Search: a111959 -id:a111959
|
| |
|
|
A046521
|
|
Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.
|
|
+10
39
|
|
|
|
1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.
As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005
The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007
The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):
(E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).
This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).
For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)
The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....
2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).
Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)
|
|
|
REFERENCES
|
Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
|
|
|
LINKS
|
|
|
|
FORMULA
|
T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1.
As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555.
The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of formal power series then
G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
The m-th power of this array has entries m^(n-k)*T(n,k). (End)
|
|
|
EXAMPLE
|
Array begins:
1, 2, 6, 20, 70, ...
1, 6, 30, 140, 630, ...
1, 10, 70, 420, 2310, ...
1, 14, 126, 924, 6006, ...
Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
As a number triangle, T(n, m) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 2 1
2: 6 6 1
3: 20 30 10 1
4: 70 140 70 14 1
5: 252 630 420 126 18 1
6: 924 2772 2310 924 198 22 1
7: 3432 12012 12012 6006 1716 286 26 1
8: 12870 51480 60060 36036 12870 2860 390 30 1
9: 48620 218790 291720 204204 87516 24310 4420 510 34 1
10: 184756 923780 1385670 1108536 554268 184756 41990 6460 646 38 1
Production matrix begins
2, 1,
2, 4, 1,
-4, 0, 4, 1,
10, 0, 0, 4, 1,
-28, 0, 0, 0, 4, 1,
84, 0, 0, 0, 0, 4, 1,
-264, 0, 0, 0, 0, 0, 4, 1,
858, 0, 0, 0, 0, 0, 0, 4, 1,
-2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - Wolfdieter Lang, Aug 10 2017
With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
-x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ).
x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End)
|
|
|
MATHEMATICA
|
t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *)
|
|
|
PROG
|
(PARI) T(i, j)=if(i<0 || j<0, 0, (2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)
(GAP) Flat(List([0..9], n->List([0..n], m->Binomial(2*n, n)*Binomial(n, m)/Binomial(2*m, m)))); # Muniru A Asiru, Jul 19 2018
(Magma) [Binomial(n+1, k+1)*Catalan(n)/Catalan(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 28 2024
(SageMath)
def A046521(n, k): return binomial(n+1, k+1)*catalan_number(n)/catalan_number(k)
|
|
|
CROSSREFS
|
Columns include: A000984 (m=0), A002457 (m=1), A002802 (m=2), A020918 (m=3), A020920 (m=4), A020922 (m=5), A020924 (m=6), A020926 (m=7), A020928 (m=8), A020930 (m=9), A020932 (m=10).
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A012150
|
|
Expansion of e.g.f. exp(tan(arcsin(x))).
|
|
+10
5
|
|
|
|
1, 1, 1, 4, 13, 76, 421, 3256, 25369, 245008, 2449801, 28441216, 346065061, 4700478784, 67243537453, 1047088053376, 17192488230961, 302112622479616, 5593309059948049, 109527844826856448, 2255588021494237501
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,4
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = n!*Sum_{k=1..n} A111959(n-1,k-1)*2^(k-n)/k!.
a(n) = n!*Sum_{k=1..n} (1+(-1)^(n-k))*C((n-2)/2,(n-k)/2)/(2*k!), n>0.
E.g.f.: exp(x/sqrt(1-x^2)). (End)
E.g.f.: S(x) = exp(x/sqrt(1-x^2)) = 1 + 2*(x/sqrt(1-x^2))/(G(0) - x/sqrt(1-x^2)), G(k) = 8*k + 2 + (x^2)/((1-x^2)*(8*k+6) + x^2/G(k+1)); (continued fraction). - Sergei N. Gladkovskii, Dec 16 2011
a(n) = (3*n^2 - 12*n + 13)*a(n-2) - 3*(n-4)*(n-3)^2*(n-2)*a(n-4) + (n-6)*(n-5)*(n-4)^2*(n-3)*(n-2)*a(n-6). - Vaclav Kotesovec, Nov 08 2013
a(n) ~ n^(n-1/3) * exp(3/2*n^(1/3)-n) / sqrt(3) * (1 - 19/(36*n^(1/3)) + 553/(2592*n^(2/3))). - Vaclav Kotesovec, Nov 08 2013
a(n) = n! * Sum_{k=0..floor(n/2)} binomial(n/2-1,k)/(n-2*k)!. - Seiichi Manyama, Jun 08 2024
|
|
|
EXAMPLE
|
exp(tan(arcsin(x))) = 1+x+1/2!*x^2+4/3!*x^3+13/4!*x^4+76/5!*x^5...
|
|
|
MAPLE
|
A012150 := proc(n) if n = 0 then 1; else add( (1+(-1)^(n-k)) *binomial((n-2)/2, (n-k)/2)/(2*k!), k=1..n) ; %*n! ; end if; end proc: # R. J. Mathar, Mar 20 2011
|
|
|
MATHEMATICA
|
Range[0, 20]! CoefficientList[Series[Exp[Tan[ArcSin[x]]], {x, 0, 20}], x] (* Or *)
f[n_] := n! Sum[(1 + (-1)^(n - k)) Binomial[(n - 2)/2, (n - k)/2]/2/k!, {k, n}]; f[0] = 1; Array[f, 21, 0] (* Robert G. Wilson v, Feb 19 2011 *)
|
|
|
PROG
|
(PARI) my(x='x+O('x^30)); Vec(serlaplace(exp(tan(asin(x))))) \\ Michel Marcus, Oct 30 2022
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
Patrick Demichel (patrick.demichel(AT)hp.com)
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
| |
|
|
|
1, 1, 1, 3, 2, 1, 7, 7, 3, 1, 19, 20, 12, 4, 1, 51, 61, 40, 18, 5, 1, 141, 182, 135, 68, 25, 6, 1, 393, 547, 441, 251, 105, 33, 7, 1, 1107, 1640, 1428, 888, 420, 152, 42, 8, 1, 3139, 4921, 4572, 3076, 1596, 654, 210, 52, 9, 1, 8953, 14762, 14535, 10456, 5880, 2652, 966, 280, 63, 10, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,4
|
|
|
COMMENTS
|
|
|
|
LINKS
|
|
|
|
FORMULA
|
Factors as (1/(1-x), x/(1-x))*(1/sqrt(1-4x^2), x/sqrt(1-4x^2)).
