A118980
Triangle read by rows: rows = inverse binomial transforms of columns of A309220.
Original entry on oeis.org
1, 2, 1, 6, 5, 2, 14, 22, 18, 6, 34, 85, 118, 84, 24, 82, 311, 660, 780, 480, 120, 198, 1100, 3380, 5964, 6024, 3240, 720, 478, 3809, 16380, 40740, 60480, 52920, 25200, 5040, 1154, 13005, 76518, 258804, 531864, 676080, 519840, 221760, 40320, 2786, 43978, 348462, 1564314, 4286880, 7444800, 8240400
Offset: 1
First few rows of the triangle:
1;
2, 1;
6, 5, 2;
14, 22, 18, 6;
34, 85, 118, 84, 24;
82, 311, 660, 780, 480, 120;
...
Column 3 of A309220 = (6, 11, 18, 27, 38, 51, ...), whose inverse binomial transform is (6, 5, 2).
-
with(transforms);
M := 12;
T := [1];
S := series((1 + x^2)/(1-x-x^2 + x*y), x, 120):
for n from 1 to M do
R2 := expand(coeff(S, x, n));
R3 := [seq(abs(coeff(R2,y,n-i)),i=0..n)];
f := k-> add( R3[i]*k^(n-i+1), i=1..nops(R3) ):
s1 := [seq(f(i),i=1..3*n)];
s2 := BINOMIALi(s1);
s3 := [seq(s2[i],i=1..n+1)];
T := [op(T), op(s3)];
od:
T; # N. J. A. Sloane, Aug 12 2019
A185263
Triangle T(n,k) read by rows: coefficients (in compressed forms) in order of decreasing exponents of polynomials p_n(t) related to Hultman numbers.
Original entry on oeis.org
1, 1, 1, 1, 1, 5, 1, 15, 8, 1, 35, 84, 1, 70, 469, 180, 1, 126, 1869, 3044, 1, 210, 5985, 26060, 8064, 1, 330, 16401, 152900, 193248, 1, 495, 39963, 696905, 2286636, 604800, 1, 715, 88803, 2641925, 18128396, 19056960, 1, 1001, 183183, 8691683, 109425316, 292271616, 68428800, 1, 1365, 355355, 25537655, 539651112, 2961802480, 2699672832
Offset: 0
Triangle begins:
n\k| 0 1 2 3 4 5 6
-----+---------------------------------------------------
0 | 1
1 | 1
2 | 1 1
3 | 1 5
4 | 1 15 8
5 | 1 35 84
6 | 1 70 469 180
7 | 1 126 1869 3044
8 | 1 210 5985 26060 8064
9 | 1 330 16401 152900 193248
10 | 1 495 39963 696905 2286636 604800
11 | 1 715 88803 2641925 18128396 19056960
12 | 1 1001 183183 8691683 109425316 292271616 68428800
...
Polynomials p_n(t):
p_0 = t;
p_1 = t^2;
p_2 = t^3 + t;
p_3 = t^4 + 5*t^2;
p_4 = t^5 + 15*t^3 + 8*t;
p_5 = t^6 + 35*t^4 + 84*t^2;
p_6 = t^7 + 70*t^5 + 469*t^3 + 180*t;
p_7 = t^8 + 126*t^6 + 1869*t^4 + 3044*t^2;
...
A(x;t) = t + t^2*x/1! + (t^3 + t)*x^2/2! + (t^4 + 5*t^2)*x^3/3! + ...
For uncompressed form of polynomial coefficients, in order of increasing powers, see
A164652.
-
T[n_, k_] := Abs[StirlingS1[n+2, n-2k+1]]/Binomial[n+2, 2];
Table[T[n, k], {n, 0, 13}, {k, 0, n/2}] // Flatten (* Jean-François Alcover, Aug 12 2018 *)
-
seq(N) = {
my(p=vector(N), t='t, v); p[1] = t^2; p[2] = t^3 + t;
for (n=3, N,
p[n] = ((2*n+1)*t*p[n-1] + (n-1)*(n^2-t^2)*p[n-2])/(n+2));
v = vector(#p, n, vector(1+n\2, k, polcoeff(p[n], n+1-2*(k-1))));
concat([[1]], v);
};
concat(seq(13))
-
N=14; x='x+O('x^(N+1));
concat(apply(p->select(a->a!=0, Vec(p)), Vec(serlaplace(((1-x)^(-t) - (1+x)^t)/x^2))))
-
T(n,k) = -stirling(n+2, n+1-2*k, 1)/binomial(n+2,2);
concat(1, concat(vector(13, n, vector(1+n\2, k, T(n, k-1)))))
\\ Gheorghe Coserea, Jan 29 2018
A214178
Triangle T(n,k) by rows: the k-th derivative of the Fibonacci Polynomial F_n(x) evaluated at x=1.
