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Determinant of rank n matrix of 1..n^2 filled successively back and forth along antidiagonals.
+10
9
1, -2, 15, -594, -5187, 23244, 122475, -279292, -1157143, 1850930, 6642839, -8529278, -27810555, 30741424, 93575187, -92784984, -268191855, 244875462, 679807583, -581798410, -1563707379, 1270245108, 3324627195, -2587197204, -6623079687, 4972012474, 12491212135
OFFSET
1,2
COMMENTS
The matrix is formed by writing numbers 1 .. n^2 in zig-zag pattern as shown in examples below. Every other antidiagonal reads backwards from A069480.
Whereas each antidiagonal of A069480 begins with one more than a triangular number and ends with the next triangular number, here every other antidiagonal begins with one more than a triangular number and the next antidiagonal begins with a triangular number.
The trace of the matrix is the sequence A006003 (proved). - Stefano Spezia, Aug 07 2018
The matrix is defined by A[i,j] = (2 - i - j)*((i + j - 1) mod 2)+(j^2 + (2*i - 1)*j + i^2 - i)/2 + (j - 1)*(1 - 2*((i + j) mod 2)) if i + j <= n + 1 and A[i,j] = n^2 - ((4*n^2 + (- 4*j - 4*i + 6)*n + j^2 + (2*i - 3)*j + i^2 - 3*i + 2)/2 + (i + j - 2*n)*((2*n - i - j + 1) mod 2)) + 1 - (n - j)*(1 - 2*((i + j) mod 2)) if i + j > n + 1 (proved). - Stefano Spezia, Aug 11 2018
LINKS
Index entries for linear recurrences with constant coefficients, signature (0,-9,0,-36,0,-84,0,-126,0,-126,0,-84,0,-36,0,-9,0,-1).
FORMULA
From Vaclav Kotesovec, Jan 08 2019: (Start)
Recurrence: (5*n^16 - 176*n^15 + 2888*n^14 - 29332*n^13 + 206454*n^12 - 1068276*n^11 + 4205934*n^10 - 12861022*n^9 + 30891328*n^8 - 58524140*n^7 + 87229074*n^6 - 101275380*n^5 + 89823673*n^4 - 58824210*n^3 + 26795412*n^2 - 7559784*n + 985608)*a(n) = 8*(n^14 - 20*n^13 + 169*n^12 - 754*n^11 + 1630*n^10 + 564*n^9 - 15184*n^8 + 52244*n^7 - 109015*n^6 + 167071*n^5 - 202816*n^4 + 191592*n^3 - 125145*n^2 + 45333*n - 5832)*a(n-1) - (5*n^16 - 96*n^15 + 848*n^14 - 4580*n^13 + 16966*n^12 - 45892*n^11 + 94310*n^10 - 151266*n^9 + 192520*n^8 - 195196*n^7 + 155666*n^6 - 94052*n^5 + 39329*n^4 - 6798*n^3 - 4572*n^2 + 5400*n - 1944)*a(n-2).
