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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. 7
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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
LINKS
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
EXAMPLE
[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)
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 *)
PROG
(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)]
CROSSREFS
Cf. A132393 (m=1), A028338 (m=2), A225470 (m=3).
Sequence in context: A107717 A143173 A000369 * A136236 A113090 A223549
KEYWORD
nonn,easy,tabl
AUTHOR
Peter Luschny, May 17 2013
STATUS
approved

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Last modified April 19 07:22 EDT 2024. Contains 371782 sequences. (Running on oeis4.)