, Mar 08 2014
  • Mathematica
    gf[n_, k_] := Module[{f, s}, f[x_, t_] := x*Exp[t*x/k]/(1-x*Exp[t*x]); s = Series[f[x, t], {t, 0, n+2}]; ((1-x)/x)^(n+1)*k^n*n!*SeriesCoefficient[s, {t, 0, n}]]; Table[Table[SeriesCoefficient[gf[n, 3], {x, 0, n-k}], {k, 0, n}], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
  • PARI
    T(n, k) = sum(j=0, n - k, (-1)^(n - k - j)*binomial(n + 1, n - k - j)*(1 + 3*j)^n);
    for(n=0, 10, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 10 2017
    
  • Python
    from sympy import binomial
    def T(n,k): return sum((-1)**(n - k - j)* binomial(n + 1, n - k - j)*(1 + 3*j)**n for j in range(n - k + 1))
    for n in range(11): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 10 2017
  • Sage
    @CachedFunction
    def EB(n, k, x):  # Modified cardinal B-splines
        if n == 1: return 0 if (x < 0) or (x >= 1) else 1
        return k*x*EB(n-1, k, x) + k*(n-x)*EB(n-1, k, x-1)
    def EulerianPolynomial(n, k): # Generalized Eulerian polynomials
        R. = ZZ[]
        if x == 0: return 1
        return add(EB(n+1, k, m+1/k)*x^m for m in (0..n))
    [EulerianPolynomial(n, 3).coefficients()[::-1] for n in (0..5)]
    

Formula

Generating function of the polynomials is gf(n, k) = k^n*n!*(1/x-1)^(n+1)[t^n](x*e^(t*x/k)*(1-x*e(t*x))^(-1)) for k = 3; here [t^n]f(t,x) is the coefficient of t^n in f(t,x).
From Wolfdieter Lang, Apr 10 2017: (Start)
T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*binomial(n+1, n-k-j)*(1+3*j)^n, 0 <= k <= n.
T(n, k) = Sum_{m=0..n-k} (-1)^(n-k-m)*binomial(n-m, k)*A284861(n, m), 0 <= k <= n.
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k are R(n, x) = (x-1)^n*Sum_{m=0} A284861(n, m)*(1/(x-1))^m, n >= 0, i.e. the row polynomials of A284861 in the variable 1/(x-1) multiplied by (x-1)^n.
The row polynomials with falling powers are P(n, x) = (1-x)^n*Sum_{m=0..n} A284861(n, m)*(x/(1-x))^m, n >= 0.
The e.g.f. of the row polynomials in falling powers of x (A_{n, 3}(x) of the name) is exp((1-x)*z)/(1 - (x/(1 - x)) * (exp(3*(1-x)*z) - 1)) = (1-x)*exp((1-x)*z)/(1 - x*exp(3*(1-x)*z)).
The e.g.f. of the row polynomials R(n, x) (rising powers of x) is then (1-x)*exp(2*(1-x)*z)/(1 - x*exp(3*(1-x)*z)).
Three term recurrence: T(n, k) = 0 if n < k , T(n, -1) = 0, T(0,0) = 1, T(n, k) = (3*(n-k)+1)*T(n-1, k-1) + (3*k+2)*T(n-1, k) for n >= 1, k=0..n. (End)

A303486 a(n) = n! * [x^n] 1/(1 - 3*x)^(n/3).

Original entry on oeis.org

1, 1, 10, 162, 3640, 104720, 3674160, 152152000, 7264216960, 392841187200, 23734494784000, 1584471003315200, 115825295634048000, 9201578813819392000, 789383453851632640000, 72728093032166347776000, 7162140885524461957120000, 750766815289210771251200000
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 24 2018

Keywords

Examples

			a(1) = 1;
a(2) = 2*5 = 10;
a(3) = 3*6*9 = 162;
a(4) = 4*7*10*13 = 3640;
a(5) = 5*8*11*14*17 = 104720, etc.
		

Crossrefs

Programs

  • Mathematica
    Table[n! SeriesCoefficient[1/(1 - 3 x)^(n/3), {x, 0, n}], {n, 0, 17}]
    Table[Product[3 k + n, {k, 0, n - 1}], {n, 0, 17}]
    Table[3^n Pochhammer[n/3, n], {n, 0, 17}]

Formula

a(n) = Product_{k=0..n-1} (3*k + n).
a(n) = 3^n*Gamma(4*n/3)/Gamma(n/3).
a(n) ~ 2^(8*n/3-1)*n^n/exp(n).

