|
| |
|
|
A084784
|
|
Binomial transform = self-convolution: first column of the triangle (A084783).
|
|
15
|
|
|
|
1, 1, 2, 6, 25, 137, 944, 7884, 77514, 877002, 11218428, 160010244, 2516742498, 43260962754, 806650405800, 16213824084864, 349441656710217, 8037981040874313, 196539809431339642, 5090276002949080318, 139202688233361310841, 4008133046329085884137
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
In the triangle (A084783), the diagonal (A084785) is the self-convolution of this sequence and the row sums (A084786) gives the differences of the diagonal and this sequence.
Ramanujan considers the continued fraction phi(x) = 1 / (x + 1 - 1^2 / (x + 3 - 2^2 / (x + 5 - 3^2 / (x + 7 - 4^2 / ...)))) and states that phi(x+1) approaches x phi(x)^2 as x gets large. The asymptotic expansion is phi(x) = 1/x - 1/x^2 + 2/x^3 - 6/x^4 + 24/x^5 - ... + (-1)^n * n! / x^(n+1) + ... but if we replace this with f(x) = a(0)/x - a(1)/x^2 + a(2)/x^3 - a(3)/x^4 + ... then formally f(x+1) = x f(x)^2 which is similar to my Feb 16 2006 formula. - Michael Somos, Jun 20 2015
|
|
|
REFERENCES
|
S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 223.
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f. satisfies A(n*x)^2 = n-th binomial transform of A(n*x).
G.f. A(x) satisfies 1 + x = A(x/(1 + x))^2 / A(x). - Michael Somos, Feb 16 2006
G.f.: A(x) = Product_{n>=1} 1/(1 - n*x)^(1/2^(n+1)). - Paul D. Hanna, Jun 16 2010
G.f.: A(x) = exp( Sum_{n>=1} A000670(n)*x^n/n ) where Sum_{n>=0} A000670(n)*x^n = Sum_{n>=0} n!*x^n/Product_{k=0..n} (1-k*x). - Paul D. Hanna, Sep 26 2011
|
|
|
EXAMPLE
|
G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 25*x^4 + 137*x^5 + 944*x^6 + ...
where
A(x) = (1-x)^(-1/4)*(1-2*x)^(-1/8)*(1-3*x)^(-1/16)*(1-4*x)^(-1/32)*...
Also,
log(A(x)) = x + 3*x^2/2 + 13*x^3/3 + 75*x^4/4 + 541*x^5/5 + 4683*x^6/6 + ... + A000670(n)*x^n/n + ...
thus, the logarithmic derivative equals the series:
A'(x)/A(x) = 1/(1-x) + 2!*x/((1-x)*(1-2*x)) + 3!*x^2/((1-x)*(1-2*x)*(1-3*x)) + 4!*x^3/((1-x)*(1-2*x)*(1-3*x)*(1-4*x)) + ...
|
|
|
MAPLE
|
a:= proc(n) option remember;
1+add(a(j)*(binomial(n, j)-a(n-j)), j=1..n-1)
end:
|
|
|
MATHEMATICA
|
a[ n_]:= If[n<1, Boole[n==0], Module[{A= 1/x - 1/x^2}, Do [A= 2 A - Normal @ Series[ (x A^2) /. x -> x-1, {x, Infinity, k+1}], {k, 2, n}]; (-1)^n Coefficient[A, x, -n-1]]]; (* Michael Somos, Jun 20 2015 *)
nn=20; CoefficientList[Series[Exp[Sum[Times[1/k, i!, StirlingS2[k, i], x^k], {k, nn}, {i, k}]], {x, 0, nn}], x] (* Gus Wiseman, Oct 18 2016 *)
|
|
|
PROG
|
(PARI) {a(n) = my(A); if( n<0, 0, A=1; for(k=1, n, A = truncate(A + O(x^k)) + x * O(x^k); A += A - 1 / subst(A^-2, x, x / (1 + x)) / (1 + x); ); polcoeff(A, n))}; /* Michael Somos, Feb 18 2006 */
(PARI) /* Using o.g.f. exp( Sum_{n>=1} A000670(n)*x^n/n ): */
{a(n)=polcoeff(exp(intformal(sum(m=1, n+1, m!*x^(m-1)/prod(k=1, m, 1-k*x+x*O(x^n))))), n)}
for(n=0, 30, print1(a(n), ", "))
(Magma)
m:=50;
f:= func< n, x | Exp((&+[(&+[Factorial(j)*StirlingSecond(k, j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R<x>:=PowerSeriesRing(Rationals(), m+1); // A084784
(SageMath)
m=40
def f(n, x): return exp(sum(sum(factorial(j)*stirling_number2(k, j) *x^k/k for j in range(1, k+1)) for k in range(1, n+2)))
P.<x> = PowerSeriesRing(QQ, prec)
return P( f(m, x) ).list()
from functools import cache
from math import comb as binomial
@cache
def a(n: int) -> int:
return 1 + sum((binomial(n, j) - a(n - j)) * a(j) for j in range(1, n))
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