a(n) ~ ((-1)^n - 3) * (cos(Pi*n/2) + sin(Pi*n/2)) * n^8 / 72. (End)
a(n) = -9*a(n-2) - 36*a(n-4) - 84*a(n-6) - 126*a(n-8) - 126*a(n-10) - 84*a(n-12) - 36*a(n-14) - 9*a(n-16) - a(n-18) for n > 18. - Stefano Spezia, Apr 25 2021, simplified by Boštjan Gec, Sep 21 2023
EXAMPLE
n=2, det=-2: {1 2 / 3 4 }
n=3, det=15: {1 2 6 / 3 5 7 / 4 8 9 }
n=4, det=-594: { 1 2 6 7 / 3 5 8 13 / 4 9 12 14 / 10 11 15 16 }
n=5, det=-5187: { 1 2 6 7 15 / 3 5 8 14 16 / 4 9 13 17 22 / 10 12 18 21 23 / 11 19 20 24 25 }
MATHEMATICA
a[i_, j_, n_] := If[i+j<=n+1, (2-i-j)*Mod[i+j-1, 2]+(j^2+(2*i-1)*j+i^2-i)/2+(j-1)*(1-2*Mod[i+j, 2]), n^2-((4*n^2+(-4*j-4*i+6)*n+j^2+(2*i-3)*j+i^2-3*i+2)/2+(i+j-2*n)*Mod[2*n-i-j+1, 2])+1-(n-j)*(1-2*Mod[i+j, 2])]; f[n_] := Det[ Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 27] (* Stefano Spezia, Aug 11 2018 *)
PROG
(PARI) A(i, j, n) = if (i + j <= n + 1, (2 - i - j)*((i + j - 1) % 2)+(j^2 + (2*i - 1)*j + i^2 - i)/2 + (j - 1)*(1 - 2*((i + j) % 2)), n^2 - ((4*n^2 + (- 4*j - 4*i + 6)*n + j^2 + (2*i - 3)*j + i^2 - 3*i + 2)/2 + (i + j - 2*n)*((2*n - i - j + 1) % 2)) + 1 - (n - j)*(1 - 2*((i + j) % 2)));
a(n) = matdet(matrix(n, n, i, j, A(i, j, n))); \\ Michel Marcus, Aug 11 2018
(MATLAB, FreeMat and Octave)
for(n=1:27)
A=zeros(n, n);
for(i=1:n)
for(j=1:n)
if(i+j<=n+1)
A(i, j)=(2-i-j)*mod(i+j-1, 2)+(j^2+(2*i-1)*j+i^2-i)/2+(j-1)*(1-2*mod(i+j, 2));
else
A(i, j)=n^2-((4*n^2+(-4*j-4*i+6)*n+j^2+(2*i-3)*j+i^2-3*i+2)/2+(i+j-2*n)*mod(2*n-i-j+1, 2))+1-(n-j)*(1-2*mod(i+j, 2));
end
end
end
fprintf('%d %0.f\n', n, det(A));
end # Stefano Spezia, Aug 12 2018
(GAP)
A078475 := function(k)
local i, j, n;
for n in [1 .. k] do
A:=NullMat(n, n);
for i in [1 .. n] do
for j in [1 .. n] do
if i+j<=n+1 then
A[i][j] := (2-i-j)*RemInt(i+j-1, 2)+(j^2+(2*i-1)*j+i^2-i)/2+(j-1)*(1-2*RemInt(i+j, 2));;
else
A[i][j] := n^2-((4*n^2+(-4*j-4*i+6)*n+j^2+(2*i-3)*j+i^2-3*i+2)/2+(i+j-2*n)*RemInt(2*n-i-j+1, 2))+1-(n-j)*(1-2*RemInt(i+j, 2));
fi;
od;
od;
Print(n, " ", Determinant(A), "\n");
od;
end;
A078475(27); # Stefano Spezia, Aug 12 2018
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Kit Vongmahadlek (kit119(AT)yahoo.com), Jan 03 2003
EXTENSIONS
Edited and extended by Robert G. Wilson v, May 08 2003
STATUS
approved
Determinant of the n X n matrix in which the entries are 1 through n^2, spiraling inward starting with 1 in the (1,1)-entry.
+10
6
1, -5, -48, 660, 11760, -257040, -6652800, 198918720, 6745939200, -255826771200, -10727081164800, 492775291008000, 24610605962342400, -1327677426915840000, -76940526008586240000, 4766815315895592960000, 314406967644177408000000, -21995911456386651463680000
OFFSET
1,2
LINKS
Sergey Sadov, Problem 11270, American Mathematical Monthly, Vol. 114, No. 1, 2007, p. 78.
FORMULA
a(n) = (-1)^[n*(n-1)/2]*2^(2*n-3)*(3*n-1)*Product_{k=0..n-2} (1/2+k) for n>=2.
