Displaying 1-10 of 21 results found.
a(n) = A000984(n)* A004981(n), the term-wise product of the coefficients in (1-4*x)^(-1/2) and (1-8*x)^(-1/4).
+20
0
1, 4, 60, 1200, 27300, 668304, 17153136, 455083200, 12372574500, 342766138000, 9638583800560, 274341178587840, 7887308884400400, 228685287180840000, 6678543795015960000, 196260140322869011200, 5798873833602270315300, 172160337343624495866000
COMMENTS
The sequences A000984 and A004981 are related by the aesthetic identity:
EXAMPLE
G.f.: A(x) = 1 + 4*x + 60*x^2 + 1200*x^3 + 27300*x^4 + 668304*x^5 +...
The terms are the term-wise products of the sequences:
A000984 = [1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, ...],
A004981 = [1, 2, 10, 60, 390, 2652, 18564, 132600, 961350, ...].
Related sequences:
A^2: [1, 8, 136, 2880, 67800, 1699008, 44368704, 1193107968, ...],
A^4: [1, 16, 336, 7936, 200176, 5266176, 142657536, 3948773376, ...],
A^8: [1, 32, 928, 26624, 767200, 22270976, 651331072, 19178651648, ...].
PROG
(PARI) { A000984(n)=polcoeff((1-4*x +x*O(x^n))^(-1/2), n)} { A004981(n)=polcoeff((1-8*x +x*O(x^n))^(-1/4), n)} for(n=0, 20, print1(a(n), ", "))
Quartic (or 4-fold) factorial numbers: a(n) = Product_{k = 0..n-1} (4*k + 1).
(Formerly M4001)
+10
81
1, 1, 5, 45, 585, 9945, 208845, 5221125, 151412625, 4996616625, 184874815125, 7579867420125, 341094033905625, 16713607661375625, 885821206052908125, 50491808745015763125, 3080000333445961550625, 200200021673987500790625, 13813801495505137554553125
COMMENTS
a(n), n >= 1, enumerates increasing quintic (5-ary) trees. See David Callan's comment on A007559 (number of increasing quarterny trees).
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
E.g.f.: (1 - 4*x)^(-1/4).
a(n) ~ 2^(5/2) * Pi^(1/2) * Gamma(1/4)^(-1) * n^(3/4) * 2^(2*n) * e^(-n) * n^n * (1 + 23/96 * n^(-1) - ...). - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
G.f.: 1/(1 - x/(1 - 4*x/(1 - 5*x/(1 - 8*x/(1 - 9*x/(1 - 12*x/(1 - 13*x/(1 - .../(1 - A042948(n+1)*x/(1 -... (continued fraction). - Paul Barry, Dec 03 2009 a(n) = (-3)^n * Sum_{k = 0..n} (4/3)^k * s(n+1, n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012 G.f.: 1/T(0), where T(k) = 1 - x * (4*k + 1)/(1 - x * (4*k + 4)/T(k+1)) (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013 G.f.: 1 + x/Q(0), where Q(k) = 1 + x + 2*(2*k - 1)*x - 4*x*(k+1)/Q(k+1) (continued fraction). - Sergei N. Gladkovskii, May 03 2013 G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x * (4*k + 1)/(x * (4*k + 1) + 1/G(k+1))) (continued fraction). - Sergei N. Gladkovskii, Jun 04 2013 0 = a(n) * (4*a(n+1) - a(n+2)) + a(n+1) * a(n+1) for all n in Z. - Michael Somos, Jan 17 2014 Let T(x) = 1/(1 - 3*x)^(1/3) be the e.g.f. for the sequence of triple factorial numbers A007559. Then the e.g.f. A(x) for the quartic factorial numbers satisfies T(int_{0..x} A(t) dt) = A(x). (Cf. A007559 and A008548.) - Peter Bala, Jan 02 2015 O.g.f.: hypergeom([1, 1/4], [], 4*x). - Peter Luschny, Oct 08 2015 D-finite with recurrence: a(n) +(-4*n+3)*a(n-1)=0, n>=1. - R. J. Mathar, Feb 14 2020 Sum_{n>=0} 1/a(n) = 1 + exp(1/4)*(Gamma(1/4) - Gamma(1/4, 1/4))/(2*sqrt(2)). - Amiram Eldar, Dec 18 2022
EXAMPLE
G.f. = 1 + x + 5*x^2 + 45*x^3 + 585*x^4 + 9945*x^5 + 208845*x^6 + ...
