Displaying 1-10 of 25 results found.
Logarithmic derivative of A112934 such that a(n)=(1/2)* A112934(n+1) for n>0, where A112934 equals the INVERT transform of double factorials A001147.
+20
9
1, 3, 13, 79, 641, 6579, 81677, 1187039, 19728193, 368562723, 7639512013, 173893382575, 4310656806977, 115569893763411, 3331588687405133, 102751933334045375, 3375782951798785921, 117693183724386637635
FORMULA
G.f.: log(1 + x + 2*x*[Sum_{n>=1} a(n)*x^n]) = Sum_{k>=1} a(n)/n*x^n.
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k-1)/(1 - x*(2*k+4)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013 G.f.: 1/(x*G(0)) - 1/(2*x), where G(k)= 1 + 1/(1 - 2*x*(2*k+2)/(2*x*(2*k+2) - 1 + 2*x*(2*k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
EXAMPLE
log(1+x + 2*x*[x + 3*x^2 + 13*x^3 + 79*x^4 + 641*x^5 +...])
= x + 3/2*x^2 + 13/3*x^3 + 79/4*x^4 + 641/5*x^5 +...
PROG
(PARI) {a(n)=local(F=1+x+x*O(x^n)); for(i=1, n, F=1+x+2*x^2*deriv(F)/F); return(n*polcoeff(log(F), n, x))}
Square table, read by antidiagonals: the g.f. for row n is given recursively by (2*n-1)*x*R(n,x) = 1 + (2*n-3)*x - 1/R(n-1,x) for n >= 1 with the initial value R(0,x) = Sum_{k >= 0} A112934(k+1)*x^k.
+20
9
1, 1, 2, 1, 2, 6, 1, 2, 10, 26, 1, 2, 14, 74, 158, 1, 2, 18, 138, 706, 1282, 1, 2, 22, 218, 1686, 8162, 13158, 1, 2, 26, 314, 3194, 24162, 110410, 163354, 1, 2, 30, 426, 5326, 53890, 394254, 1708394, 2374078, 1, 2, 34, 554, 8178, 102722, 1019250, 7191018, 29752066, 39456386
COMMENTS
Compare with A111528, which has a similar definition.
FORMULA
Let d(n) = Product_{k = 1..n} 2*k-1 = A001147(n) denote the double factorial of odd numbers. O.g.f. for row n: R(n,x) = ( Sum_{k >= 0} d(n+k)/d(n)*x^k )/( Sum_{k >= 0} d(n-1+k)/d(n-1)*x^k ).
R(n,x)/(1 - (2*n-1)*x*R(n,x)) = Sum_{k >= 0} d(n+k)/d(n)*x^k.
R(n,x) = 1/(1 + (2*n-1)*x - (2*n+1)*x/(1 + (2*n+1)*x - (2*n+3)*x/(1 + (2*n+3)*x - (2*n+5)*x/(1 + (2*n+5)*x - ... )))).
R(n,x) satisfies the Riccati differential equation 2*x^2*d/dx(R(n,x)) + (2*n-1)*x*R(n,x)^2 - (1 + (2*n-3)*x)*R(n,x) + 1 = 0 with R(n,0) = 1.
Applying Stokes 1982 gives A(x) = 1/(1 - 2*x/(1 - (2*n+1)*x/(1 - 4*x/(1 - (2*n+3)*x/(1 - 6*x/(1 - (2*n+5)*x/(1 - ... - 2*m*x/(1 - (2*n+2*m-1)*x/(1 - ... ))))))))), a continued fraction of Stieltjes type.
EXAMPLE
Square array begins
1, 2, 6, 26, 158, 1282, 13158, 163354, 2374078, 39456386, ...
1, 2, 10, 74, 706, 8162, 110410, 1708394, 29752066, 576037442, ...
1, 2, 14, 138, 1686, 24162, 394254, 7191018, 144786006, 3188449602, ...
1, 2, 18, 218, 3194, 53890, 1019250, 21256090, 483426010, 11895873410, ...
1, 2, 22, 314, 5326, 102722, 2197558, 51355514, 1297759918, 35208930050, ...
1, 2, 26, 426, 8178, 176802, 4206618, 108577674, 3011332338, 89141101506, ...
1, 2, 30, 554, 11846, 283042, 7396830, 208569034, 6288011206, 201404591042, ...
...
