

A038207


Triangle whose (i,j)th entry is binomial(i,j)*2^(ij).


83



1, 2, 1, 4, 4, 1, 8, 12, 6, 1, 16, 32, 24, 8, 1, 32, 80, 80, 40, 10, 1, 64, 192, 240, 160, 60, 12, 1, 128, 448, 672, 560, 280, 84, 14, 1, 256, 1024, 1792, 1792, 1120, 448, 112, 16, 1, 512, 2304, 4608, 5376, 4032, 2016, 672, 144, 18, 1, 1024, 5120, 11520, 15360, 13440
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OFFSET

0,2


COMMENTS

This infinite matrix is the square of the Pascal matrix (A007318) whose rows are [ 1,0,... ], [ 1,1,0,... ], [ 1,2,1,0,... ], ...
Number of different partial sums of 1+[1,1,2]+[2,2,3]+[3,3,4]+[4,4,5]+... with entries that are zero removed.  Jon Perry, Jan 01 2004
Row sums are powers of 3 (A000244), antidiagonal sums are Pell numbers (A000129).  Gerald McGarvey, May 17 2005
Riordan array (1/(12x),x/(12x)).  Paul Barry, Jul 28 2005
T(n,k) is the number of elements of the Coxeter group B_n with descent set contained in {s_k}, 0<=k<=n1. For T(n,n), we interpret this as the number of elements of B_n with empty descent set (since s_n does not exist).  Elizabeth Morris (epmorris(AT)math.washington.edu), Mar 01 2006
Let S be a binary relation on the power set P(A) of a set A having n = A elements such that for every element x, y of P(A), xSy if x is a subset of y. Then T(n,k) = the number of elements (x,y) of S for which y has exactly k more elements than x.  Ross La Haye, Oct 12 2007
T(n,k) is number of paths in the first quadrant going from (0,0) to (n,k) using only steps B=(1,0) colored blue, R=(1,0) colored red and U=(1,1). Example: T(3,2)=6 because we have BUU, RUU, UBU, URU, UUB and UUR.  Emeric Deutsch, Nov 04 2007
T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (0,1), and two kinds of step (1,0).  Joerg Arndt, Jul 01 2011
T(i,j) is the number of ipermutations of {1,2,3} containing j 1's. Example: T(2,1)=4 because we have 12, 13, 21 and 31; T(3,2)=6 because we have 112, 113, 121, 131, 211 and 311.  Zerinvary Lajos, Dec 21 2007
Triangle of coefficients in expansion of (2+x)^n.  NE. Fahssi, Apr 13 2008
Sum of diagonals are Jacobsthalnumbers: A001045.  Mark Dols, Aug 31 2009
Triangle T(n,k), read by rows, given by [2,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.  Philippe Deléham, Dec 15 2009
Eigensequence of the triangle = A004211: (1, 3, 11, 49, 257, 1539, ...).  Gary W. Adamson, Feb 07 2010
fvectors ("face"vectors) for ndimensional cubes [see e.g., Hoare]. (This is a restatement of Bottomley's above.)  Tom Copeland, Oct 19 2012
With P = Pascal matrix, the sequence of matrices I, A007318, A038207, A027465, A038231, A038243, A038255, A027466 ... = P^0, P^1, P^2, ... are related by Copeland's formula below to the evolution at integral time steps n= 0, 1, 2, ... of an exponential distribution exp(x*z) governed by the FokkerPlanck equation as given in the Dattoli et al. ref. below.  Tom Copeland, Oct 26 2012
The matrix elements of the inverse are T^(1)(n,k) = (1)^(n+k)*T(n,k).  R. J. Mathar, Mar 12 2013
Unsigned diagonals of A133156 are rows of this array.  Tom Copeland, Oct 11 2014
Omitting the first row, this is the production matrix for A039683, where an equivalent differential operator can be found.  Tom Copeland, Oct 11 2016
T(n,k) is the number of functions f:[n]>[3] with exactly k elements mapped to 3. Note that there are C(n,k) ways to choose the k elements mapped to 3, and there are 2^(nk) ways to map the other (nk) elements to {1,2}. Hence, by summing T(n,k) as k runs from 0 to n, we obtain 3^n = Sum_{k=0..n} T(n,k).  Dennis P. Walsh, Sep 26 2017


REFERENCES

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 155.
H. S. M. Coxeter, Regular Polytopes, Dover Publications, New York (1973), p. 122.
S. J. Cyvin et al., Unbranched catacondensed polygonal systems containing hexagons and tetragons, Croatica Chem. Acta, 69 (1996), 757774.


