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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.

A225471 Triangle read by rows, s_4(n, k) where s_m(n, k) are the Stirling-Frobenius cycle numbers of order m; n >= 0, k >= 0.

Original entry on oeis.org

1, 3, 1, 21, 10, 1, 231, 131, 21, 1, 3465, 2196, 446, 36, 1, 65835, 45189, 10670, 1130, 55, 1, 1514205, 1105182, 290599, 36660, 2395, 78, 1, 40883535, 31354119, 8951355, 1280419, 101325, 4501, 105, 1, 1267389585, 1012861224, 308846124, 48644344, 4421494, 240856, 7756, 136, 1
Offset: 0

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Author

Peter Luschny, May 17 2013

Keywords

Comments

The Stirling-Frobenius cycle numbers are defined in A225470.
Triangle T(n,k), read by rows, given by (3, 4, 7, 8, 11, 12, 15, 16, 19, 20, ... (A014601)) DELTA (1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, May 14 2015

Examples

			[n\k][    0,       1,      2,     3,    4,  5,  6 ]
[0]       1,
[1]       3,       1,
[2]      21,      10,      1,
[3]     231,     131,     21,     1,
[4]    3465,    2196,    446,    36,    1,
[5]   65835,   45189,  10670,  1130,   55,  1,
[6] 1514205, 1105182, 290599, 36660, 2395, 78,  1.
...
From _Wolfdieter Lang_, Aug 11 2017: (Start)
Recurrence: T(4, 2) = T(3, 1) + (4*4 - 1)*T(3, 2) = 131 +15*21 = 446.
Boas-Buck recurrence for column k=2 and n=4: T(4, 2) = (4!/2)*(4*(3+8*(5/12)) *T(2, 2)/2! + 1*(3 + 8*(1/2))*T(3,2)/3!) = (4!/2)*(4*(19/3)/2  + 7*21/3!) =  446.
(End)
		

Crossrefs

Columns k=0..3 give A008545, A286723(n-1), A383702, A383703.
Cf. A132393 (m=1), A028338 (m=2), A225470 (m=3).

Programs

  • Mathematica
    T[0, 0] = 1; T[n_, k_] := Sum[Binomial[n - j, k]*Abs[StirlingS1[n, n - j]]* 3^(n - k - j)*4^j, {j, 0, n - k}];
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 30 2018, after Wolfdieter Lang *)
  • Sage
    @CachedFunction
    def SF_C(n, k, m):
        if k > n or k < 0 : return 0
        if n == 0 and k == 0: return 1
        return SF_C(n-1, k-1, m) + (m*n-1)*SF_C(n-1, k, m)
    for n in (0..8): [SF_C(n, k, 4) for k in (0..n)]

Formula

For a recurrence see the Sage program.
T(n, 0) ~ A008545; T(n, n) ~ A000012; T(n, n-1) = A014105.
Row sums ~ A047053; alternating row sums ~ A001813.
From Wolfdieter Lang, May 29 2017: (Start)
This is the Sheffer triangle (1/(1 - 4*x)^{-3/4}, -(1/4)*log(1-4*x)). See the P. Bala link where this is called exponential Riordan array, and the signed version is denoted by s_{(4,0,3)}.
E.g.f. of row polynomials in the variable x (i.e., of the triangle): (1 - 4*z)^{-(3+x)/4}.
E.g.f. of column k: (1-4*x)^(-3/4)*(-(1/4)*log(1-4*x))^k/k!, k >= 0.
Recurrence for row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k: R(n, x) = (x+3)*R(n-1,x+4), with R(0, x) = 1.
R(n, x) = risefac(4,3;x,n) := Product_{j=0..(n-1)} (x + (3 + 4*j)). (See the P. Bala link, eq. (16) for the signed s_{4,0,3} row polynomials.)
T(n, k) = Sum_{j=0..(n-m)} binomial(n-j, k)* S1p(n, n-j)*3^(n-k-j)*4^j, with S1p(n, m) = A132393(n, m).
T(n, k) = sigma[4,3]^{(n)}_{n-k}, with the elementary symmetric functions sigma[4,3]^{(n)}_m of degree m in the n numbers 3, 7, 11, ..., 3+4*(n-1), with sigma[4,3]^{(n)}_0 := 1. (End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!/(n - k)) * Sum_{p=k..n-1} 4^(n-1-p)*(3 + 8*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), beginning with {1/2, 5/12, 3/8, 251/720, ...}. See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017