MAPLE
T := (n, k) -> coeff(series(hypergeom([n+1/2, 1], [], 2*x)/ hypergeom([n-1/2, 1], [], 2*x), x, 21), x, k):
# display as a sequence
seq(seq(T(n-k, k), k = 0..n), n = 0..10);
# display as a square array
seq(print(seq(T(n, k), k = 0..10)), n = 0..10);
Expansion of (1 - x + x^2) / (1 - x)^2 in powers of x.
+10
93
1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71
COMMENTS
1 followed by the natural numbers.
Molien series for ring of Hamming weight enumerators of self-dual codes (with respect to Euclidean inner product) of length n over GF(4).
The right-shifted sequence (with a(0)=0) is an autosequence (of the first kind - see definition in links). - Jean-François Alcover, Mar 14 2017
LINKS
E. M. Rains and N. J. A. Sloane, Self-dual codes, pp. 177-294 of Handbook of Coding Theory, Elsevier, 1998 ( Abstract, pdf, ps).
FORMULA
G.f.: (1 - x + x^2) / (1 - x)^2 = (1 - x^6) /((1 - x) * (1 - x^2) * (1 - x^3)) = (1 + x^3) / ((1 - x) * (1 - x^2)). a(0) = 1, a(n) = n if n>0.
Euler transform of length 6 sequence [ 1, 1, 1, 0, 0, -1]. - Michael Somos Jul 30 2006 G.f.: 1 / (1 - x / (1 - x / (1 + x / (1 - x)))). - Michael Somos, Apr 05 2012 E.g.f.: 1-x + x*E(0), where E(k) = 2 + x/(2*k+1 - x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
EXAMPLE
G.f. = 1 + x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 + 6*x^6 + 7*x^7 + 8*x^8 + 9*x^9 + ...
MAPLE
a:= n-> `if`(n=0, 1, n):
seq(a(n), n=0..60);
MATHEMATICA
Denominator@ CoefficientList[Series[Log[1+x], {x, 0, 75}], x] (* or *)
CoefficientList[ Series[(1 -x +x^2)/(1-x)^2, {x, 0, 75}], x] (* Robert G. Wilson v, Aug 14 2015 *) LinearRecurrence[{2, -1}, {1, 1, 2}, 80] (* Harvey P. Dale, Jan 29 2025 *)
PROG
(PARI) {a(n) = (n==0) + max(n, 0)} /* Michael Somos, Feb 02 2004 */ (Haskell)
a028310 n = 0 ^ n + n
(Python)
(Magma) [n eq 0 select 1 else n: n in [0..75]]; // G. C. Greubel, Jan 05 2024 (SageMath) [n + int(n==0) for n in range(76)] # G. C. Greubel, Jan 05 2024
Double-superfactorials: a(n) = Product_{k=1..n} (2k)!.
+10
19
1, 2, 48, 34560, 1393459200, 5056584744960000, 2422112183371431936000000, 211155601241022491077587763200000000, 4417964278440225627098723475313498521600000000000
FORMULA
a(n) = Product_{k=0..n} (2*(k+1)*(2*k+1))^(n-k). - Paul Barry, Jan 28 2008 G.f.: G(0)/(2*x)-1/x, where G(k)= 1 + 1/(1 - 1/(1 + 1/(2*k+2)!/x/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013 a(n) ~ 2^(n^2+2*n+17/24) * n^(n^2+3*n/2+11/24) * Pi^((n+1)/2) / (A^(1/2) * exp(3*n^2/2+3*n/2-1/24)), where A = 1.2824271291... is the Glaisher-Kinkelin constant (see A074962). - Vaclav Kotesovec, Nov 13 2014 a(n) = A^(3/2)*2^(n^2+n-1/24)*Pi^(-n/2-1/4)*G(n+3/2)*G(n+2)/exp(1/8), where G(n) is the Barnes G-function and A is the Glaisher-Kinkelin constant. - Ilya Gutkovskiy, Dec 11 2016 For n > 0, a(n) = 2^((n+1)/2) * n * sqrt(BarnesG(2*n)*Gamma(n)) * Gamma(2*n). - Vaclav Kotesovec, Nov 27 2024
MATHEMATICA
Table[Product[(2k)!, {k, 1, n}], {n, 0, 10}] (* Vaclav Kotesovec, Nov 13 2014 *)
PROG
(PARI) a(n) = prod(k=1, n, (2*k)!); \\ Michel Marcus, Dec 11 2016 (Magma) [&*[ Factorial(2*k): k in [0..n] ]: n in [0..10]]; // Vincenzo Librandi, Dec 11 2016 (Python)
from math import prod
def A098694(n): return prod(((k+1)*((k<<1)+1)<<1)**(n-k) for k in range(1, n+1))<<n # Chai Wah Wu, Nov 26 2023
INVERT transform (with offset) of triple factorials ( A008544), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^3]/A(x)^3.