E.g.f.: (((-16*x^2-1)*sqrt(2*sqrt(16*x^2+1)+2)-8*sqrt(16*x^2+1)*x^2+16*x^2 + sqrt(16*x^2+1)+1)*sqrt(2*sqrt(16*x^2+1)-2)+(8*(sqrt(16*x^2+1)*x^2+2*x^2-(1/8) * sqrt(16*x^2+1)+1/8))*sqrt(2*sqrt(16*x^2+1)+2))/(512*x^3+32*x). - Robert Israel, Apr 20 2017
EXAMPLE
For n = 2, the 2 X 2 (spiral) matrix A is
[1, 2]
[4, 3]
Then a(2) = -5 because det(A) = 1*3 - 2*4 = -5.
MAPLE
a:=n->(-1)^(n*(n-1)/2)*2^(2*n-3)*(3*n-1)*product(1/2+k, k=0..n-2): seq(a(n), n=1..20);
# second Maple program:
a:= proc(n) option remember; `if`(n<2, (3*n+1)/4,
4*(1-3*n)*(2*n-5)*(2*n-3) *a(n-2) /(3*n-7))
end:
seq(a(n), n=1..20); # Alois P. Heinz, Jan 21 2014
MATHEMATICA
a[n_] := (-1)^(n*(n-1)/2)*2^(2n-3)*(3n-1)*Pochhammer[1/2, n-1]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, May 26 2015 *)
CROSSREFS
Cf. A023999 (absolute values). - Alois P. Heinz, Jan 21 2014
KEYWORD
sign
AUTHOR
Emeric Deutsch, Dec 31 2006
STATUS
approved
Absolute value of determinant of n X n matrix whose entries are the integers from 1 to n^2 spiraling outward, ending in a corner.
+10
3
1, 5, 72, 1380, 31920, 861840, 26611200, 925404480, 35805369600, 1526139014400, 71066912716800, 3590219977344000, 195589552648089600, 11430978821982720000, 713448513897799680000, 47363888351558338560000
OFFSET
1,2
COMMENTS
If n == 0 or 1 (mod 4), the sign of the determinant will be independent of the orientation of the spiral. For n == 2 or 3 (mod 4), the sign will be reversed when the orientation is rotated by 1/4 or flipped on the horizontal or vertical axis. - Franklin T. Adams-Watters, Dec 31 2013
This distribution of the integers is sometimes known as Ulam's spiral, although that is sometimes reserved for when the primes are marked out in some way. - Franklin T. Adams-Watters, Dec 31 2013
LINKS
Gaurav Bhatnagar, Christian Krattenthaler, Spiral determinants, arXiv:1704.02859 [math.CO], 2017.
FORMULA
a(n) = (2*n^2-3*n+3) (2n-2)!/(2 (n-1)!) = A096376(n-1)*A000407(n-2), n>1. - Conjectured by Dean Hickerson, Jan 30 2003. Proved in the article by Bhatnagar and Krattenthaler.
D-finite with recurrence (2*n^2-7*n+8)*a(n) -2*(2*n-3)*(2*n^2-3*n+3)*a(n-1)=0. - R. J. Mathar, May 03 2019
EXAMPLE
n=2, det=-5: {1 2 / 4 3 }.
n=3, det=72: {7 8 9 / 6 1 2 / 5 4 3 }.
n=4, det=-1380: { 7 8 9 10 / 6 1 2 11 /5 4 3 12 / 16 15 14 13 }.
