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.

Showing 1-5 of 5 results.

A132393 Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

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Author

Philippe Deléham, Nov 10 2007, Oct 15 2008, Oct 17 2008

Keywords

Comments

Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.
A094645*A007318 as infinite lower triangular matrices.
Row sums are the factorial numbers. - Roger L. Bagula, Apr 18 2008
Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014
Also the Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015
This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017
T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017
From Wolfdieter Lang, Jul 20 2017: (Start)
The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End)
For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019

Examples

			Triangle T(n,k) begins:
  1;
  0,    1;
  0,    1,     1;
  0,    2,     3,     1;
  0,    6,    11,     6,    1;
  0,   24,    50,    35,   10,    1;
  0,  120,   274,   225,   85,   15,   1;
  0,  720,  1764,  1624,  735,  175,  21,  1;
  0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
  ...
---------------------------------------------------
Production matrix is
  0, 1
  0, 1, 1
  0, 1, 2,  1
  0, 1, 3,  3,  1
  0, 1, 4,  6,  4,  1
  0, 1, 5, 10, 10,  5,  1
  0, 1, 6, 15, 20, 15,  6, 1
  0, 1, 7, 21, 35, 35, 21, 7, 1
  ...
From _Wolfdieter Lang_, May 09 2017: (Start)
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - _Wolfdieter Lang_, May 28 2017
O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - _Wolfdieter Lang_, Jul 20 2017
Boas-Buck recurrence for column k = 2 and n = 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*(3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - _Wolfdieter Lang_, Aug 11 2017
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
  • Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150

Crossrefs

Essentially a duplicate of A048994. Cf. A008275, A008277, A112007, A130534, A288874, A354795.

Programs

  • Haskell
    a132393 n k = a132393_tabl !! n !! k
    a132393_row n = a132393_tabl !! n
    a132393_tabl = map (map abs) a048994_tabl
    -- Reinhard Zumkeller, Nov 06 2013
    
  • Maple
    a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x,n)),x,k),k=0..n) end: # Peter Luschny, Nov 28 2010
  • Mathematica
    p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *)
    Flatten[Table[Abs[StirlingS1[n,i]],{n,0,10},{i,0,n}]] (* Harvey P. Dale, Feb 04 2014 *)
  • Maxima
    create_list(abs(stirling1(n,k)),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    column(n,k) = my(v1, v2); v1 = vector(n-1, i, 0); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, v1[i] = (i+k)*(i+k-1)/2*v2[i]; for(j=1, i-1, v1[j] *= (i-j)*(i+k)/(i-j+2)); v2[i+1] = vecsum(v1)/i); v2 \\ generates n first elements of the k-th column starting from the first nonzero element. - Mikhail Kurkov, Mar 05 2025

Formula

T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007
Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008
Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008
a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008
Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008
Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008
From Wolfdieter Lang, Feb 21 2017: (Start)
E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/A027642. The e.g.f. for the z-sequence is 0. (End)
From Wolfdieter Lang, May 28 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1)}_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} beta(n-1-p)*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)*A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023
n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(-k,j)*T(n,k+j-1)*(-1)^j for 1 <= k < n with T(n,n) = 1.
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(n,j)*T(n-j+1,k) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link). (End)

A256890 Triangle T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.

Original entry on oeis.org

1, 2, 2, 4, 12, 4, 8, 52, 52, 8, 16, 196, 416, 196, 16, 32, 684, 2644, 2644, 684, 32, 64, 2276, 14680, 26440, 14680, 2276, 64, 128, 7340, 74652, 220280, 220280, 74652, 7340, 128, 256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172, 256, 512, 72076, 1637860, 10978444, 27227908, 27227908, 10978444, 1637860, 72076, 512
Offset: 0

Views

Author

Dale Gerdemann, Apr 12 2015

Keywords

Comments

Related triangles may be found by varying the function f(x). If f(x) is a linear function, it can be parameterized as f(x) = a*x + b. With different values for a and b, the following triangles are obtained:
a\b 1.......2.......3.......4.......5.......6
The row sums of these, and similarly constructed number triangles, are shown in the following table:
a\b 1.......2.......3.......4.......5.......6.......7.......8.......9
The formula can be further generalized to: t(n,m) = f(m+s)*t(n-1,m) + f(n-s)*t(n,m-1), where f(x) = a*x + b. The following table specifies triangles with nonzero values for s (given after the slash).
a\b 0 1 2 3
-2 A130595/1
-1
0
With the absolute value, f(x) = |x|, one obtains A038221/3, A038234/4, A038247/5, A038260/6, A038273/7, A038286/8, A038299/9 (with value for s after the slash).
If f(x) = A000045(x) (Fibonacci) and s = 1, the result is A010048 (Fibonomial).
In the notation of Carlitz and Scoville, this is the triangle of generalized Eulerian numbers A(r, s | alpha, beta) with alpha = beta = 2. Also the array A(2,1,4) in the notation of Hwang et al. (see page 31). - Peter Bala, Dec 27 2019