+10
16
1, 1, 3, 15, 111, 1131, 14943, 243915, 4742391, 106912131, 2739347103, 78569371275, 2492748594471, 86650852740531, 3274367635513263, 133625238021647835, 5856377114106629751, 274320168321004350531
FORMULA
G.f. satisfies: A(x) = 1+x + 3*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +3*x^2/(1-5*x -3*2*2*x^2/(1-11*x -3*3*5*x^2/(1-17*x -3*4*8*x^2/(1-23*x -... -3*n*(3*n-4)*x^2/(1-(6*n-1)*x -...)))) (continued fraction).
G.f.: A(x) = 1/(1-x/(1 -2*x/(1-3*x/(1 -5*x/(1-6*x/(1 -8*x/(1-9*x/(1 -...)))))))) (continued fraction).
a(n) = (3*n - 2) * a(n-1) - Sum_{k=1..n-1} a(k) * a(n-k) if n>1. - Michael Somos, Jul 23 2011 G.f.: Q(0) where Q(k) = 1 - x*(3*k-1)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013 G.f.: 2/G(0)+4*x, where G(k)= 1 + 1/(1 - x*(3*k+3)/(x*(3*k+5) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013 G.f.: 2/G(0), where G(k)= 1 + 1/(1 - x*(3*k-1)/(x*(3*k-1) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013 Given g.f. A(x), then y = x * A(x^3) satisfies y^2 = x*y + x^5*y'. - Michael Somos, Oct 17 2016
EXAMPLE
A(x) = 1 + x + 3*x^2 + 15*x^3 + 111*x^4 + 1131*x^5 + 14943*x^6 +...
1/A(x) = 1 - x - 2*x^2 - 10*x^3 - 80*x^4 - 880*x^5 -...- A008544(n)*x^(n+1)-...
MATHEMATICA
a = ConstantArray[0, 20]; a[[1]]=1; Do[a[[n]] = (3*n-2)*a[[n-1]] - Sum[a[[k]]*a[[n-k]], {k, 1, n-1}], {n, 2, 20}]; Flatten[{1, a}] (* Vaclav Kotesovec after Michael Somos, Feb 22 2014 *) CoefficientList[Series[1/(1+(1/3*ExpIntegralE[2/3, -1/(3*x)])/E^(1/(3*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)
PROG
(PARI) {a(n)=local(F=1+x+x*O(x^n)); for(i=1, n, F=1+x+3*x^2*deriv(F)/F); return(polcoeff(F, n, x))}
(PARI) {a(n) = my(A); if( n<1, n==0, A = vector(n, k, 1); for(k=2, n, A[k] = (3*k - 2)*A[k-1] - sum(j=1, k-1, A[j] * A[k-j])); A[n])}; /* Michael Somos, Jul 23 2011 */ (PARI) {a(n) = if( n<1, n==0, polcoeff( 1 / sum(k=0, n, x^k * prod(i=1, k, 3*i - 4), x * O(x^n)), n))}; /* Michael Somos, Oct 17 2016 */ (PARI) {a(n) = my(A); if( n<0, 0, A = O(x); for(k=0, n, A = (x + sqrt(x^2 + 4*x^5*A')) / 2); polcoeff(A, 3*n + 1))}; /* Michael Somos, Oct 17 2016 */ (PARI) {a(n) = my(A); if( n<1, n==0, A = x; for(k=1, n, A = truncate(A) + O(x^(3*k + 4)); A += A + x^4*A' - A^2/x); polcoeff(A, 3*n + 1))}; /* Michael Somos, Oct 17 2016 */
Triangle T(n,k), read by rows, given by [0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, ...] DELTA [1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is the operator defined in A084938.
+10
14
1, 0, 1, 0, 0, 2, 0, 0, 1, 4, 0, 0, 2, 5, 8, 0, 0, 6, 15, 17, 16, 0, 0, 24, 62, 68, 49, 32, 0, 0, 120, 322, 359, 243, 129, 64, 0, 0, 720, 2004, 2308, 1553, 756, 321, 128, 0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256, 0, 0, 40320, 119664
COMMENTS
Let R(m,n,k), 0<=k<=n, the Riordan array (1, x*g(x)) where g(x) is g.f. of the m-fold factorials . Then Sum_{k, 0<=k<=n} = R(m,n,k) = Sum_{k, 0<=k<=n} T(n,k)*m^(n-k).