LINKS

T. D. Noe, Rows n=0..100 of triangle, flattened
P. Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, On the fMatrices of Pascallike Triangles Defined by Riordan Arrays, arXiv:1805.02274 [math.CO], 2018.
John Cartan, Starmaze: Cartan's Triangle.
T. Copeland, Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras.
B. N. Cyvin et al., Isomer enumeration of unbranched catacondensed polygonal systems with pentagons and heptagons, Match, No. 34 (Oct 1996), pp. 109121.
G. Dattoli, A. Mancho, M. Quattromini and A. Torre, Exponential operators, generalized polynomials and evolution problems, Radiation Physics and Chemistry 61 (2001) 99108. [From Tom Copeland, Oct 25 2012]
Filippo Disanto, Some Statistics on the Hypercubes of Catalan Permutations, Journal of Integer Sequences, Vol. 18 (2015), Article 15.2.2.
W. G. Harter, Representations of multidimensional symmetries in networks, J. Math. Phys., 15 (1974), 20162021.
Graham Hoare, Hypercubes and Chebyshev, Math. Gaz. 74 (470) (1990) 375377
Ross La Haye, Binary Relations on the Power Set of an nElement Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.


FORMULA

T(n, k) = Sum_{i=0..n} binomial(n,i)*binomial(i,k).
T(n, k) = (1)^k*A065109(n,k).
G.f.: 1/(12*zt*z).  Emeric Deutsch, Nov 04 2007
Rows of the triangle are generated by taking successive iterates of (A135387)^n * [1, 0, 0, 0, ...].  Gary W. Adamson, Dec 09 2007
From the formalism of A133314, the e.g.f. for the row polynomials of A038207 is exp(x*t)*exp(2x). The e.g.f. for the row polynomials of the inverse matrix is exp(x*t)*exp(2x). p iterates of the matrix give the matrix with e.g.f. exp(x*t)*exp(p*2x). The results generalize for 2 replaced by any number.  Tom Copeland, Aug 18 2008
Sum_{k=0..n} T(n,k)*x^k = (2+x)^n.  Philippe Deléham, Dec 15 2009
nth row is obtained by taking pairwise sums of triangle A112857 terms starting from the right.  Gary W. Adamson, Feb 06 2012
T(n,n) = 1 and T(n,k) = T(n1,k1) + 2*T(n1,k) for k<n.  Jon Perry, Oct 11 2012
The e.g.f. for the nth row is given by umbral composition of the normalized Laguerre polynomials A021009 as p(n,x) = L(n,L(.,x))/n! = 2^n L(n,x/2)/n!. E.g., L(2,x) = 24x+x^2, so p(2,x)= (1/2)L(2,L(.,x)) = (1/2)[2 L(0,x) + 4 L(1,x) + L(2,x)]= (1/2)[2 + 4(1+x) + (2+4x+x^2)] = 4 + 4x + x^2/2.  Tom Copeland, Oct 20 2012
From Tom Copeland, Oct 26 2012: (Start)
From the formalism of A132440 and A218272:
Let P and P^T be the Pascal matrix and its transpose and H= P^2= A038207.
Then with D the derivative operator,
exp[x*z/(12z)]/(12z)= exp(2z D_z z) e^(x*z)= exp(2D_x x D_x) e^(z*x)
= (1 z z^2 z^3 ...) H (1 x x^2/2! x^3/3! ...)^T
= (1 x x^2/2! x^3/3! ...) H^T (1 z z^2 z^3 ...)^T
= sum(n=0 to infn) z^n * 2^n Lag_n(x/2)= exp[z*EF(.,x)], an o.g.f. for the fvectors (rows) of A038207 where EF(n,x) is an e.g.f. for the nth fvector. (Lag_n(x) are the unnormalized Laguerre polynomials.)
Conversely,
exp(z*(2+x))= exp(2D_x) exp(x*z)= exp(2x) exp(x*z)
= (1 x x^2 x^3 ...) H^T (1 z z^2/2! z^3/3! ...)^T
= (1 z z^2/2! z^3/3! ...) H (1 x x^2 x^3 ...)^T
= exp[z*OF(.,x)], an e.g.f for the fvectors of A038207 where
OF(n,x)= (2+x)^n is an o.g.f. for the nth fvector.
(End)
G.f.: R(0)/2, where R(k) = 1 + 1/(1  (2*k+1+ (1+y))*x/((2*k+2+ (1+y))*x + 1/R(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Nov 09 2013
A038207 = exp[M*B(.,2)] where M = A238385I and (B(.,x))^n = B(n,x) are the Bell polynomials (cf. A008277). B(n,2) = A001861(n).  Tom Copeland, Apr 17 2014
T = (A007318)^2 = A112857*A167374 = A118801*A167374 = A118801*A167374 = P*A167374*P^(1)*A167374 = P*NpdP*A167374. Cf. A118801.  Tom Copeland, Nov 17 2016
E.g.f. for the nth subdiagonal, n = 0,1,2,..., equals exp(x)*P(n,x), where P(n,x) is the polynomial 2^n*Sum_{k = 0..n} binomial(n,k)*x^k/k!. For example, the e.g.f. for the third subdiagonal is exp(x)*(8 + 24*x + 12*x^2 + 4*x^3/3) = 8 + 32*x + 80*x^2/2! + 160*x^3/3! + ....  Peter Bala, Mar 05 2017