A051607 a(n) = (3*n+7)!!!/7!!!.

Original entry on oeis.org

1, 10, 130, 2080, 39520, 869440, 21736000, 608608000, 18866848000, 641472832000, 23734494784000, 949379791360000, 40823331028480000, 1877873227310080000, 92015788138193920000, 4784820983186083840000, 263165154075234611200000, 15263578936363607449600000
Offset: 0

Views

Author

Keywords

Comments

Related to A007559(n+1) ((3*n+1)!!! triple factorials).
Row m=7 of the array A(4; m,n) := ((3*n+m)(!^3))/m(!^3), m >= 0, n >= 0.

Crossrefs

Cf. A032031, A007559(n+1), A034000(n+1), A034001(n+1), A051604, A051605, A051606, A051608, A051609 (rows m=0..9).

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1/(1-3*x)^(10/3))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Aug 15 2018
  • Mathematica
    With[{nn = 30}, CoefficientList[Series[1/(1 - 3*x)^(10/3), {x, 0, nn}], x]*Range[0, nn]!] (* G. C. Greubel, Aug 15 2018 *)
  • PARI
    x='x+O('x^30); Vec(serlaplace(1/(1-3*x)^(10/3))) \\ G. C. Greubel, Aug 15 2018
    

Formula

a(n) = ((3*n+7)(!^3))/7(!^3).
E.g.f.: 1/(1-3*x)^(10/3).
Sum_{n>=0} 1/a(n) = 1 + 9*(3*e)^(1/3)*(Gamma(10/3) - Gamma(10/3, 1/3)). - Amiram Eldar, Dec 23 2022

A284861 Triangle read by rows: T(n, k) = S2[3,1](n, k)*k! with the Sheffer triangle S2[3,1] = (exp(x), exp(3*x) -1) given in A282629.

Original entry on oeis.org

1, 1, 3, 1, 15, 18, 1, 63, 216, 162, 1, 255, 1890, 3564, 1944, 1, 1023, 14760, 52650, 68040, 29160, 1, 4095, 109458, 659340, 1516320, 1487160, 524880, 1, 16383, 790776, 7578522, 27624240, 46539360, 36741600, 11022480, 1, 65535, 5633730, 82902204, 450057384, 1158993360, 1535798880, 1014068160, 264539520
Offset: 0

Views

Author

Wolfdieter Lang, Apr 09 2017

Keywords

Comments

This is a generalization of triangle A131689(n, k) = Stirling2(n, k)*k!, because S2[3,1] is a generalization of the Stirling2 triangle written as S2[1,0].
This triangle appears in the o.g.f. G(3,1;n,x) of the powers {(1+3*m)^n}{m>=0} as G(3,1;n,x) = Sum{k>=0..n} T(n, k)*x^k / (1-x)^k.
This triangle is also related to the generalized row reversed Euler triangle rEu[3,1] with row polynomial rEu(3,1;n,x) = Sum_{m=0..n} rEu(3,1;n,m)*x^m with rEu(3,1;n,m) = Sum_{j=0..m} (-1)^(m-j)*binomial(n-j, m-j)*T(n, m). This follows from the above given o.g.f. of powers G(3,1;n,x) = rEu(3,1;n,x)/(1-x)^(n+1). The Euler triangle E[3,1] (row reversed rEu[3,1] is given in A225117. See a formula below.
The e.g.f. of the row polynomials R(3,1;n,x) = Sum_{m=0..n} T(n, m)*x^m follows from the e.g.f. of the row polynomials of the Sheffer triangle A282629. See the formula section.
The diagonal sequence is A032031(k) = k!*3^k.
The row sums give unsigned A151919, and the alternating row sums give A122803.
The first column k sequences divided by A032031(k) are A000012, A002450 (with a leading 0), A016223, A021874. For the e.g.f.s and o.g.f.s see below.

