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Revision History for A138737

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The n-th term of the n-th inverse binomial transform of this sequence equals (n+1)^(n-1) for n>=0.
(history; published version)
#6 by Vaclav Kotesovec at Mon Oct 30 08:17:14 EDT 2017
STATUS

editing

approved

#5 by Vaclav Kotesovec at Mon Oct 30 08:17:00 EDT 2017
FORMULA

a(n) ~ (1 + LambertW(exp(-1)))^(3/2) * n^(n-1) / (exp(n-1) * LambertW(exp(-1))^n). - Vaclav Kotesovec, Oct 30 2017

STATUS

approved

editing

#4 by Vaclav Kotesovec at Mon Oct 30 07:55:58 EDT 2017
STATUS

editing

approved

#3 by Vaclav Kotesovec at Mon Oct 30 07:55:53 EDT 2017
LINKS

Vaclav Kotesovec, <a href="/A138737/b138737.txt">Table of n, a(n) for n = 0..300</a>

STATUS

approved

editing

#2 by Russ Cox at Fri Mar 30 18:37:10 EDT 2012
AUTHOR

_Paul D. Hanna (pauldhanna(AT)juno.com), _, Apr 05 2008

Discussion
Fri Mar 30
18:37
OEIS Server: https://oeis.org/edit/global/213
#1 by N. J. A. Sloane at Sun Jun 29 03:00:00 EDT 2008
NAME

The n-th term of the n-th inverse binomial transform of this sequence equals (n+1)^(n-1) for n>=0.

DATA

1, 2, 7, 52, 541, 7446, 127939, 2641192, 63746169, 1762380010, 54938528191, 1906911695580, 72949449568021, 3049813346508670, 138352912908850683, 6769028553912294736, 355311287187804226033, 19918243846821103623378

OFFSET

0,2

COMMENTS

Related to LambertW(-x)/(-x) = Sum_{n>=0} (n+1)^(n-1)*x^n/n!.

FORMULA

O.g.f. satisfies: [x^n] A( x/(1+n*x) )/(1+n*x) = (n+1)^(n-1) for n>=0.

E.g.f. satisfies: [x^n] A(x)*exp(-n*x) = (n+1)^(n-1)/n! for n>=0.

EXAMPLE

If the successive inverse binomial transforms are placed in a table,

then we see that the diagonal consists of terms (n+1)^(n-1):

n=0:[(1),2,7,52,541,7446,127939,2641192,63746169,1762380010,...];

n=1:[1,(1),4,36,368,5200,90432,1884736,45817088,1273874688,...];

n=2:[1, 0,(3),26,245,3684,64087,1349214,33003945,922386824,...];

n=3:[1,-1, 4,(16),160,2688,45184,970240,23814144,668975104,...];

n=4:[1,-2,7, 0,(125),2002,31203,705268,17177273,486100710,...];

n=5:[1,-3,12,-28, 176,(1296),21184,524352,12305664,354510080,...];

n=6:[1,-4,19,-74,373, 0,(16807),395866,8645673,260994628,...];

n=7:[1,-5,28,-144,800,-2816, 24192,(262144),5980160,195969024,...];

n=8:[1,-6,39,-244,1565,-8562,56419, 0,(4782969),149083874,...];

n=9:[1,-7,52,-380,2800,-19248,136768,-638912, 6966528,(100000000),..];

n=10:[1,-8,67,-558,4661,-37604,302679,-2112938,17204009, 0,...].

Notice the occurrence of zeros in the secondary diagonal = A138734.

PROG

(PARI) {a(n)=local(A=[1]); for(k=1, n, A=concat(A, 0); A[k+1]=(k+1)^(k-1)-Vec(subst(Ser(A), x, x/(1+k*x+x*O(x^k)))/(1+k*x))[k+1]); A[n+1]}

CROSSREFS

Cf. A138736 (inverse binomial transform), A138734; variants: A138909, A138911.

KEYWORD

nonn

AUTHOR

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 05 2008

STATUS

approved