n=5, det=31920: { 21 22 23 24 25 / 20 7 8 9 10 / 19 6 1 2 11 /18 5 4 3 12 / 17 16 15 14 13 }
MATHEMATICA
M[0, 0] = 1;
M[i_, j_] := If[i <= j,
If[i + j >= 0, If[i != j, M[i + 1, j] + 1, M[i, j - 1] + 1],
M[i, j + 1] + 1],
If[i + j > 1, M[i, j - 1] + 1, M[i - 1, j] + 1]
]
M[n_] := If[EvenQ[n],
Table[M[i, j], {j, n/2, -n/2 + 1, -1}, {i, -n/2 + 1, n/2}],
Table[M[i, j], {j, (n - 1)/2, -(n - 1)/2, -1}, {i, -(n - 1)/2, (n - 1)/2}]]
a[n_]:=Det[M[n]] (* Christian Krattenthaler, Apr 19 2017 *)
PROG
(Maxima) A079340(n):=if n=1 then 1 else (2*n^2-3*n+3)*(2*n-2)!/(2*(n-1)!)$
makelist(A079340(n), n, 1, 30); /* Martin Ettl, Nov 05 2012 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Kit Vongmahadlek (kit119(AT)yahoo.com), Jan 03 2003
EXTENSIONS
Extended by Robert G. Wilson v, Jan 25 2003
STATUS
approved
Array read by antidiagonals: a(i,j) is the number of ways of labeling a tableau of shape (i,1^j) with the integers 1, 2, ... i+j-2 (each label being used once) such that the first row is decreasing, and the first column has m-1 labels.
+10
2
1, 3, 1, 12, 5, 1, 60, 27, 7, 1, 360, 168, 48, 9, 1, 2520, 1200, 360, 75, 11, 1, 20160, 9720, 3000, 660, 108, 13, 1, 181440, 88200, 27720, 6300, 1092, 147, 15, 1, 1814400, 887040, 282240, 65520, 11760, 1680, 192, 17, 1, 19958400, 9797760, 3144960, 740880, 136080, 20160, 2448, 243, 19, 1
OFFSET
1,2
COMMENTS
For an arbitrary composition c, let F_c^p denote the linear transformation of NSym that is adjoint to multiplication by the fundamental quasi-symmetric function indexed by c. Then a(i,j) equals the coefficient of H_(1,1) in (F_(1)^p)^(i+j-2)(H_(i,1^j)) (see below SAGE program, and Corollary 2.7 in the below link).
Let M(n) = [a(i,j)]_{n x n}. Then det(M(n))=A000178(n)=the n-th superfactorial.
Let p_n(x) denote the polynomial such that a(x,n)=p_n(x). Then the coefficient of x in p_n(x) is |A009575(n)|. For example, p_4(x)=4x^3+18x^2+26x+12, and the coefficient of x in p_4(x) is |A009575(4)|=26.
First row is A001710. Second row is A138772. Fourth row is A136659.
LINKS
C. Berg, N. Bergeron, F. Saliola, L. Serrano, and M. Zabrocki, A Lift of the Schur and Hall-Littlewood Bases to Non-Commutative Symmetric Functions, 10-11.
FORMULA
a(i,j) = (i+j-2)!/i!*(2*i+j-1)*j/2.
EXAMPLE
There are a(3,2) = 7 ways of labeling the tableau of shape (3,1,1) with 1, 2 and 3 (with each label being used once) such that the first row is decreasing and the first column has 1 label:
1 2 3 X X X X
X X X 1 2 3 X
X32 X31 X21 X32 X31 X21 321
The matrix [a(i,j)]_(6 x 6) is given below:
[1 3 12 60 360 2520]
[1 5 27 168 1200 9720]
[1 7 48 360 3000 27720]
[1 9 75 660 6300 65520]
[1 11 108 1092 11760 136080]
[1 13 147 1680 20160 257040]
MAPLE
a:= (i, j)-> (i+j-2)!/i!*(2*i+j-1)*j/2:
seq(seq(a(i, 1+d-i), i=1..d), d=1..12); # Alois P. Heinz, Jan 21 2014