Examples

			The triangle T(n, k) begins
n\k 0     1      2       3        4        5        6        7 ...
0:  1
1:  1     3
2:  1    15     18
3:  1    63    216     162
4:  1   255   1890    3564     1944
5:  1  1023  14760   52650    68040    29160
6:  1  4095 109458  659340  1516320  1487160   524880
7:  1 16383 790776 7578522 27624240 46539360 36741600 11022480
...
row n=8: 1 65535 5633730 82902204 450057384 1158993360 1535798880 1014068160 264539520,
row n=9: 1 262143 39829320 879725610 6845572440 25294754520 50042059200 54561276000 30951123840 7142567040,
row n=10: 1 1048575 280378098 9155719980 99549149040 507399658920 1406104706160 2251231315200 2083248720000 1035672220800 214277011200.
------------------------------------------------------------------
T(2, 1) =  -1 + 4^2 = 15 = 2*A225117(2,2) + 1*A225117(2,1) = 2*1 + 1*13.
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[k, m] (-1)^(k - m) (1 + 3m)^n, {m, 0, k}], {n, 0, 10}, {k, 0, n}]// Flatten (* Indranil Ghosh, Apr 09 2017 *)
  • PARI
    for(n=0, 10, for(k=0, n, print1(sum(m=0, k, binomial(k, m) * (-1)^(k - m)*(1 + 3*m)^n),", "); ); print();) \\ Indranil Ghosh, Apr 09 2017
    
  • Python
    # Indranil Ghosh, Apr 09 2017
    from sympy import binomial
    for n in range(11):
        print([sum([binomial(k, m)*(-1)**(k - m)*(1 + 3*m)**n for m in range(k + 1)]) for k in range(n + 1)])

Formula

E.g.f. of the row polynomials R(n, x) (see a comment above) is exp(z)/(1 - x*(exp(3*z) - 1)). This is the e.g.f. for the triangle.
T(n, k) = Sum_{m=0..k} binomial(k, m)*(-1)^(k-m)*(1+ 3*m)^n, 0 <= k <= n.
T(n, k) = Sum_{m=0..k} binomial(n-m, k-m)*A225117(n,n-m), 0 <= k <= n.
Three term recurrence: T(n, k) = 0 if n < k, T(n,-1) = 0, T(0, 0) = 1, T(n, k) = 3*k*T(n-1, k-1) + (1+3*k)*T(n-1, k) for n >= 1. See A282629.
The column k sequence has e.g.f. exp(x)*(exp(3*x) - 1)^k (from the Sheffer property of A282629).
The o.g.f. is A032031(k)*x^k/Product_{j=0..k} (1 - (1+3*j)*x).
From Peter Bala, Jan 12 2018: (Start)
n-th row polynomial R(n,x) = (1 + 3*x) o (1 + 3*x) o ... o (1 + 3*x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E14 in the Bala link. Cf. A145901.
R(n,x) = Sum_{k = 0..n} binomial(n,k)*3^k*F(k,x) where F(k,x) is the Fubini polynomial of order k, the k-th row polynomial of A019538. (End)

A051608 a(n) = (3*n+8)!!!/8!!!.

Original entry on oeis.org

1, 11, 154, 2618, 52360, 1204280, 31311280, 908027120, 29056867840, 1016990374400, 38645634227200, 1584471003315200, 69716724145868800, 3276686034855833600, 163834301742791680000, 8683217992367959040000, 486260207572605706240000, 28689352246783736668160000
Offset: 0

Views

Author

Keywords

Comments

Related to A008544(n+1) ((3*n+2)!!! triple factorials).
Row m=8 of the array A(4; m,n) := ((3*n+m)(!^3))/m(!^3), m >= 0, n >= 0.

Crossrefs

Cf. A032031, A007559(n+1), A034000(n+1), A034001(n+1), A051604, A051605, A051606, A051607, A051609 (rows m=0..9).
Cf. A008544.

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1/(1-3*x)^(11/3))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Aug 15 2018
  • Mathematica
    With[{nn = 30}, CoefficientList[Series[1/(1 - 3*x)^(11/3), {x, 0, nn}], x]*Range[0, nn]!] (* G. C. Greubel, Aug 15 2018 *)
  • PARI
    x='x+O('x^30); Vec(serlaplace(1/(1-3*x)^(11/3))) \\ G. C. Greubel, Aug 15 2018
    

Formula

a(n) = ((3*n+8)(!^3))/8(!^3).
E.g.f.: 1/(1-3*x)^(11/3).
Sum_{n>=0} 1/a(n) = 1 + 9*(9*e)^(1/3)*(Gamma(11/3) - Gamma(11/3, 1/3)). - Amiram Eldar, Dec 23 2022

