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 A120919 Cascadence of (1+x)^3; a triangle, read by rows of 3n+1 terms, that retains its original form upon convolving each row with [1,3,3,1] and then letting excess terms spill over from each row into the initial positions of the next row such that only 3n+1 terms remain in row n for n>=0. 6
 1, 3, 3, 1, 3, 12, 19, 18, 15, 10, 3, 12, 55, 111, 138, 128, 96, 66, 55, 39, 12, 55, 276, 636, 930, 1005, 876, 669, 498, 360, 263, 240, 177, 55, 276, 1464, 3666, 5979, 7317, 7242, 6138, 4737, 3506, 2607, 2046, 1569, 1212, 1170, 883, 276, 1464, 8058, 21369, 37716 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 LINKS FORMULA G.f.: A(x,y) = ( x*H(x) - y*H(x*y^3) )/( x*(1+y)^3 - y ), where H(x) satisfies: H(x) = G*H(x^4*G^3) and G(x) is g.f. of A001764: G(x) = 1 + x*G(x)^3. EXAMPLE Triangle begins: 1; 3, 3, 1, 3; 12, 19, 18, 15, 10, 3, 12; 55, 111, 138, 128, 96, 66, 55, 39, 12, 55; 276, 636, 930, 1005, 876, 669, 498, 360, 263, 240, 177, 55, 276; Convolution of [1,3,3,1] with each row produces: [1,3,3,1]*[1] = [1,3,3,1]; [1,3,3,1]*[3,3,1,3] = [3,12,19,18,15,10,3]; [1,3,3,1]*[12,19,18,15,10,3,12] = [12,55,111,138,128,96,66,55,39,12]; [1,3,3,1]*[55,111,138,128,96,66,55,39,12,55] = [55,276,636,930,1005,876,669,498,360,263,240,177,55]; These convoluted rows, when concatenated, yield the sequence: 1,3,3,1, 3,12,19,18,15,10,3, 12,55,111,138,128,96,66,55,39,12, 55,... which equals the concatenated rows of this original triangle: 1, 3,3,1,3, 12,19,18,15,10,3,12, 55,111,138,128,96,66,55,39,12,55, ... MATHEMATICA T[n_, k_] := T[n, k] = If[3*n < k || k < 0, 0, If[n == 0 && k == 0, 1, If[k == 3*n, T[n, 0], T[n - 1, k + 1] + 3*T[n - 1, k] + 3*T[n - 1, k - 1] + T[n - 1, k - 2]]]]; Table[T[n, k], {n, 0, 10}, {k, 0, 3 n}] // Flatten (* Jean-François Alcover, Jan 24 2018, translated from PARI *) PROG (PARI) /* Generate Triangle by the Recurrence: */ {T(n, k)=if(3*n

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Last modified September 28 16:38 EDT 2021. Contains 347716 sequences. (Running on oeis4.)