For m = -1, R(-1,n,k) is A026729(n,k). For m = 0, R(0,n,k) is A097805(n,k). For m = 1, R(1,n,k) is A084938(n,k). For m = 2, R(2,n,k) is A111106(n,k).
FORMULA
Sum_{k, 0<=k<=n} (-1)^(n-k)*T(n, k) = A000045(n+1), Fibonacci numbers. Sum_{k, 0<=k<=n} T(n, k) = A051295(n). Sum_{k, 0<=k<=n} 2^(n-k)*T(n, k) = A112934(n). T(0, 0) = 1, T(n, n) = 2^(n-1).
G.f.: A(x, y) = 1/(1 - x*y*Sum_{j>=0} (y-1+j)!/(y-1)!*x^j ). - Paul D. Hanna, Oct 26 2005
EXAMPLE
Triangle begins:
.1;
.0, 1;
.0, 0, 2;
.0, 0, 1, 4;
.0, 0, 2, 5, 8;
.0, 0, 6, 15, 17, 16;
.0, 0, 24, 62, 68, 49, 32;
.0, 0, 120, 322, 359, 243, 129, 64;
.0, 0, 720, 2004, 2308, 1553, 756, 321, 128;
.0, 0, 5040, 14508, 17332, 11903, 5622, 2151, 769, 256;
.0, 0, 40320, 119664, 148232, 105048, 49840, 18066, 5756, 1793, 512;
....................................................................
At y=2: Sum_{k=0..n} 2^k*T(n,k) = A113327(n) where (1 + 2*x + 8*x^2 + 36*x^3 +...+ A113327(n)*x^n +..) = 1/(1 - 2/1!*x*(1! + 2!*x + 3!*x^2 + 4!*x^3 +..) ). At y=3: Sum_{k=0..n} 3^k*T(n,k) = A113328(n) where (1 + 3*x + 18*x^2 + 117*x^3 +...+ A113328(n)*x^n +..) = 1/(1 - 3/2!*x*(2! + 3!*x + 4!*x^2 + 5!*x^3 +..) ). At y=4: Sum_{k=0..n} 4^k*T(n,k) = A113329(n) where (1 + 4*x + 32*x^2 + 272*x^3 +...+ A113329(n)*x^n +..) = 1/(1 - 4/3!*x*(3! + 4!*x + 5!*x^2 + 6!*x^3 +..) ).
MATHEMATICA
T[n_, k_] := Module[{x = X + X*O[X]^n, y = Y + Y*O[Y]^k}, A = 1/(1 - x*y*Sum[x^j*Product[y + i, {i, 0, j - 1}], {j, 0, n}]); Coefficient[ Coefficient[A, X, n], Y, k]];
PROG
(PARI) {T(n, k)=local(x=X+X*O(X^n), y=Y+Y*O(Y^k)); A=1/(1-x*y*sum(j=0, n, x^j*prod(i=0, j-1, y+i))); return(polcoeff(polcoeff(A, n, X), k, Y))} (Hanna)
a(n) = Product_{k=0..n} ((2*k+2)*(2*k+3))^(n-k).
+10
13
1, 6, 720, 3628800, 1316818944000, 52563198423859200000, 327312129899898454671360000000, 428017682605583614976547335700480000000000, 152240508705590071980086429193304853792686080000000000000
COMMENTS
The sequence 1,1,6,720,... with general term Product_{k=0..n, ((2k+1)(2k+0^k))^(n-k)} is the Hankel transform of A112934. - Paul Barry, Dec 04 2009 a(n) is also the determinant of the n X n matrix M(i,j) = i^(2*j)*sinh(2*j*arccsch(i))/(2*sqrt(i^2+1)), with i and j from 1 to n, which is the same matrix generated by sequences of length n by the linear recurrences with kernel { 2*(k^2 + z), -k^4 }, and initial conditions { 1, 2*(k^2 + z) }, with k from 1 to n, and z = 2. Regardless of the value of z, for every n, the determinant of the n X n matrix of polynomials generated gives always a(n) as result. - Federico Provvedi, Feb 01 2021
FORMULA
G.f.: Q(0)/(2*x) -1/x, where Q(k) = 1 + 1/(1 -(2*k+1)!*x/((2*k+1)!*x + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 17 2013 a(n) ~ A^(-1/2) * 2^(n^2 + 3*n + 53/24) * exp((-3/2)*n^2 + (-5/2)*n + 1/24) * n^(n^2 + (5/2)*n + 35/24) * Pi^((n+1)/2), where A = A074962 is the Glaisher-Kinkelin constant. - Vladimir Reshetnikov, Sep 06 2016 For n > 0, a(n) = n * (2*n+1) * sqrt(BarnesG(2*n)) * Gamma(2*n)^2 / (sqrt(Gamma(n)) * 2^((n-3)/2)). - Vaclav Kotesovec, Nov 27 2024
EXAMPLE
From both formulas in the comment above and in particular with z=2 from the linear recurrences, the determinant of the 5 X 5 matrix: ( (1,6,35,204,1189), (1,12,128,1344,14080),(1,22,403,7084,123205), (1,36,1040,28224,749824), (1,54,2291,89964,3426181) ) = 1316818944000 = a(5).