A371077 Square array read by ascending antidiagonals: A(n, k) = 3^n*Pochhammer(k/3, n).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 4, 2, 1, 0, 28, 10, 3, 1, 0, 280, 80, 18, 4, 1, 0, 3640, 880, 162, 28, 5, 1, 0, 58240, 12320, 1944, 280, 40, 6, 1, 0, 1106560, 209440, 29160, 3640, 440, 54, 7, 1, 0, 24344320, 4188800, 524880, 58240, 6160, 648, 70, 8, 1
Offset: 0

Views

Author

Werner Schulte and Peter Luschny, Mar 10 2024

Keywords

Examples

			The array starts:
  [0] 1,    1,     1,     1,     1,      1,      1,      1,      1, ...
  [1] 0,    1,     2,     3,     4,      5,      6,      7,      8, ...
  [2] 0,    4,    10,    18,    28,     40,     54,     70,     88, ...
  [3] 0,   28,    80,   162,   280,    440,    648,    910,   1232, ...
  [4] 0,  280,   880,  1944,  3640,   6160,   9720,  14560,  20944, ...
  [5] 0, 3640, 12320, 29160, 58240, 104720, 174960, 276640, 418880, ...
.
Seen as the triangle T(n, k) = A(n - k, k):
  [0] 1;
  [1] 0,       1;
  [2] 0,       1,      1;
  [3] 0,       4,      2,     1;
  [4] 0,      28,     10,     3,    1;
  [5] 0,     280,     80,    18,    4,   1;
  [6] 0,    3640,    880,   162,   28,   5,  1;
  [7] 0,   58240,  12320,  1944,  280,  40,  6, 1;
  [8] 0, 1106560, 209440, 29160, 3640, 440, 54, 7, 1;
.
Illustrating the LU decomposition of A:
    / 1                \   / 1 1 1 1 1 ... \   / 1   1   1    1    1 ... \
    | 0   1            |   |   1 2 3 4 ... |   | 0   1   2    3    4 ... |
    | 0   4   2        | * |     1 3 6 ... | = | 0   4  10   18   28 ... |
    | 0  28  24   6    |   |       1 4 ... |   | 0  28  80  162  280 ... |
    | 0 280 320 144 24 |   |         1 ... |   | 0 280 880 1944 3640 ... |
    | . . .            |   | . . .         |   | . . .                   |
		

Crossrefs

Family m^n*Pochhammer(k/m, n): A094587 (m=1), A370419 (m=2), this sequence (m=3), A370915 (m=4).
Cf. A303486 (main diagonal), A371079 (row sums of triangle), A371076, A371080.

Programs

  • Maple
    A := (n, k) -> 3^n*pochhammer(k/3, n):
    A := (n, k) -> local j; mul(3*j + k, j = 0..n-1):
    # Read by antidiagonals:
    T := (n, k) -> A(n - k, k): seq(seq(T(n, k), k = 0..n), n = 0..9);
    seq(lprint([n], seq(T(n, k), k = 0..n)), n = 0..9);
    # Using the generating polynomials of the rows:
    P := (n, x) -> local k; add(Stirling1(n, k)*(-3)^(n - k)*x^k, k=0..n):
    seq(lprint([n], seq(P(n, k), k = 0..9)), n = 0..5);
    # Using the exponential generating functions of the columns:
    EGFcol := proc(k, len) local egf, ser, n; egf := (1 - 3*x)^(-k/3);
    ser := series(egf, x, len+2): seq(n!*coeff(ser, x, n), n = 0..len) end:
    seq(lprint([k], EGFcol(k, 8)), k = 0..6);
    # As a matrix product:
    with(LinearAlgebra):
    L := Matrix(7, 7, (n, k) -> A371076(n - 1,  k - 1)):
    U := Matrix(7, 7, (n, k) -> binomial(n - 1, k - 1)):
    MatrixMatrixMultiply(L, Transpose(U));
  • Mathematica
    Table[3^(n-k)*Pochhammer[k/3, n-k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Mar 14 2024 *)
  • SageMath
    def A(n, k): return 3**n * rising_factorial(k/3, n)
    def A(n, k): return (-3)**n * falling_factorial(-k/3, n)

Formula

A(n, k) = Product_{j=0..n-1} (3*j + k).
A(n, k) = A(n+1, k-3) / (k - 3) for k > 3.
A(n, k) = Sum_{j=0..n} Stirling1(n, j)*(-3)^(n - j)* k^j.
A(n, k) = k! * [x^k] (exp(x) * p(n, x)), where p(n, x) are the row polynomials of A371080.
E.g.f. of column k: (1 - 3*t)^(-k/3).
E.g.f. of row n: exp(x) * (Sum_{k=0..n} A371076(n, k) * x^k / (k!)).
Sum_{n>=0, k>=0} A(n, k) * x^k * t^n / (n!) = 1/(1 - x/(1 - 3*t)^(1/3)).
Sum_{n>=0, k>=0} A(n, k) * x^k * t^n /(n! * k!) = exp(x/(1 - 3*t)^(1/3)).
The LU decomposition of this array is given by the upper triangular matrix U which is the transpose of A007318 and the lower triangular matrix L = A371076, i.e., A(n, k) = Sum_{i=0..k} A371076(n, i) * binomial(k, i).