MAPLE
x:='x'; G(x):=(1-4*x)^(-1/4): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1], x) od: seq(eval(f[n], x=0), n=0..17); # Zerinvary Lajos, Apr 03 2009
MATHEMATICA
a[ n_]:= Pochhammer[ 1/4, n] 4^n; (* Michael Somos, Jan 17 2014 *) a[ n_]:= If[n < 0, 1 / Product[ -k, {k, 3, -4n-1, 4}], Product[ k, {k, 1, 4n-3, 4}]]; (* Michael Somos, Jan 17 2014 *) Range[0, 19]! CoefficientList[Series[((1-4x)^(-1/4)), {x, 0, 19}], x] (* Vincenzo Librandi, Oct 08 2015 *)
PROG
(PARI) {a(n) = if( n<0, 1 / prod(k=1, -n, 1 - 4*k), prod(k=1, n, 4*k - 3))}; /* Michael Somos, Jan 17 2014 */ (Maxima) A007696(n):=prod(4*k+1, k, 0, n-1)$ (Magma) [n le 2 select 1 else (4*(n-1)-7)*(Self(n-1) + 4*Self(n-2)): n in [1..20]]; // G. C. Greubel, Aug 15 2019 (Sage) [4^n*rising_factorial(1/4, n) for n in (0..20)] # G. C. Greubel, Aug 15 2019 (GAP) a:=[1, 1];; for n in [3..20] do a[n]:=(4*(n-1)-7)*(a[n-1]+4*a[n-2]); od; a; # G. C. Greubel, Aug 15 2019
CROSSREFS
a(n) = A049029(n, 1) for n >= 1 (first column of triangle).
a(n) = 2^n * (2*n)! / (n!)^2.
+10
29
1, 4, 24, 160, 1120, 8064, 59136, 439296, 3294720, 24893440, 189190144, 1444724736, 11076222976, 85201715200, 657270374400, 5082890895360, 39392404439040, 305870434467840, 2378992268083200, 18531097667174400
COMMENTS
Number of lattice paths from (0,0) to (n,n) using steps (0,1), and two kinds of steps (1,0). - Joerg Arndt, Jul 01 2011 The convolution square root of this sequence is A004981. - T. D. Noe, Jun 11 2002 Also main diagonal of array: T(i,1)=2^(i-1), T(1,j)=1, T(i,j) = T(i,j-1) + 2*T(i-1,j). - Benoit Cloitre, Feb 26 2003 The Hankel transform (see A001906 for definition) of this sequence with interpolated zeros(1, 0, 4, 0, 24, 0, 160, 0, 1120, ...) = is A036442: 1, 4, 32, 512, 16384, ... . - Philippe Deléham, Jul 03 2005 Equals number of permutations whose reverse shares the same recording tableau in the Robinson-Schensted correspondence with n=(k-1)/2 for k odd. - Dang-Son Nguyen, Jul 02 2024
FORMULA
a(n) = C(2*n,n) * 2^n.
D-finite with recurrence a(n) = a(n-1)*(8-4/n).
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+1)/(4*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013 E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 4*x/(4*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013 G.f.: Q(0)/(1+2*sqrt(x)), where Q(k) = 1 + 2*sqrt(x)/(1 - 2*sqrt(x)*(2*k+1)/(2*sqrt(x)*(2*k+1) + (k+1)/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 09 2013 a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(3*n-2*k,n)* binomial(n+k,n). - Peter Bala, Aug 04 2016 Sum_{n>=0} 1/a(n) = 8/7 + 8*sqrt(7)*arcsin(1/sqrt(8))/49.