Examples

			Array, t(n, k), begins as:
   1,    2,      4,        8,        16,         32,          64, ...;
   2,   12,     52,      196,       684,       2276,        7340, ...;
   4,   52,    416,     2644,     14680,      74652,      357328, ...;
   8,  196,   2644,    26440,    220280,    1623964,    10978444, ...;
  16,  684,  14680,   220280,   2643360,   27227908,   251195000, ...;
  32, 2276,  74652,  1623964,  27227908,  381190712,  4677894984, ...;
  64, 7340, 357328, 10978444, 251195000, 4677894984, 74846319744, ...;
Triangle, T(n, k), begins as:
    1;
    2,     2;
    4,    12,      4;
    8,    52,     52,       8;
   16,   196,    416,     196,      16;
   32,   684,   2644,    2644,     684,      32;
   64,  2276,  14680,   26440,   14680,    2276,     64;
  128,  7340,  74652,  220280,  220280,   74652,   7340,   128;
  256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172,   256;
		

Crossrefs

Programs

  • Magma
    A256890:= func< n,k | (&+[(-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n: j in [0..k]]) >;
    [A256890(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Oct 18 2022
    
  • Mathematica
    Table[Sum[(-1)^(k-j)*Binomial[j+3, j] Binomial[n+4, k-j] (j+2)^n, {j,0,k}], {n,0, 9}, {k,0,n}]//Flatten (* Michael De Vlieger, Dec 27 2019 *)
  • PARI
    t(n,m) = if ((n<0) || (m<0), 0, if ((n==0) && (m==0), 1, (m+2)*t(n-1, m) + (n+2)*t(n, m-1)));
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n-k, k), ", ");); print(););} \\ Michel Marcus, Apr 14 2015
    
  • SageMath
    def A256890(n,k): return sum((-1)^(k-j)*Binomial(j+3,j)*Binomial(n+4,k-j)*(j+2)^n for j in range(k+1))
    flatten([[A256890(n,k) for k in range(n+1)] for n in range(11)]) # G. C. Greubel, Oct 18 2022

Formula

T(n,k) = t(n-k, k); t(0,0) = 1, t(n,m) = 0 if n < 0 or m < 0 else t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.
Sum_{k=0..n} T(n, k) = A001715(n).
T(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(j+3,j)*binomial(n+4,k-j)*(j+2)^n. - Peter Bala, Dec 27 2019
Modified rule of Pascal: T(0,0) = 1, T(n,k) = 0 if k < 0 or k > n else T(n,k) = f(n-k) * T(n-1,k-1) + f(k) * T(n-1,k), where f(x) = x + 2. - Georg Fischer, Nov 11 2021
From G. C. Greubel, Oct 18 2022: (Start)
T(n, n-k) = T(n, k).
T(n, 0) = A000079(n). (End)

A035017 One quarter of 9-factorial numbers.

Original entry on oeis.org

1, 13, 286, 8866, 354640, 17377360, 1007886880, 67528420960, 5132159992960, 436233599401600, 41005958343750400, 4223613709406291200, 473044735453504614400, 57238412989874058342400, 7440993688683627584512000, 1034298122727024234247168000, 153076122163599586668580864000
Offset: 1

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (9*n-5)*Self(n-1): n in [1..40]]; // G. C. Greubel, Oct 18 2022
    
  • Mathematica
    s=1;lst={s};Do[s+=n*s;AppendTo[lst, s], {n, 12, 2*5!, 9}];lst (* Vladimir Joseph Stephan Orlovsky, Nov 08 2008 *)
    Table[9^n*Pochhammer[4/9, n]/4, {n,40}] (* G. C. Greubel, Oct 18 2022 *)
  • SageMath
    [9^n*rising_factorial(4/9,n)/4 for n in range(1,40)] # G. C. Greubel, Oct 18 2022

Formula

4*a(n) = (9*n-5)(!^9) := Product_{j=1..n} (9*j-5).
E.g.f.: (-1+(1-9*x)^(-4/9))/4.
From G. C. Greubel, Oct 18 2022: (Start)
a(n) = (1/4) * 9^n * Pochhammer(n, 4/9).
a(n) = (9*n-5)*a(n-1). (End)
From Amiram Eldar, Dec 21 2022: (Start)
a(n) = A144829(n)/4.
Sum_{n>=1} 1/a(n) = 4*(e/9^5)^(1/9)*(Gamma(4/9) - Gamma(4/9, 1/9)). (End)

A257619 Triangle read by rows: T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = 9*x + 2.

