A213505 Rectangular array: (row n) = b**c, where b(h) = h^2, c(h) = (n-1+h)^2, n>=1, h>=1, and ** = convolution.
1, 8, 4, 34, 25, 9, 104, 88, 52, 16, 259, 234, 170, 89, 25, 560, 524, 424, 280, 136, 36, 1092, 1043, 899, 674, 418, 193, 49, 1968, 1904, 1708, 1384, 984, 584, 260, 64, 3333, 3252, 2996, 2555, 1979, 1354, 778, 337, 81, 5368, 5268, 4944, 4368, 3584
Offset: 1
Examples
Northwest corner (the array is read by falling antidiagonals): 1....8.....34....104...259....560 4....25....88....234...524....1043 9....52....170...424...899....1708 16...89....280...674...1384...2555 25...136...418...984...1979...3584 ... T(5,1) = (1)**(25) = 25 T(5,2) = (1,4)**(25,36) = 1*36+4*25 = 136 T(5,3) = (1,4,9)**(25,36,49) = 1*49+4*36+9*25 = 418
Links
- Clark Kimberling, Antidiagonals n = 1..60, flattened
- Henri Muehle, Proper Mergings of Stars and Chains are Counted by Sums of Antidiagonals in Certain Convolution Arrays -- The Details, arXiv preprint arXiv:1301.1654, 2013.
Crossrefs
Cf. A213500.
Programs
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Mathematica
b[n_] := n^2; c[n_] := n^2 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}] (* A213505 *) d = Table[t[n, n], {n, 1, 40}] (* A213546 *) s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}] s1 = Table[s[n], {n, 1, 50}] (* A213547 *)
Formula
T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) - T(n,k-6).
G.f. for row n: f(x)/g(x), where f(x) = n^2 - (n^2 - 2*n - 1)*x - (n^2 - 2)*x^2 - ((n - 1)^2)*x^3 and g(x) = (1 - x)^6.
Comments