Sum_{n>=0} (-1)^n/a(n) = (8/27)*(3 - arcsinh(1/sqrt(8))). (End)
a(n) = Sum_{k = n..2*n} binomial(2*n,k)*binomial(k,n). In general, for m >= 1, Sum_{k = n..m*n} binomial(m*n,k)*binomial(k,n) = 2^((m-1)*n)*binomial(m*n,n). - Peter Bala, Mar 25 2023 Conjecture: a(n) = Sum_{0 <= j, k <= n} binomial(n, j)*binomial(n, k)* binomial(k+j, n). - Peter Bala, Jul 16 2024
MATHEMATICA
Table[2^n Binomial[2n, n], {n, 0, 30}] (* Harvey P. Dale, Dec 16 2014 *)
PROG
(PARI) {a(n)=if(n<0, 0, 2^n*(2*n)!/n!^2)} /* Michael Somos, Jan 31 2007 */ (PARI) { for (n = 0, 200, write("b059304.txt", n, " ", 2^n * (2*n)! / n!^2); ) } \\ Harry J. Smith, Jun 25 2009 (PARI) /* as lattice paths: same as in A092566 but use */ steps=[[1, 0], [1, 0], [0, 1]]; /* note the double [1, 0] */
(Magma) [2^n*Factorial(2*n)/Factorial(n)^2: n in [0..25]]; // Vincenzo Librandi, Oct 08 2015
Related to triple factorial numbers A007559(n+1).
+10
12
1, 6, 42, 315, 2457, 19656, 160056, 1320462, 11003850, 92432340, 781473420, 6642524070, 56716936290, 486145168200, 4180848446520, 36059817851235, 311811366125385, 2702365173086670, 23467908082068450, 204170800313995515, 1779202688450532345, 15527587099204645920
COMMENTS
Working with an offset of 1, we conjecture a(p*n) = a(n) (mod p^2) for prime p = 1 (mod 3) and all positive integers n except those n of the form n = m*p + k for 0 <= m <= (p-1)/3 and 1 <= k <= (p-1)/3. Cf. A298799, A004981 and A004982. - Peter Bala, Dec 23 2019
FORMULA
G.f.: (-1+(1-9*x)^(-1/3))/(3*x).
a(n)= A035529(n+1, 1) (first column of triangle). D-finite with recurrence: (n+1)*a(n) +3*(-3*n-1)*a(n-1)=0. - R. J. Mathar, Jan 28 2020 Sum_{n>=0} 1/a(n) = 3/8 + 3*sqrt(3)*Pi/32 + 9*log(3)/32. - Amiram Eldar, Dec 22 2022
MATHEMATICA
CoefficientList[Series[(-1 + (1 - 9 x)^(-1/3))/(3 x), {x, 0, 19}], x] (* Michael De Vlieger, Oct 13 2019 *)
a(n) = (2^n/n!) * Product_{k=0..n-1} (4*k + 3).
+10
9
1, 6, 42, 308, 2310, 17556, 134596, 1038312, 8046918, 62587140, 488179692, 3816677592, 29897307804, 234578876616, 1843119744840, 14499208659408, 114181268192838, 900017055167076, 7100134546318044, 56053693786721400, 442824180915099060, 3500419715805068760, 27685137752276452920
COMMENTS
Conjecture: a(p*n) = a(n) (mod p^2) for prime p == 1 (mod 4) and all positive integers n. Cf. A004981. - Peter Bala, Dec 22 2019
FORMULA
G.f.: (1 - 8*x)^(-3/4).
a(n) ~ Gamma(3/4)^-1*n^(-1/4)*2^(3*n)*{1 - 3/32*n^-1 + ...}
a(n) = (-8)^n*binomial(-3/4,n).
E.g.f.: is the hypergeometric function of type 1F1, in Maple notation hypergeom([3/4], [1], 8*x).