Original entry on oeis.org
0, 1, 0, 1, 1, 0, 2, 2, 2, 0, 3, 5, 6, 6, 0, 5, 10, 18, 24, 24, 0, 8, 20, 44, 84, 120, 120, 0, 13, 38, 102, 240, 480, 720, 720, 0, 21, 71, 222, 630, 1560, 3240, 5040, 5040, 0, 34, 130, 466, 1536, 4560, 11760, 25200, 40320, 40320, 0, 55, 235, 948, 3564, 12264
Offset: 0
The triangle begins:
. 0: [0]
. 1: [1, 0]
. 2: [1, 1, 0]
. 3: [2, 2, 2, 0]
. 4: [3, 5, 6, 6, 0]
. 5: [5, 10, 18, 24, 24, 0]
. 6: [8, 20, 44, 84, 120, 120, 0]
. 7: [13, 38, 102, 240, 480, 720, 720, 0]
. 8: [21, 71, 222, 630, 1560, 3240, 5040, 5040, 0]
. 9: [34, 130, 466, 1536, 4560, 11760, 25200, 40320, 40320, 0]
. 10: [55, 235, 948, 3564, 12264, 37800, 100800, 221760, 362880, 362880, 0]
...
-
a214178 n k = a214178_tabl !! n !! k
a214178_row n = a214178_tabl !! n
a214178_tabl = [0] : map f a037027_tabl where
f row = (zipWith (*) a000142_list row) ++ [0]
-
T[n_, k_] := D[Fibonacci[n, x], {x, k}] /. x -> 1;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 20 2021 *)
A230056
G.f.: Sum_{n>=0} (n+3)^n * x^n / (1 + (n+3)*x)^n.
Original entry on oeis.org
1, 4, 9, 30, 132, 720, 4680, 35280, 302400, 2903040, 30844800, 359251200, 4550515200, 62270208000, 915372057600, 14384418048000, 240612083712000, 4268249137152000, 80029671321600000, 1581386305314816000, 32844177110384640000, 715273190403932160000, 16298010552775311360000
Offset: 0
O.g.f.: A(x) = 1 + 4*x + 9*x^2 + 30*x^3 + 132*x^4 + 720*x^5 + 4680*x^6 +...
where
A(x) = 1 + 4*x/(1+4*x) + 5^2*x^2/(1+5*x)^2 + 6^3*x^3/(1+6*x)^3 + 7^4*x^4/(1+7*x)^4 + 8^5*x^5/(1+8*x)^5 +...
E.g.f.: E(x) = 1 + 4*x + 9*x^2/2! + 30*x^3/3! + 132*x^4/4! + 720*x^5/5! +...
where
E(x) = 1 + 4*x + 9/2*x^2 + 5*x^3 + 11/2*x^4 + 6*x^5 + 13/2*x^6 + 7*x^7 +...
which is the expansion of: (2 + 4*x - 5*x^2) / (2 - 4*x + 2*x^2).
-
a:=series(add((n+3)^n*x^n/(1+(n+3)*x)^n,n=0..100),x=0,23): seq(coeff(a,x,n),n=0..22); # Paolo P. Lava, Mar 27 2019
-
a[n_] := (n + 7)*n!/2; a[0] = 1; Array[a, 25, 0] (* Amiram Eldar, Dec 11 2022 *)
-
{a(n)=polcoeff( sum(m=0, n, ((m+3)*x)^m / (1 + (m+3)*x +x*O(x^n))^m), n)}
for(n=0, 20, print1(a(n), ", "))
-
{a(n)=if(n==0, 1, (n+7) * n!/2 )}
for(n=0, 20, print1(a(n), ", "))
Comments