Original entry on oeis.org

1, 2, 2, 4, 44, 4, 8, 564, 564, 8, 16, 6436, 22560, 6436, 16, 32, 71404, 637844, 637844, 71404, 32, 64, 786948, 15470232, 36994952, 15470232, 786948, 64, 128, 8660012, 346391196, 1660722424, 1660722424, 346391196, 8660012, 128
Offset: 0

Views

Author

Dale Gerdemann, May 09 2015

Keywords

Examples

			Triangle begins as:
    1;
    2,       2;
    4,      44,         4;
    8,     564,       564,          8;
   16,    6436,     22560,       6436,         16;
   32,   71404,    637844,     637844,      71404,        32;
   64,  786948,  15470232,   36994952,   15470232,    786948,      64;
  128, 8660012, 346391196, 1660722424, 1660722424, 346391196, 8660012, 128;
		

Crossrefs

Cf. A000079, A144829 (row sums), A257608.
Similar sequences listed in A256890.

Programs

  • Mathematica
    T[n_, k_, a_, b_]:= T[n, k, a, b]= If[k<0 || k>n, 0, If[n==0, 1, (a*(n-k)+b)*T[n-1, k-1, a, b] + (a*k+b)*T[n-1, k, a, b]]];
    Table[T[n,k,9,2], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 24 2022 *)
  • PARI
    f(x) = 9*x + 2;
    t(n, m) = if ((n<0) || (m<0), 0, if ((n==0) && (m==0), 1, f(m)*t(n-1,m) + f(n)*t(n,m-1)));
    tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n-k, k), ", "); ); print(); ); } \\ Michel Marcus, May 23 2015
    
  • Sage
    def T(n,k,a,b): # A257619
        if (k<0 or k>n): return 0
        elif (n==0): return 1
        else: return  (a*k+b)*T(n-1,k,a,b) + (a*(n-k)+b)*T(n-1,k-1,a,b)
    flatten([[T(n,k,9,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 24 2022

Formula

T(n,k) = t(n-k, k); t(0,0) = 1, t(n,m) = 0 if n < 0 or m < 0, else t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = 9*x + 2.
Sum_{k=0..n} T(n, k) = A144829(n).
From G. C. Greubel, Mar 24 2022: (Start)
T(n, k) = (a*k + b)*T(n-1, k) + (a*(n-k) + b)*T(n-1, k-1), with T(n, 0) = 1, a = 9, and b = 2.
T(n, n-k) = T(n, k).
T(n, 0) = A000079(n).
T(n, 1) = (1/9)*(4*11^n - 2^n*(9*n + 4)).
T(n, 2) = (1/81)*(26*20^n - 4*(4+9*n)*11^n - 2^(n-1)*(20 + 9*n - 81*n^2)). (End)

A020022 Nearest integer to Gamma(n + 4/9)/Gamma(4/9).

Original entry on oeis.org

1, 0, 1, 2, 5, 24, 131, 843, 6275, 52988, 500442, 5226842, 59818300, 744405515, 10008118595, 144561713037, 2232675345794, 36715105686391, 640474621418148, 11813198572823623, 229701083360459334, 4696111037591613052
Offset: 0

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Author

Keywords

Comments

Gamma(n + 4/9)/Gamma(4/9) = 1, 4/9, 52/81, 1144/729, 35464/6561, 1418560/59049, 69509440/531441, 4031547520/4782969, ... - R. J. Mathar, Sep 04 2016

Crossrefs

Programs

  • Magma
    [Round(Gamma(n +4/9)/Gamma(4/9)): n in [0..30]]; // G. C. Greubel, Feb 03 2018
  • Maple
    Digits := 64:f := proc(n,x) round(GAMMA(n+x)/GAMMA(x)); end;
  • Mathematica
    Table[Round[Gamma[n + 4/9]/Gamma[4/9]], {n, 0, 50}] (* G. C. Greubel, Feb 03 2018 *)
  • PARI
    for(n=0,30, print1(round(gamma(n+4/9)/gamma(4/9)), ", ")) \\ G. C. Greubel, Feb 03 2018
    
Showing 1-5 of 5 results.