EXAMPLE

Triangle begins:
1;
2, 1;
4, 4, 1;
8, 12, 6, 1;
16, 32, 24, 8, 1;
32, 80, 80, 40, 10, 1;
...  corrected by Clark Kimberling, Aug 05 2011


MAPLE

for i from 0 to 12 do seq(binomial(i, j)*2^(ij), j = 0 .. i) end do; # yields sequence in triangular form  Emeric Deutsch, Nov 04 2007


MATHEMATICA

Table[CoefficientList[Expand[(y + x + x^2)^n], y] /. x > 1, {n, 0, 10}] // TableForm (* Geoffrey Critzer, Nov 20 2011 *)


PROG

(PARI) {T(n, k) = polcoeff((x+2)^n, k)}; /* Michael Somos, Apr 27 2000 */
(PARI) { n=13; v=vector(n); for (i=1, n, v[i]=vector(3^(i1))); v[1][1]=1; for (i=2, n, k=length(v[i1]); for (j=1, k, v[i][j]=v[i1][j]+i; v[i][j+k]=v[i1][j]+i; v[i][j+k+k]=v[i1][j]+i+1)); c=vector(n); for (i=1, n, for (j=1, 3^(i1), if (v[i][j]<=n, c[v[i][j]]++))); c } ( Jon Perry )
(Haskell)
a038207 n = a038207_list !! n
a038207_list = concat $ iterate ([2, 1] *) [1]
instance Num a => Num [a] where
fromInteger k = [fromInteger k]
(p:ps) + (q:qs) = p + q : ps + qs
ps + qs = ps ++ qs
(p:ps) * qs'@(q:qs) = p * q : ps * qs' + [p] * qs
_ * _ = []
 Reinhard Zumkeller, Apr 02 2011
(Haskell)
a038207' n k = a038207_tabl !! n !! k
a038207_row n = a038207_tabl !! n
a038207_tabl = iterate f [1] where
f row = zipWith (+) ([0] ++ row) (map (* 2) row ++ [0])
 Reinhard Zumkeller, Feb 27 2013
(Sage)
def A038207_triangle(dim):
M = matrix(SR, dim, dim)
for n in range(dim): M[n, n] = 1
for n in (1..dim1):
for k in (0..n1):
M[n, k] = M[n1, k1]+2*M[n1, k]
return M
A038207_triangle(9) # Peter Luschny, Sep 20 2012


CROSSREFS

Cf. A007318, A013609, A013610, etc.
See also A000079, A001787, A001788, A001789, A003472, A054849, A002409, A054851, A062715, A099089.
Cf. A135387.
Cf. A004211.  Gary W. Adamson, Feb 07 2010
Cf. A001861, A008277, A027465, A027466,A038231, A038243, A038255, A039683, A112857, A118801, A132440, A133156, A133314, A167374, A218272, A238385.
Sequence in context: A134397 A134395 * A065109 A113988 A134308 A209240
Adjacent sequences: A038204 A038205 A038206 * A038208 A038209 A038210


KEYWORD

nonn,tabl,easy,nice


AUTHOR

N. J. A. Sloane


STATUS

approved



