

A049611


a(n) = T(n,2), array T as in A049600.


18



0, 1, 4, 13, 38, 104, 272, 688, 1696, 4096, 9728, 22784, 52736, 120832, 274432, 618496, 1384448, 3080192, 6815744, 15007744, 32899072, 71827456, 156237824, 338690048, 731906048, 1577058304, 3388997632, 7264534528, 15535702016
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,3


COMMENTS

Refer to A089378 and A075729 for the definition of hierarchies, subhierarchies and onestep transitions.  Thomas Wieder, Feb 28 2004
We may ask for the number of onestep transitions (NOOST) between all unlabeled hierarchies of n elements with the restriction that no subhierarchies are allowed. As an example, consider n = 4 and the hierarchy H1 = [[2,2]] with two elements on level 1 and two on level 2. Starting from H1 the hierarchies [[1, 3]], [[2, 1, 1]], [[1, 2, 1]] can be reached by moving one element only, but [[1, 1, 2]] cannot be reached in a onestep transitition. The solution is n = 1, NOOST = 0; n = 2, NOOST = 1; n = 3, NOOST = 4; n = 4, NOOST = 13; n = 5, NOOST = 38; n = 6, NOOST = 104; n = 7, NOOST = 272; n = 8, NOOST = 688; n = 9, NOOST = 1696. This is sequence A049611.  Thomas Wieder, Feb 28 2004
If X_1,X_2,...,X_n are 2blocks of a (2n+2)set X then, for n>=1, a(n+1) is the number of (n+2)subsets of X intersecting each X_i, (i=1,2,...,n).  Milan Janjic, Nov 18 2007
In each composition (ordered partition) of the integer n, circle the first summand once, circle the second summand twice, etc. a(n) is the total number of circles in all compositions of n (that is, add k*(k+1)/2 for each composition into k parts). Note the O.g.f. is B(A(x)) where A(x)= x/(1x) and B(x)= x/(1x)^3.
This is the Riordan transform with the Riordan matrix A097805 (of the associated type) of the triangular number sequence A000217. See a Feb 17 2017 comment on A097805.  Wolfdieter Lang, Feb 17 2017


REFERENCES

Sergey Kitaev, JB Remmel A note on pAscent Sequences, Preprint, 2016, https://pure.strath.ac.uk/portal/files/46917816/Kitaev_Remmel_JC2016_a_note_on_p_ascent_sequences.pdf


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions
M. Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.  From N. J. A. Sloane, Feb 13 2013
S. Kitaev, J. Remmel, pAscent Sequences, arXiv:1503.00914 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (6,12,8).


FORMULA

G.f.: x*(1x)^2/(12*x)^3.
Binomial transform of quarter squares A002620(n+1).  Paul Barry, May 27 2003
a(n) = Sum_{k=0..n} binomial(n, k)floor((k+1)^2/4).  Paul Barry, May 27 2003
a(n) = 2^(n4)(n^2+5n+2)  0^n/8.  Paul Barry, Jun 09 2003
a(n+2) = A001787(n+2) + A001788(n). Floretion Algebra Multiplication Program, FAMP Code: 1vessum(pos)seq[A] (= (a(n)), from 2nd term), 1vessum(neg)seq[A] and 1vessumseq[A] with A = + .5'i + .5i' + .5'ij' + .5'ki' + 2e. Sumtype is set to: default (ver. f).  Creighton Dement, Aug 02 2005
Row sums of triangle A133729 = (1, 4, 13, 38, ...).  Gary W. Adamson, Sep 21 2007
a(n) = Hyper2F1([n+1, 3], [1], 1) for n>0.  Peter Luschny, Aug 02 2014


MATHEMATICA

CoefficientList[Series[x (1x)^2/(12x)^3, {x, 0, 40}], x] (* Harvey P. Dale, Sep 24 2013 *)


PROG

(PARI) Vec(x*(1x)^2/(12*x)^3+O(x^99)) \\ Charles R Greathouse IV, Jun 12 2015


CROSSREFS

a(n+1)= A055252(n, 0), n >= 0. Row sums of triangle A055249.
Cf. A001793, A058396, A075729, A089378, A133729, A000217.
Sequence in context: A277974 A089092 A181527 * A084851 A094706 A056014
Adjacent sequences: A049608 A049609 A049610 * A049612 A049613 A049614


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling


STATUS

approved