Representation as n-th moment of a positive function on (0, 8): a(n) = Integral_{x=0..8} ( x^n*(2^(1/4)/(8*Pi*x^(1/4)*(1-x/8)^(3/4)) ) dx, n >= 0. This function is the solution of the Hausdorff moment problem on (0, 8) with moments equal to a(n). As a consequence this representation is unique. (End)
D-finite with recurrence: n*a(n) +2*(-4*n+1)*a(n-1)=0. - R. J. Mathar, Jan 16 2020
MAPLE
A004982 := n -> (-8)^n*binomial(-3/4, n):
MATHEMATICA
Table[2^n/n! Product[4k+3, {k, 0, n-1}], {n, 0, 30}] (* Harvey P. Dale, Oct 03 2011 *) Table[Sum[2^k*Binomial[2*n-2*k, n-k]*Binomial[n+k, n], {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Sep 15 2013 *) FullSimplify[Table[8^n*Gamma[n+3/4]/(n!*Gamma[3/4]), {n, 0, 25}]] (* Vaclav Kotesovec, Sep 15 2013 *) max = 30; s = Hypergeometric1F1[3/4, 1, 8x] + O[x]^(max+1);
PROG
(PARI) a(n)=2^n/n!*prod(k=0, n-1, 4*k+3)
for(n=0, 25, print(a(n)))
(PARI) x='x+O('x^66); Vec((1-8*x)^(-3/4)) \\ Joerg Arndt, Apr 20 2013 (Magma) [1] cat [2^n*&*[4*k+3: k in [0..n-1]]/Factorial(n): n in [1..25]]; // G. C. Greubel, Aug 22 2019 (Sage) [8^n*rising_factorial(3/4, n)/factorial(n) for n in (0..25)] # G. C. Greubel, Aug 22 2019 (GAP) List([0..25], n-> 2^n*Product([0..n-1], k-> 4*k+3)/Factorial(n) ); # G. C. Greubel, Aug 22 2019
AUTHOR
Joe Keane (jgk(AT)jgk.org)
a(n) = (2^n/n!)*Product_{k=0..n-1} (4*k - 1).
+10
8
1, -2, -6, -28, -154, -924, -5852, -38456, -259578, -1788204, -12517428, -88759944, -636112932, -4599585816, -33511268088, -245749299312, -1812401082426, -13433090375628, -100001895018564, -747382583822952, -5605369378672140, -42173731515723720
FORMULA
G.f.: (1 - 8*x)^(1/4).
a(n) ~ -1/4*Gamma(3/4)^-1*n^(-5/4)*2^(3*n)*{1 + 5/32*n^-1 + ...}
a(n) = -1/n*(if n=1 then (2)^n else Sum_{k=1..n-1} 2^n*binomial(k+n-1, n-1) * (Sum_{j=0..k} (binomial(k,j)*binomial(j,n-1-3*k+2*j)*(3/2)^(3*k-n+1-j)*(-1)^(n-1-3*k)*(1/4)^(k-j) ) ), n>0. - Vladimir Kruchinin, Sep 14 2010 a(n) = 8^n*Pochhammer(-1/4, n)/n! = -(1/4)*8^n*Gamma(n-1/4)/(Gamma(3/4)*n!). - Robert Israel, Sep 29 2014 D-finite with recurrence: n*a(n) +2*(-4*n+5)*a(n-1)=0. - R. J. Mathar, Jan 16 2020
MAPLE
seq(-(1/4)*8^n*GAMMA(n-1/4)/(GAMMA(3/4)*n!), n=0..30); # Robert Israel, Sep 29 2014
MATHEMATICA
Table[8^n*Pochhammer[-1/4, n]/n!, {n, 0, 30}] (* G. C. Greubel, Aug 22 2019 *) CoefficientList[Series[Surd[1-8x, 4], {x, 0, 30}], x] (* Harvey P. Dale, Dec 08 2019 *)
PROG
(PARI) for(n=0, 28, print1(2^n/n!*prod(k=0, n-1, (4*k-1)), ", "))
(Maxima) a(n):=-1/n*(if n=1 then (2)^n else sum(sum(binomial(k, j)* binomial(j, n-1-3*k+2*j)*(3/2)^(3*k-n+1-j)*(-1)^(n-1-3*k)*(1/4)^(k-j), j, 0, k)*binomial(k+n-1, n-1)*(2)^n, k, 1, n-1)); (for ) /* Vladimir Kruchinin, Sep 14 2010 */ (Maxima) a(n):=binomial(1/4, n)*(-8)^n; /* Tani Akinari, Sep 28 2014 */ (GAP) List([0..25], n->(2^n/Factorial(n))*Product([0..n-1], k->4*k-1)); # Muniru A Asiru, Apr 28 2018 (Magma) [1] cat [2^n*(&*[4*k-1: k in [0..n-1]])/Factorial(n): n in [1..30]]; // G. C. Greubel, Aug 22 2019 (Sage) [8^n*rising_factorial(-1/4, n)/factorial(n) for n in (0..30)] # G. C. Greubel, Aug 22 2019
AUTHOR
Joe Keane (jgk(AT)jgk.org)
G.f. satisfies: A(x) = 1/A(-x*A(x)^4).