A144758 Partial products of successive terms of A017197.

Original entry on oeis.org

1, 3, 36, 756, 22680, 884520, 42456960, 2420046720, 159723083520, 11979231264000, 1006255426176000, 93581754634368000, 9545338972705536000, 1059532625970314496000, 127143915116437739520000, 16401565050020468398080000, 2263415976902824638935040000
Offset: 0

Views

Author

Philippe Deléham, Sep 20 2008

Keywords

Examples

			a(0)=1, a(1)=3, a(2)=3*12=36, a(3)=3*12*21=756, a(4)=3*12*21*30=22680, ...
		

Crossrefs

Programs

  • Magma
    [Round(9^n*Gamma(n+1/3)/Gamma(1/3)): n in [0..20]]; // G. C. Greubel, Dec 03 2019
    
  • Maple
    seq(9^n*pochhammer(1/3, n), n = 0..20); # G. C. Greubel, Dec 03 2019
  • Mathematica
    Table[9^n*Pochhammer[1/3, n], {n, 0, 20}] (* G. C. Greubel, Dec 03 2019 *)
    Join[{1},FoldList[Times,NestList[#+9&,3,20]]] (* Harvey P. Dale, Mar 09 2025 *)
  • PARI
    a(n)=3^n*prod(i=1,n,3*i-2) \\ Charles R Greathouse IV, Jan 17 2012
    
  • Sage
    [9^n*rising_factorial(1/3, n) for n in (0..20)] # G. C. Greubel, Dec 03 2019

Formula

a(n) = Sum_{k=0..n} A132393(n,k)*3^k*9^(n-k).
a(n) = (-6)^n*Sum_{k=0..n} (3/2)^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
Sum_{n>=0} 1/a(n) = 1 + (e/9^6)^(1/9)*(Gamma(1/3) - Gamma(1/3, 1/9)). - Amiram Eldar, Dec 21 2022

A196347 Triangle T(n, k) read by rows, T(n, k) = n!*binomial(n, k).

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 6, 18, 18, 6, 24, 96, 144, 96, 24, 120, 600, 1200, 1200, 600, 120, 720, 4320, 10800, 14400, 10800, 4320, 720, 5040, 35280, 105840, 176400, 176400, 105840, 35280, 5040, 40320, 322560, 1128960, 2257920, 2822400, 2257920, 1128960, 322560, 40320
Offset: 0

Views

Author

Philippe Deléham, Oct 28 2011

Keywords

Comments

Unsigned version of A021012.
Equal to A136572*A007318.

Examples

			Triangle begins:
    1;
    1,   1;
    2,   4,    2;
    6,  18,   18,    6;
   24,  96,  144,   96,  24;
  120, 600, 1200, 1200, 600, 120;
  ...
		

Crossrefs

Programs

  • Magma
    /* As triangle */ [[Factorial(n)*Binomial(n, k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 28 2015
  • Mathematica
    Table[n!*Binomial[n, j], {n, 0, 30}, {j, 0, n}] (* G. C. Greubel, Sep 27 2015 *)
  • Sage
    factorial(n)*binomial(n,k) # Danny Rorabaugh, Sep 27 2015
    

Formula

T(n,k) is given by (1,1,2,2,3,3,4,4,5,5,6,6,...) DELTA (1,1,2,2,3,3,4,4,5,5,6,6, ...) where DELTA is the operator defined in A084938.
Sum_{k>=0} T(m,k)*T(n,k) = (m+n)!.
T(2n,n) = A122747(n).
Sum_{k>=0} T(n,k)^2 = A010050(n) = (2n)!.
Sum_{k>=0} T(n,k)*x^k = A000007(n), A000142(n), A000165(n), A032031(n), A047053(n), A052562(n), A047058(n), A051188(n), A051189(n), A051232(n), A051262(n), A196258(n), A145448(n) for x = -1,0,1,2,3,4,5,6,7,8,9,10,11 respectively.
The row polynomials have the form (x + 1) o (x + 2) o ... o (x + n), where o denotes the black diamond multiplication operator of Dukes and White. See example E10 in the Bala link. - Peter Bala, Jan 18 2018

