A331514
Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/(1 - 2*k*x + ((k-2)*x)^2)^(3/2).
Original entry on oeis.org
1, 1, 0, 1, 3, -6, 1, 6, 6, 0, 1, 9, 30, 10, 30, 1, 12, 66, 140, 15, 0, 1, 15, 114, 450, 630, 21, -140, 1, 18, 174, 1000, 2955, 2772, 28, 0, 1, 21, 246, 1850, 8430, 18963, 12012, 36, 630, 1, 24, 330, 3060, 18855, 69384, 119812, 51480, 45, 0
Offset: 0
Square array begins:
1, 1, 1, 1, 1, 1, ...
0, 3, 6, 9, 12, 15, ...
-6, 6, 30, 66, 114, 174, ...
0, 10, 140, 450, 1000, 1850, ...
30, 15, 630, 2955, 8430, 18855, ...
0, 21, 2772, 18963, 69384, 187425, ...
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T[n_, k_] = 1/2 * Sum[If[k == 2 && n == j - 1, 1, (k - 2)^(n + 1 - j)] * j * Binomial[n + 1, j] * Binomial[n + 1 + j, j], {j, 1, n + 1}]; Table[Table[T[n, k - n], {n, 0, k}], {k, 0, 9}] //Flatten (* Amiram Eldar, Jan 20 2020 *)
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T(n,k) = (1/2)*sum(j=1,n+1,(k-2)^(n+1-j)*j*binomial(n+1,j)*binomial(n+1+j,j));
matrix(7, 7, n, k, T(n-1, k-1)) \\ Michel Marcus, Jan 20 2020
A331792
Expansion of ((1 - 4*x)/sqrt(1 - 8*x + 4*x^2) - 1)/(6*x^2).
Original entry on oeis.org
1, 8, 57, 400, 2810, 19824, 140497, 999968, 7143966, 51206320, 368094122, 2652720096, 19159794004, 138658606688, 1005231020865, 7299082678336, 53074479789878, 386419850997552, 2816685368479342, 20553133273532000, 150120362670452076
Offset: 0
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a[n_] := Sum[3^k * Binomial[n + 1, k] * Binomial[n + 1, k + 1], {k, 0, n}]; Array[a, 21, 0] (* Amiram Eldar, May 05 2021 *)
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N=20; x='x+O('x^N); Vec(((1-4*x)/sqrt(1-8*x+4*x^2)-1)/(6*x^2))
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{a(n) = sum(k=0, n, 3^k*binomial(n+1, k)*binomial(n+1, k+1))}
A385728
Expansion of 1/((1-2*x) * (1-6*x))^(3/2).
Original entry on oeis.org
1, 12, 102, 760, 5310, 35784, 235788, 1530288, 9824310, 62557000, 395797908, 2491381776, 15616141996, 97537784400, 607391245080, 3772617319008, 23379854507046, 144605546475336, 892834113930180, 5504041611527760, 33883431379007364, 208327771987901808
Offset: 0
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R := PowerSeriesRing(Rationals(), 34); f := 1 / ((1 - 2*x) * (1 - 6*x))^(3/2); coeffs := [ Coefficient(f, n) : n in [0..33] ]; coeffs; // Vincenzo Librandi, Aug 22 2025
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Module[{a, n}, RecurrenceTable[{a[n] == ((8*n+4)*a[n-1] - 12*(n+1)*a[n-2])/n, a[0] == 1, a[1] == 12}, a, {n, 0, 25}]] (* Paolo Xausa, Aug 21 2025 *)
CoefficientList[Series[ 1/((1-2*x)*(1-6*x))^(3/2),{x,0,33}],x] (* Vincenzo Librandi, Aug 22 2025 *)
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my(N=30, x='x+O('x^N)); Vec(1/((1-2*x)*(1-6*x))^(3/2))
A387343
Expansion of 1/(1 - 8*x + 4*x^2)^(5/2).
Original entry on oeis.org
1, 20, 270, 3080, 31990, 312984, 2937900, 26751120, 237977190, 2078447800, 17884238372, 152002796400, 1278603975740, 10660760170480, 88213513627800, 725107271106336, 5925674432448390, 48175954959638520, 389871795632108020, 3142078444590396080, 25228464363569709396
Offset: 0
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R := PowerSeriesRing(Rationals(), 34); f := 1/(1 - 8*x + 4*x^2)^(5/2); coeffs := [ Coefficient(f, n) : n in [0..33] ]; coeffs; // Vincenzo Librandi, Aug 28 2025
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CoefficientList[Series[1/(1-8*x+4*x^2)^(5/2),{x,0,33}],x] (* Vincenzo Librandi, Aug 28 2025 *)
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my(N=30, x='x+O('x^N)); Vec(1/(1-8*x+4*x^2)^(5/2))
Showing 1-4 of 4 results.