For a generic z, the determinant doesn't change as shown in this example, where the determinant of the 3 X 3 square matrix:
( ( 1, 2*(z+1), (2*z + 1)*(2*z+3) ),
( 1, 2*(z+4), 4*(z+6)*(z+2) ),
( 1, 2*(z+9), (2*z + 9)(2*z + 27)) ) = 720 = a(3). (End)
MATHEMATICA
Table[2^(n^2 + 2*n + 23/24) Glaisher^(3/2) Pi^(-n/2 - 3/4) BarnesG[n + 2] BarnesG[n + 5/2]/E^(1/8), {n, 0, 10}] (* Vladimir Reshetnikov, Sep 06 2016 *) Table[Product[((2k+2)(2k+3))^(n-k), {k, 0, n}], {n, 0, 10}] (* Harvey P. Dale, Dec 26 2019 *) Table[Det@Table[LinearRecurrence[{2*k^2, -k^4}, {1, 2*k^2}, n], {k, 1, n}], {n, 1, 20}] (* Federico Provvedi, Feb 01 2021 *)
Det@Expand@Array[(#1^(2 #2))/(4 Sqrt[1 + #1^2])((Sqrt[1+1/#1^2]+1/#1)^(2 #2)-(Sqrt[1+1/#1^2]-1/#1)^(2 #2))&, {#, #}]&/@Range[20] (* Federico Provvedi, Apr 01 2021 *)
PROG
(Python)
from math import prod
def A168467(n): return prod(((m:=k+1<<1)*(m+1))**(n-k) for k in range(1, n+1))*3**n<<n # Chai Wah Wu, Nov 26 2023
INVERT transform (with offset) of quadruple factorials ( A008545), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^4]/A(x)^4.
+10
12
1, 1, 4, 28, 292, 4156, 75844, 1694812, 44835172, 1369657468, 47422855300, 1834403141788, 78377228106148, 3664969183404220, 186134931067171012, 10201887125268108508, 600142156513333537252, 37713563573426417361148
FORMULA
G.f. satisfies: A(x) = 1+x + 4*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x +4*x^2/(1-7*x -4*2*3*x^2/(1-15*x -4*3*7*x^2/(1-23*x -4*4*11*x^2/(1-31*x -... -4*n*(4*n-5)*x^2/(1-(8*n-1)*x -...)))) (continued fraction).
G.f.: A(x) = 1/(1-1*x/(1 -3*x/(1-4*x/(1 -7*x/(1-8*x/(1 -11*x/(1-12*x/(1 -...)))))))) (continued fraction).
G.f.: Q(0) where Q(k) = 1 - x*(4*k-1)/(1 - x*(4*k+4)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013 G.f.: 1 + 2*x/G(0), where G(k)= 1 + 1/(1 - 2*x*(4*k+4)/(2*x*(4*k+4) - 1 + 2*x*(4*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013 a(n) ~ (n-1)! * 4^(n-1) / (GAMMA(3/4) * n^(1/4)). - Vaclav Kotesovec, Feb 22 2014
EXAMPLE
A(x) = 1 + x + 4*x^2 + 28*x^3 + 292*x^4 + 4156*x^5 + ...
1/A(x) = 1 - x - 3*x^2 - 21*x^3 - 231*x^4 -... - A008545(n)*x^(n+1)-...