+10
8
1, 2, 10, 60, 390, 2660, 18772, 138984, 1107686, 9576100, 87944188, 830857464, 7876505340, 73967614584, 685644645896, 6289047266480, 57465415636166, 528315307772004, 4947263762389484, 47785581838822232, 480797992896880788, 5058812497153271912
COMMENTS
Compare to: W(x) = 1/W(-x*W(x)^4) when W(x) = Sum_{n>=0} (2*n+1)^(n-1)*x^n/n!.
Compare to: B(x) = 1/B(-x*B(x)^4) when B(x) = 1/(1-8*x)^(1/4) = g.f. of A004981. An infinite number of functions G(x) satisfy (*) G(x) = 1/G(-x*G(x)^4); for example, (*) is satisfied by G(x) = W(m*x), where W(x) = Sum_{n>=0} (2*n+1)^(n-1)*x^n/n!.
FORMULA
The g.f. of this sequence is the limit of the recurrence:
(*) G_{n+1}(x) = (G_n(x) + 1/G_n(-x*G_n(x)^4))/2 starting at G_0(x) = 1+2*x.
EXAMPLE
G.f.: A(x) = 1 + 2*x + 10*x^2 + 60*x^3 + 390*x^4 + 2660*x^5 + 18772*x^6 +...
A(x)^4 = 1 + 8*x + 64*x^2 + 512*x^3 + 4096*x^4 + 32800*x^5 + 263168*x^6 +...
PROG
(PARI) {a(n)=local(A=1+2*x); for(i=0, n, A=(A+1/subst(A, x, -x*A^4+x*O(x^n)))/2); polcoeff(A, n)}
for(n=0, 31, print1(a(n), ", "))
Numerators in expansion of (1-x)^{-1/4}.
+10
6
1, 1, 5, 15, 195, 663, 4641, 16575, 480675, 1762475, 13042315, 48612265, 729183975, 2748462675, 20809788825, 79077197535, 4823709049635, 18443593425075, 141400882925575, 543277076503525, 8366466978154285, 32270658344309385
COMMENTS
Numerators in expansion of sqrt(1/sqrt(1-4x)). - Paul Barry, Jul 12 2005
FORMULA
a(n) = 2^(e_2((2n)!)-n)/n! Product[4k+1,{k,0,n-1}], where e_2((2n)!) is the highest power of 2 that divides (2n)! (sequence A005187). - Emanuele Munarini, Jan 25 2011 Numerators in (1-4t)^(-1/4) = 1 + t + (5/2)t^2 + (15/2)t^3 + (195/8)t^4 + (663/8)t^5 + (4641/16)t^6 + (16575/16)t^7 + ... = 1 + t + 5*t^2/2! + 45*t^3/3! + 585*t^4/4! + ... = e.g.f. for the quartic factorials A007696 (cf. A094638). - Tom Copeland, Dec 04 2013
MATHEMATICA
Table[Numerator[Binomial[-1/4, n] (-1)^n], {n, 0, 20}]
PROG
(PARI) {a(n) = if( n<0, 0, numerator( polcoeff( (1 - x +x*O(x^n))^(-1/4), n ) ) ) } /* Michael Somos, Aug 23 2007 */
a(n) = ( (2^n / n!) * Product_{k=0..n-1} (4*k + 1) )^2.
+10
5
1, 4, 100, 3600, 152100, 7033104, 344622096, 17582760000, 924193822500, 49701090010000, 2721631688947600, 151241747739534400, 8507348310348810000, 483459012855561960000, 27715027900230072360000, 1600820011517288979513600, 93072675982122379574532900
COMMENTS
Cayley (1878) refers to Gauss, Werke, t. iii, p. 424 for a slightly different form of a square of a hypergeometric series being hypergeometric. - Michael Somos, Jun 25 2012
REFERENCES
A. Cayley, An Identity, Messenger of Mathematics, 7 (1878), p. 69
FORMULA
Expansion of K(k) / (Pi/2) in powers of (k * k'/4)^2, where K(k) is the complete elliptic integral of first kind evaluated at modulus k.
