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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 (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

This is also the Euler transform of A060223. - Gus Wiseman, Oct 16 2016

REFERENCES

S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 223.

LINKS

Paul D. Hanna, Table of n, a(n) for n = 0..200

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

a(n) ~ (n-1)! / (2 * (log(2))^(n+1)). - Vaclav Kotesovec, Nov 18 2014

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)) +...

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, 1, nn}, {i, 1, 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), ", "))

CROSSREFS

Cf. A084783, A084785, A084786, A195983, A000670, A060223.

Sequence in context: A102812 A020100 A128230 * A255841 A197772 A135881

Adjacent sequences:  A084781 A084782 A084783 * A084785 A084786 A084787

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Jun 13 2003

STATUS

approved

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Last modified October 20 15:44 EDT 2018. Contains 316389 sequences. (Running on oeis4.)