A213500
Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.
Original entry on oeis.org
1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90
Offset: 1
Northwest corner (the array is read by southwest falling antidiagonals):
1, 4, 10, 20, 35, 56, 84, ...
2, 7, 16, 30, 50, 77, 112, ...
3, 10, 22, 40, 65, 98, 140, ...
4, 13, 28, 50, 80, 119, 168, ...
5, 16, 34, 60, 95, 140, 196, ...
6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.
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b[n_] := n; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}] (* A213500 *)
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t(n,k) = sum(i=0, k - 1, (k - i) * (n + i));
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k,n - k + 1),", ");); print(););};
tabl(12) \\ Indranil Ghosh, Mar 26 2017
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def t(n, k): return sum((k - i) * (n + i) for i in range(k))
for n in range(1, 13):
print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017
A067418
Triangle A067330 with rows read backwards.
Original entry on oeis.org
1, 2, 1, 5, 3, 2, 10, 7, 5, 3, 20, 15, 12, 8, 5, 38, 30, 25, 19, 13, 8, 71, 58, 50, 40, 31, 21, 13, 130, 109, 96, 80, 65, 50, 34, 21, 235, 201, 180, 154, 130, 105, 81, 55, 34, 420, 365, 331, 289, 250, 210, 170, 131, 89, 55, 744, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1308, 1164, 1075, 965
Offset: 0
{1}; {2,1}; {5,3,2}; {10,7,5,3}; ...; p(2,n)=5+3*x+2*x^2.
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Reverse /@ Table[Sum[Fibonacci[k + 1] Fibonacci[n - k + 1], {k, 0, m}], {n, 0, 11}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)
A104509
Matrix inverse of triangle A104505, which is the right-hand side of triangle A084610 of coefficients in (1 + x - x^2)^n.
Original entry on oeis.org
1, 1, -1, 3, -2, 1, 4, -6, 3, -1, 7, -12, 10, -4, 1, 11, -25, 25, -15, 5, -1, 18, -48, 60, -44, 21, -6, 1, 29, -91, 133, -119, 70, -28, 7, -1, 47, -168, 284, -296, 210, -104, 36, -8, 1, 76, -306, 585, -699, 576, -342, 147, -45, 9, -1, 123, -550, 1175, -1580, 1485, -1022, 525, -200, 55, -10, 1, 199, -979, 2310, -3454, 3641
Offset: 0
Rows begin:
1;
1, -1;
3, -2, 1;
4, -6, 3, -1;
7, -12, 10, -4, 1;
11, -25, 25, -15, 5, -1;
18, -48, 60, -44, 21, -6, 1;
29, -91, 133, -119, 70, -28, 7, -1;
47, -168, 284, -296, 210, -104, 36, -8, 1;
76, -306, 585, -699, 576, -342, 147, -45, 9, -1; ...
- Robert Israel, Table of n, a(n) for n = 0..10152 (rows 0 to 141, flattened).
- P. Peart and W.-J. Woan, A divisibility property for a subgroup of Riordan matrices, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
- Wikipedia, Lucas polynomials.
Leftmost column is
A000204 (Lucas numbers). Other columns include:
A045925,
A067988. Row sums are: {1,0,2,0,2,0,2,...}. Absolute row sums form:
A099425. Antidiagonal sums are: {1,1,2,2,2,2,2,...}. Absolute antidiagonal sums are:
A084214.
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S:= series((1 + x^2)/(1-x-x^2 + x*y),x, 20):
for n from 0 to 19 do R[n]:= coeff(S,x,n) od:
seq(seq(coeff(R[n],y,j),j=0..n), n=0..19); # Robert Israel, Jun 30 2015
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nmax = 11;
T[n_, k_] := Coefficient[(1 + x - x^2)^n, x, n + k];
M = Table[T[n, k], {n, 0, nmax}, {k, 0, nmax}] // Inverse;
Table[M[[n+1, k+1]], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 27 2019 *)
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{ T(n,k) = my(X=x+x*O(x^n),Y=y+y*O(y^k)); polcoeff(polcoeff((1 + X^2)/(1-X-X^2 + X*Y),n,x),k,y); }
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{ tabl(nn) = my(m = matrix(nn, nn, n, k, n--; k--; if((nMichel Marcus, Jun 30 2015
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{ A104509(n,k) = if(n==0, k==0, (-1)^k * sum(i=0, (n-k)\2, n/(n-i) * binomial(n-k-i,i) * binomial(n-i,k) )); } \\ Max Alekseyev, Oct 11 2021
Showing 1-3 of 3 results.
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