

A141765


Triangle T, read by rows, such that row n equals column 0 of matrix power M^n where M is a triangular matrix defined by M(k+m,k) = binomial(k+m,k) for m=0..2 and zeros elsewhere. Width2restricted finite functions.


1



1, 1, 1, 1, 1, 2, 4, 6, 6, 1, 3, 9, 24, 54, 90, 90, 1, 4, 16, 60, 204, 600, 1440, 2520, 2520, 1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400, 1, 6, 36, 210, 1170, 6120, 29520, 128520, 491400, 1587600, 4082400, 7484400, 7484400, 1, 7, 49, 336, 2226
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OFFSET

0,6


COMMENTS

T(k,n) is the number of distinct ways in which n labeled objects can be distributed in k labeled urns allowing at most 2 objects to fall in each urn.  NE. Fahssi, Apr 22 2009
T(k,n) is the number of functions f:[n]>[k] such that the preimage set under f of any element of [k] has size 2 or less.  Dennis P. Walsh, Feb 15 2011


LINKS



FORMULA

T(k,n) = n!*Sum_{i=ceiling(n/2)..k} binomial(k,i)*binomial(i,ni)*2^(in).  Dennis P. Walsh, Feb 15 2011
T(n,2*n) = (2n)!/2^n; thus the rightmost border of T equals A000680.
Main diagonal (central terms) equals A012244.
Row sums of triangle T equals A003011, the number of permutations of up to n kinds of objects, where each kind of object can occur at most two times.
T(k,n) = n![x^n](1+x+x^2/2)^k. Double e.g.f.: Sum_{k,n} T(k,n)*(z^k/k!)*(x^n/n!) = exp(z(1+x+x^2/2)).  NE. Fahssi, Apr 22 2009
T(j+k,n) = Sum_{i=0..n} binomial(n,i)*T(j,i)*T(k,ni).  Dennis P. Walsh, Feb 15 2011


EXAMPLE

This triangle T begins:
1;
1, 1, 1;
1, 2, 4, 6, 6;
1, 3, 9, 24, 54, 90, 90;
1, 4, 16, 60, 204, 600, 1440, 2520, 2520;
1, 5, 25, 120, 540, 2220, 8100, 25200, 63000, 113400, 113400;
1, 6, 36, 210, 1170, 6120, 29520, 128520, 491400, 1587600, 4082400, 7484400, 7484400;
1, 7, 49, 336, 2226, 14070, 83790, 463680, 2346120, 10636920, 42071400, 139708800, 366735600, 681080400, 681080400,
1, 8, 64, 504, 3864, 28560, 201600, 1345680, 8401680, 48444480, 254016000, 1187524800, 4819953600, 16345929600, 43589145600, 81729648000, 81729648000,
1, 9, 81, 720, 6264, 52920, 430920, 3356640, 24811920, 172504080, 1116536400, 6646147200, 35835307200, 171632260800, 711047937600, 2451889440000, 6620101488000, 12504636144000, 12504636144000,
...
Let M be the triangular matrix that begins:
1;
1, 1;
1, 2, 1;
0, 3, 3, 1;
0, 0, 6, 4, 1;
0, 0, 0, 10, 5, 1; ...
where M(k+m,k) = C(k+m,k) for m=0,1,2 and zeros elsewhere.
Illustrate that row n of T = column 0 of M^n for n >= 0 as follows.
The matrix square M^2 begins:
1;
2, 1;
4, 4, 1;
6, 12, 6, 1;
6, 24, 24, 8, 1;
0, 30, 60, 40, 10, 1; ...
with column 0 of M^2 forming row 2 of T.
The matrix cube M^3 begins:
1;
3, 1;
9, 6, 1;
24, 27, 9, 1;
54, 96, 54, 12, 1;
90, 270, 240, 90, 15, 1;
90, 540, 810, 480, 135, 18, 1; ...
with column 0 of M^3 forming row 3 of T.
T(2,3)=6 because there are 6 ways to lodge 3 distinguishable balls, labeled by numbers 1,2 and 3, in 2 distinguishable boxes, each of which can hold at most 2 balls.  NE. Fahssi, Apr 22 2009
T(5,8)=63000 because there are 63000 ways to assign 8 students to a dorm room when there are 5 different twobed dorm rooms that are available. (See link for details of the count.)  Dennis P. Walsh, Feb 15 2011


MAPLE

seq(seq(n!*sum(binomial(k, j)*binomial(j, nj)*2^(jn), j=ceil(n/2)..k), n=0..2*k), k=1..10); # Dennis P. Walsh, Feb 15 2011


MATHEMATICA

T[k_, n_] := If[n == 0, 1, n! Coefficient[(1 + x + x^2/2)^k, x^n]]; TableForm[Table[T[k, n], {k, 0, 10}, {n, 0, 2 k}]] (* NE. Fahssi, Apr 22 2009 *)


PROG

(PARI) {T(n, k)=local(M=matrix(n+1, n+1, n, k, if(n>=k, if(nk<=2, binomial(n1, k1))))); if(k>2*n, 0, (M^n)[k+1, 1])}


CROSSREFS



KEYWORD

nonn,tabf


AUTHOR



STATUS

approved