MATHEMATICA
a[n_, k_]:=(n+k-2)!/n!*(2*n+k-1)*k/2 ;
Print[Array[a[#1, #2]&, {50, 50}]//MatrixForm]
(* A program which gives a list of tableaux *)
a[i_, j_] := Module[{f, list1, el, emptylist, n},
f[q_] := StringReplace[StringReplace[StringReplace[ StringReplace[ToString[q], ToString[i + j - 1] -> "X"], ", " -> ""], "{" -> ""], "}" -> ""]; list1 = Permutations[Join[Table[q, {q, 1, i + j - 2}], {i + j - 1, i + j - 1}]]; el[q_] := First[Take[list1, {q, q}]]; emptylist = {}; n = 1; While[n < 1 + Length[list1], If[Take[el[n], {j + 1, i + j}] == Sort[Take[el[n], {j + 1, i + j}], Greater] && Count[Take[el[n], {1, j + 1}], i + j - 1] == 2, emptylist = Append[emptylist, f[el[n]]], Null]; n++]; Print[emptylist]]
PROG
(Sage)
NSym = NonCommutativeSymmetricFunctions(QQ) ;
QSym = QuasiSymmetricFunctions(QQ) ;
F = QSym.Fundamental() ;
H = NSym.complete() ;
def a(n, m):
expr = H([n]+[1 for q in range(m)]) ;
w=1 ;
while w<n+m-1:
expr = expr.skew_by(F[1])
w+=1
return(expr.coefficients()[0])
print(matrix([[a(j+1, i+1) for i in range(7)] for j in range (7)]))
list1=[] ;
n=0 ;
while n<10:
list1 = list1 + [a(i+1, n+1-i) for i in range(n+1)]
n+=1
print(list1)
CROSSREFS
Main diagonal gives: A023999. - Alois P. Heinz, Jan 21 2014
KEYWORD
nonn,tabl
AUTHOR
John M. Campbell, May 29 2013
STATUS
approved
Number of support Tau-tilting modules for some algebras.
+10
1
3, 5, 12, 33, 98, 306, 990, 3289, 11154, 38454, 134368, 474810, 1693812, 6091780, 22064130, 80410185, 294647250, 1084922190, 4012165080, 14895504030, 55496654460, 207431394300, 777601790940, 2922867908298, 11013796950228, 41596652545756, 157434454904160, 597029454416724, 2268232385053096
OFFSET
0,1
COMMENTS
See Prop. A.6 in Wang's reference for the table counting Tau-tilting modules for the linear quiver modulo the relation alpha*beta = 0.
LINKS
Qi Wang, Tau-tilting finite simply connected algebras, arXiv:1910.01937 [math.RT], 2019-2022.
FORMULA
a(n) = 3*(3*n+2)*binomial(2*n+4,n+2)/(4*(2*n+1)*(2*n+3)).
a(n) = A329533(n)/(n + 1).
From Peter Luschny, Sep 13 2024: (Start)
a(n) = (3*n + 2) * [x^n] ((1 - 4*x)^(3/2) + 12*x - 2)/(4*x^2).
a(n) = A016789(n)*(3/2)*(2*n)! * [x^(2*n)] hypergeom([], [3], x^2).
a(n) = CatalanNumber(n)*(9*n + 6)/(n + 2).
a(n) = -(3*n + 2)*(-4)^(n + 1)*binomial(3/2, n + 2).
a(n) = 2^n*(9*n + 6)*(2*n - 1)!! / (n + 2)!.
a(n) = A007054(n) * (3*n + 2) / 2.
a(n) = 6*A023999(n + 1)/(n + 2)!. (End)
MAPLE
a := n -> -(3*n + 2)*(-4)^(n + 1)*binomial(3/2, n + 2):
seq(a(n), n = 0..28) # Peter Luschny, Sep 13 2024
MATHEMATICA
A376161[n_] := CatalanNumber[n]*(9*n + 6)/(n + 2);
Array[A376161, 30, 0] (* Paolo Xausa, Sep 14 2024 *)
PROG
(Sage)
def a(n):
return 3*(3*n+2)*binomial(2*n+4, n+2)/4/(2*n+1)/(2*n+3)
KEYWORD
nonn
AUTHOR
F. Chapoton, Sep 13 2024
STATUS
approved

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