Expansion of 1 / AGM( 1, (1 - 16*x)^(1/2) ) in powers of x * (1 - 16*x) where AGM() is the arithmetic-geometric mean.
G.f.: F(1/4, 1/4; 1; 64*x).
n^2 * a(n) = 4*(4*n-3)^2 * a(n-1), with a(0) = 1.
0 = 16*x*(x+64)*y'' + 8*(3*x+128)*y' + y, where y(x) = A(x/-4096). (End)
EXAMPLE
G.f. = 1 + 4*x + 100*x^2 + 3600*x^3 + 152100*x^4 + 7033104*x^5 + ...
MATHEMATICA
a[ n_] := If[ n < 0, 0, (Pochhammer[ 1/4, n] 8^n / n!)^2];
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/4, 1/4, 1, 64 x], {x, 0, n}];
a[ n_] := SeriesCoefficient[ HypergeometricPFQ[ {1/2, 1/2, 1/2}, {1, 1}, 64 x]^(1/2), {x, 0, n}];
(End)
Join[{1}, RecurrenceTable[{a[1] == 4, a[n] == (4 (4 n - 3)^2 a[n-1] / n^2)}, a, {n, 20}]] (* Vincenzo Librandi, Aug 26 2016 *)
PROG
(PARI) {a(n) = if( n<0, 0, prod( k=1, n, (8*k - 6) / k)^2)};
(PARI) {a(n) = my(A); if( n<1, n==0, A = x * O(x^n); polcoeff( subst( 1 / agm(1, sqrt(1 - 16*x + A) ), x, serreverse( x*(1 - 16*x) + A )), n))};
(PARI)
seq(N) = {
my(a = vector(N)); a[1] = 4;
for (n=2, N, a[n] = 4*(4*n-3)^2*a[n-1]/n^2);
concat(1, a);
};
(Magma) [1] cat [n le 1 select 4 else (4*(4*n-3)^2*Self(n-1)/ n^2): n in [1..30]]; // Vincenzo Librandi, Aug 26 2016
Expansion of 1/(1 - 16*x)^(1/8).
+10
4
1, 2, 18, 204, 2550, 33660, 460020, 6440280, 91773990, 1325624300, 19354114780, 285033326760, 4227994346940, 63094684869720, 946420273045800, 14259398780556720, 215673406555920390, 3273161111260438860, 49824785804742235980, 760483572809223601800
FORMULA
a(n) = (2^n/n!) * Product_{k=0..n-1} (8*k + 1).
EXAMPLE
G.f.: A(x) = 1 + 2*x + 18*x^2 + 204*x^3 + 2550*x^4 + 33660*x^5 + ...
where
A(x)^8 = 1 + 16*x + 256*x^2 + 4096*x^3 + 65536*x^4 + ... + 16^n*x^n + ...
Also,
A(x)^4 = 1 + 8*x + 96*x^2 + 1280*x^3 + 17920*x^4 + 258048*x^5 + ... + 4^n* A000984(n)*x^n + ... A(x)^2 = 1 + 4*x + 40*x^2 + 480*x^3 + 6240*x^4 + 84864*x^5 + ... + 2^n* A004981(n)*x^n + ...
MAPLE
seq(coeff(series(1/(1-16*x)^(1/8), x, 50), x, n+1), n=0..20); # Muniru A Asiru, Jun 23 2018
MATHEMATICA
CoefficientList[Series[1/(1-16*x)^(1/8), {x, 0, 20}], x] (* Vaclav Kotesovec, Jul 24 2013 *)
PROG
(PARI) {a(n)=polcoeff(1/(1-16*x +x*O(x^n))^(1/8), n)}
for(n=0, 30, print1(a(n), ", "))
(PARI) {a(n)=(2^n/n!)*prod(k=0, n-1, 8*k + 1)}
for(n=0, 30, print1(a(n), ", "))
(GAP) List([0..20], n->(2^n/Factorial(n))*Product([0..n-1], k->8*k+1)); # Muniru A Asiru, Jun 23 2018
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