A377442
Square array read by rising antidiagonals: T(n, k) = A377441(-n, k), an extension of A377441 into the domain of negative n.
Original entry on oeis.org
1, 1, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 2, 4, 14, 1, 1, 2, 3, 9, 42, 1, 1, 2, 2, 6, 22, 132, 1, 1, 2, 1, 5, 12, 57, 429, 1, 1, 2, 0, 6, 6, 26, 154, 1430, 1, 1, 2, -1, 9, -2, 15, 59, 429, 4862, 1, 1, 2, -2, 14, -18, 24, 24, 138, 1223, 16796, 1, 1, 2, -3, 21, -48, 77, -23, 53, 332, 3550, 58786, 1, 1, 2, -4, 30, -98, 222, -226, 102, 107, 814, 10455, 208012, 1, 1, 2
Offset: 0
The array begins:
[ 0] 1, 1, 2, 5, 14, 42, 132, 429, 1430, ... = A000108
[-1] 1, 1, 2, 4, 9, 22, 57, 154, 429, ... = A105633
[-2] 1, 1, 2, 3, 6, 12, 26, 59, 138, ... = A152172
[-3] 1, 1, 2, 2, 5, 6, 15, 24, 53, ...
[-4] 1, 1, 2, 1, 6, -2, 24, -23, 102, ...
[-5] 1, 1, 2, 0, 9, -18, 77, -226, 765, ...
[-6] 1, 1, 2, -1, 14, -48, 222, -921, 3914, ...
[-7] 1, 1, 2, -2, 21, -98, 531, -2756, 14373, ...
Row index written as [m] is corresponding to A377441(m, k).
Cf.
A377441 (The main entry for this sequence).
Cf.
A000012 (Hankel transform of row 0),
A006720 (Hankel transform of row 1).
Cf.
A330025 (Hankel transform of row -1),
A328380 (Hankel transform of row -2).
A377443
Triangular array T(n,k) read by rows, satisfies A377441(n, k+2) = Sum_{m=0..k} T(k, m)*n^m.
Original entry on oeis.org
2, 5, 1, 14, 6, 1, 42, 27, 8, 1, 132, 111, 45, 10, 1, 429, 441, 222, 67, 12, 1, 1430, 1728, 1029, 382, 93, 14, 1, 4862, 6733, 4608, 2005, 599, 123, 16, 1, 16796, 26181, 20199, 10018, 3495, 881, 157, 18, 1, 58786, 101763, 87270, 48445, 19188, 5641, 1236, 195, 20, 1
Offset: 0
Triangle T(n, k) starts:
[0] 2
[1] 5, 1
[2] 14, 6, 1
[3] 42, 27, 8, 1
[4] 132, 111, 45, 10, 1
[5] 429, 441, 222, 67, 12, 1
[6] 1430, 1728, 1029, 382, 93, 14, 1
[7] 4862, 6733, 4608, 2005, 599, 123, 16, 1
[8] 16796, 26181, 20199, 10018, 3495, 881, 157, 18, 1
[9] 58786, 101763, 87270, 48445, 19188, 5641, 1236, 195, 20, 1
-
A377441(n, max_k) = Vec(-2*((n+1)*x-1)/((x-1)*(n*x-1)+((n*x^2-(n+1)*x+1)^2-4*x*(x-1)*((n+1)*x-1)+O(x^max_k))^(1/2)))
T(n, k) = Vec(A377441(y, n+5)[n+3])[n-k+1]
-
A091894(n, k) = 2^(n-2*k-1)*binomial(n-1, 2*k)*(binomial(2*k, k)/(k + 1))
A175136(n, k) = sum(m=0,(n - k)/2,A091894(n-k, m)*binomial(n-m-1, n-k))
T(n, k) = sum(m=1, n+1-k, A175136(n+2-k, n-m+2-k)*binomial(m+k-1, m-1))+(k==0)
A378125
Triangle T(n, k) read by rows. Let m be a nonzero rational number then T(n, m mod (n+1)) is the n-th coefficient in the Hasse-Weil L-series (q^(n+1) in the q-expansion) associated to the elliptic equation -4*x^3 + ((m+1)^2 + 8)*x^2 - 2*(m+3)*x + 1 - y^2 = 0.
Original entry on oeis.org
1, -1, -2, 0, -3, -1, 1, 2, 1, 2, -1, -2, -1, -3, 1, 0, 6, 1, 0, 3, 2, 1, -1, -3, 1, -2, -2, -2, -1, 0, -1, 0, -1, 0, -1, 0, 0, 6, -2, 0, 6, -2, 0, 6, -2, 1, 4, 1, 6, -1, 2, 2, 2, 3, -2, -1, -5, 4, 3, 1, -2, -4, -5, -3, -1, 1, 0, -6, -1, 0, -3, -2, 0, -6, -1, 0, -3, -2, 1, -2, -7, 0, 2, -2, -1, 0, -5, -2, -5, 3, 4, -1, 2, 3, -2, 2, 4, 2, -2, 1, 6, -1, 4, 2, 4
Offset: 0
The triangle T(n, k) begins:
q^(n+1) 0, 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 sum A001615
--------------------------------------------------------------------
[q^1] 1 1 1
[q^2] -1,-2 -3 3
[q^3] 0,-3,-1 -4 4
[q^4] 1, 2, 1, 2 6 6
[q^5] -1,-2,-1,-3, 1 6 6
[q^6] 0, 6, 1, 0, 3, 2 12 12
[q^7] 1,-1,-3, 1,-2,-2,-2 -8 8
[q^8] -1, 0,-1, 0,-1, 0,-1, 0 -4 12 <- not equal
[q^9] 0, 6,-2, 0, 6,-2, 0, 6,-2 12 12
[q^10] 1, 4, 1, 6,-1, 2, 2, 2, 3,-2 18 18
[q^11] -1,-5, 4, 3, 1,-2,-4,-5,-3,-1, 1 12 12
[q^12] 0,-6,-1, 0,-3,-2, 0,-6,-1, 0,-3,-2 -24 24
[q^13] 1,-2,-7, 0, 2,-2,-1, 0,-5,-2,-5, 3, 4 -14 14
[q^14] -1, 2, 3,-2, 2, 4, 2,-2, 1, 6,-1, 4, 2, 4 24 24
[q^15] 0, 6, 1, 0,-3, 1, 0, 3, 3, 0, 3, 2, 0, 9,-1 24 24
[q^16] 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4, 1,-4 24 24
Showing 1-3 of 3 results.
Comments