Extensions

Name exchanged with a formula by Peter Luschny, Feb 01 2015

A225472 Triangle read by rows, k!*S_3(n, k) where S_m(n, k) are the Stirling-Frobenius subset numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 2, 3, 4, 21, 18, 8, 117, 270, 162, 16, 609, 2862, 4212, 1944, 32, 3093, 26550, 72090, 77760, 29160, 64, 15561, 230958, 1031940, 1953720, 1662120, 524880, 128, 77997, 1941030, 13429962, 39735360, 57561840, 40415760, 11022480, 256, 390369, 15996222, 165198852
Offset: 0

Views

Author

Peter Luschny, May 17 2013

Keywords

Comments

The Stirling-Frobenius subset numbers are defined in A225468 (see also the Sage program).

Examples

			[n\k][0,     1,      2,       3,       4,       5,      6 ]
[0]   1,
[1]   2,     3,
[2]   4,    21,     18,
[3]   8,   117,    270,     162,
[4]  16,   609,   2862,    4212,    1944,
[5]  32,  3093,  26550,   72090,   77760,   29160,
[6]  64, 15561, 230958, 1031940, 1953720, 1662120, 524880.
		

Crossrefs

Cf. A131689 (m=1), A145901 (m=2), A225473 (m=4).
Cf. A225466, A225468, columns: A000079, 3*A016127, 3^2*2!*A016297, 3^3*3!*A025999.

Programs

  • Maple
    SF_SO := proc(n, k, m) option remember;
    if n = 0 and k = 0 then return(1) fi;
    if k > n or k < 0 then return(0) fi;
    m*k*SF_SO(n-1, k-1, m) + (m*(k+1)-1)*SF_SO(n-1, k, m) end:
    seq(print(seq(SF_SO(n, k, 3), k=0..n)), n = 0..5);
  • Mathematica
    EulerianNumber[n_, k_, m_] := EulerianNumber[n, k, m] = (If[ n == 0, Return[If[k == 0, 1, 0]]]; Return[(m*(n-k)+m-1)*EulerianNumber[n-1, k-1, m] + (m*k+1)*EulerianNumber[n-1, k, m]]); SFSO[n_, k_, m_] := Sum[ EulerianNumber[n, j, m]*Binomial[j, n-k], {j, 0, n}]; Table[ SFSO[n, k, 3], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 29 2013, translated from Sage *)
  • Sage
    @CachedFunction
    def EulerianNumber(n, k, m) :
        if n == 0: return 1 if k == 0 else 0
        return (m*(n-k)+m-1)*EulerianNumber(n-1, k-1, m)+ (m*k+1)*EulerianNumber(n-1, k, m)
    def SF_SO(n, k, m):
        return add(EulerianNumber(n, j, m)*binomial(j, n - k) for j in (0..n))
    for n in (0..6): [SF_SO(n, k, 3) for k in (0..n)]

Formula

For a recurrence see the Maple program.
T(n, 0) ~ A000079; T(n, 1) ~ A005057; T(n, n) ~ A032031.
From Wolfdieter Lang, Apr 10 2017: (Start)
E.g.f. for sequence of column k: exp(2*x)*(exp(3*x) - 1)^k, k >= 0. From the Sheffer triangle S2[3,2] = A225466 with column k multiplied with k!.
O.g.f. for sequence of column k is 3^k*k!*x^k/Product_{j=0..k} (1 - (2+3*j)*x), k >= 0.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k, j)*(2+3*j)^n, 0 <= k <= n.
Three term recurrence (see the Maple program): T(n, k) = 0 if n < k , T(n, -1) = 0, T(0,0) = 1, T(n, k) = 3*k*T(n-1, k-1) + (2 + 3*k)*T(n-1, k) for n >= 1, k=0..n.
For the column scaled triangle (with diagonal 1s) see A225468, and the Bala link with (a,b,c) = (3,0,2), where Sheffer triangles are called exponential Riordan triangles.
(End)
The e.g.f. of the row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k is exp(2*z)/(1 - x*(exp(3*z) - 1)). - Wolfdieter Lang, Jul 12 2017
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