MATHEMATICA
CoefficientList[Series[1/(1 + 1/4*ExpIntegralE[3/4, -1/(4*x)]/E^(1/(4*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)
PROG
(PARI) {a(n)=local(F=1+x+x*O(x^n)); for(i=1, n, F=1+x+4*x^2*deriv(F)/F); return(polcoeff(F, n, x))}
INVERT transform (with offset) of quintuple factorials ( A008546), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^5]/A(x)^5.
+10
10
1, 1, 5, 45, 605, 11045, 257005, 7288245, 243870205, 9401560645, 410141056205, 19966451812245, 1072718714991005, 63033317759267045, 4020725747388170605, 276661592017425909045, 20424931173615717011005
FORMULA
G.f. satisfies: A(x) = 1+x + 5*x^2*[d/dx A(x)]/A(x) (log derivative). G.f.: A(x) = 1+x +5*x^2/(1-9*x -5*2*4*x^2/(1-19*x -5*3*9*x^2/(1-29*x -5*4*13*x^2/(1-39*x -... -5*n*(5*n-6)*x^2/(1-(10*n-1)*x -...)))) (continued fraction). G.f.: A(x) = 1/(1-1*x/(1 -4*x/(1-5*x/(1 -9*x/(1-10*x/(1 -14*x/(1-15*x/(1 -...)))))))) (continued fraction).
G.f.: 1 + x/( Q(0) - x ) where Q(k) = 1 - x*(5*k+4)/(1 - x*(5*k+5)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013 a(n) ~ (n-1)! * 5^(n-1) / (GAMMA(4/5) * n^(1/5)). - Vaclav Kotesovec, Feb 22 2014
EXAMPLE
A(x) = 1 + x + 5*x^2 + 45*x^3 + 605*x^4 + 11045*x^5 +...
1/A(x) = 1 - x - 4*x^2 - 36*x^3 - 504*x^4 -... - A008546(n)*x^(n+1) -...
MATHEMATICA
CoefficientList[Series[1/(1 + 1/5*ExpIntegralE[4/5, -1/(5*x)]/E^(1/(5*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)
PROG
(PARI) {a(n)=local(F=1+x+x*O(x^n)); for(i=1, n, F=1+x+5*x^2*deriv(F)/F); return(polcoeff(F, n, x))}
INVERT transform (with offset) of sextuple factorials ( A008543), where g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^6]/A(x)^6.
+10
10
1, 1, 6, 66, 1086, 24186, 684006, 23506626, 951191646, 44281107066, 2330310876486, 136747268000706, 8851092668419326, 626304664252772346, 48092138192079689766, 3982448437177141451586, 353746119265020213643806
COMMENTS
Generally, if g.f. satisfies: A(x) = 1 + x*[d/dx x*A(x)^p]/A(x)^p, then a(n) ~ (n-1)! * p^(n-1) / (Gamma((p-1)/p) * n^(1/p)). - Vaclav Kotesovec, Feb 22 2014
FORMULA
G.f. satisfies: A(x) = 1+x + 6*x^2*[d/dx A(x)]/A(x) (log derivative).
G.f.: A(x) = 1+x+6*x^2/(1-11*x-6*2*5*x^2/(1-23*x-6*3*11*x^2/(1-35*x -6*4*17*x^2/(1-47*x- ... -6*n*(6*n-7)*x^2/(1-(12*n-1)*x - ...)))) (continued fraction).
G.f.: A(x) = 1/(1-1*x/(1-5*x/(1-6*x/(1-11*x/(1-12*x/(1-17*x/(1-18*x/(1 -...)))))))) (continued fraction).
G.f.: G(0) where G(k) = 1 - x*(6*k-1)/( 1 - 6*x*(k+1)/G(k+1) ); (continued fraction ). - Sergei N. Gladkovskii, Mar 24 2013 a(n) ~ (n-1)! * 6^(n-1) / (Gamma(5/6) * n^(1/6)). - Vaclav Kotesovec, Feb 22 2014
EXAMPLE
A(x) = 1 + x + 6*x^2 + 66*x^3 + 1086*x^4 + 24186*x^5 +...
1/A(x) = 1 - x - 5*x^2 - 55*x^3 - 935*x^4 -... - A008543(n)*x^(n+1)-...
MATHEMATICA
CoefficientList[Series[1/(1 + 1/6*ExpIntegralE[5/6, -1/(6*x)]/E^(1/(6*x))), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 22 2014 *)
PROG
(PARI) {a(n)=local(F=1+x+x*O(x^n)); for(i=1, n, F=1+x+6*x^2*deriv(F)/F); return(polcoeff(F, n, x))}
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