Equals ((1-x^2)/(1+x+x^2),x/(1+x+x^2))^{-1}*(1,x/(1-x^2))=A094531*(1,x/(1-x^2)).
Riordan array (1/sqrt(1-2x-3x^2), x/sqrt(1-2x-3x^2));
T(n,k) = Sum_{j=0..n} C(n,j)*C((j-1)/2,(j-k)/2)*2^(j-k)*(1+(-1)^(j-k))/2.
G.f.: 1/(1-xy-x-2x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). (End)
|
|
|
EXAMPLE
|
Triangle T(n,k) begins:
1;
1, 1;
3, 2, 1;
7, 7, 3, 1;
19, 20, 12, 4, 1;
51, 61, 40, 18, 5, 1;
...
Production matrix is
1, 1,
2, 1, 1,
0, 2, 1, 1,
-2, 0, 2, 1, 1,
0, -2, 0, 2, 1, 1,
4, 0, -2, 0, 2, 1, 1. (End)
|
|
|
MAPLE
|
# Uses function PMatrix from A357368. Adds a row and column above and to the left.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A277604
|
|
Array of coefficients T(k,n) of the formal power series A(k,x) read by upwards antidiagonals, where A(k,x) = sqrt(1 + 2*x*A(k,x) + (4*k+1)*x^2*(A(k,x))^2), k >= 0.
|
|
+10
0
|
|
|
|
1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 5, 1, 1, 1, 7, 9, 13, 1, 1, 1, 9, 13, 37, 25, 1, 1, 1, 11, 17, 73, 81, 61, 1, 1, 1, 13, 21, 121, 169, 301, 125, 1, 1, 1, 15, 25, 181, 289, 841, 729, 295, 1, 1, 1, 17, 29, 253, 441, 1801, 2197, 2549, 625, 1, 1, 1, 19, 33, 337, 625, 3301, 4913, 10123, 6561, 1447, 1
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,9
|
|
|
COMMENTS
|
It will be interesting using the formulae for k < 0 (attention: signed terms!). Especially for k = -1 see A157674.
If G is the g.f. of central binomial coefficients (see A000984) and B(k,x) = G(k*x^2), then B(k,x) = A(k,x)/(1+x*A(k,x)) and A(k,x) = B(k,x) / (1-x*B(k,x)) for k >= 0. - Werner Schulte, Aug 07 2017
|
|
|
LINKS
|
|
|
|
FORMULA
|
A(k,x) = (x + sqrt(1 - 4*k*x^2))/(1 - (4*k+1)*x^2) for k >= 0.
T(k,0) = 1 and T(k,2*n+2) = (4*k+1)^(n+1)-2*(Sum_{i=0..n} A000108(i)*k^(i+1)* (4*k+1)^(n-i)), and T(k,2*n+1) = (4*k+1)^n for k >= 0 and n >= 0.
A(k,x) = 1/(1 - x - 2*k*x^2*C(k*x^2)), k >= 0, where C is the g.f. of A000108.
Conjecture: If B(k,n) satisfy B(k,0) = B(k,1) = 1 and B(k,n+2) = B(k,n+1) + k*B(k,n) for k >= 0 and n >= 0 (generalized Fibonacci numbers, see A015441) and G(k,x) = Sum_{n>=0} A000108(n)*B(k,n)*x^n for k >= 0, then you will have (1): A(k,x*G(k,x)) = G(k,x) and (2): G(k,x/A(k,x)) = A(k,x) for k >= 0. Especially for k = 1 see A098615 and for k = 2 see A200376.
Recurrence: T(k,2*n+2) = (4*k+1)*T(k,2*n)-2*k^(n+1)*A000108(n) with initial value T(k,0) = 1 for k >= 0 and n >= 0. - Werner Schulte, Aug 09 2017
|
|
|
EXAMPLE
|
The terms define the array T(k,n) for k >= 0 and n >= 0, i.e.,
k\n 0 1 2 3 4 5 6 7 8 9 . . .
0: 1 1 1 1 1 1 1 1 1 1 . . .
1: 1 1 3 5 13 25 61 125 295 625 . . .
2: 1 1 5 9 37 81 301 729 2549 6561 . . .
3: 1 1 7 13 73 169 841 2197 10123 28561 . . .
4: 1 1 9 17 121 289 1801 4913 28057 83521 . . .
5: 1 1 11 21 181 441 3301 9261 63071 194481 . . .
6: 1 1 13 25 253 625 5461 15625 123565 390625 . . .
7: 1 1 15 29 337 841 8401 24389 219619 707281 . . .
8: 1 1 17 33 433 1089 12241 35937 362993 1185921 . . .
9: 1 1 19 37 541 1369 17101 50653 567127 1874161 . . .
etc.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
Search completed in 0.013 seconds
|
