cp's OEIS Frontend

A321201 Irregular triangle T with the nontrivial solutions of 2*e2 + 3*e3 = n, for n >= 2, with nonnegative e2 and e3, ordered as pairs with increasing e2 values.

%I A321201 #13 May 28 2025 08:54:56
%S A321201 1,0,0,1,2,0,1,1,0,2,3,0,2,1,1,2,4,0,0,3,3,1,2,2,5,0,1,3,4,1,0,4,3,2,
%T A321201 6,0,2,3,5,1,1,4,4,2,7,0,0,5,3,3,6,1,2,4,5,2,8,0,1,5,4,3,7,1,0,6,3,4,
%U A321201 6,2,9,0,2,5,5,3,8,1,1,6,4,4,7,2,10,0
%N A321201 Irregular triangle T with the nontrivial solutions of 2*e2 + 3*e3 = n, for n >= 2, with nonnegative e2 and e3, ordered as pairs with increasing e2 values.
%C A321201 The length of row n is 2*A(n), with A(n) = A008615(n+2) for n >= 2: 2*[1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, ...].
%C A321201 The trivial solution for n = 0 is [0, 0]. There is no solution for n = 1.
%C A321201 The row sums are given in A321202.
%C A321201 If a partition of n with parts 2 or 3 (with inclusive or) is written as 2^{e2} 3^{e3}, where e2 and e3 are nonnegative numbers, then in row n, all pairs [e2, e3] are given, for n >= 2, ordered with increasing values of e2.
%C A321201 The corresponding irregular triangle with the multinomial numbers n!/((n - (e2 + e3))!*e2!*e3!) is given in A321203. It gives the coefficients of x^n = x^{2*{e2} + 3*{e3}} of (1 + x^2 + x^3)^n, for n >= 2.
%F A321201 T(n, k) gives all pairs [e2, e3] solving 2*e2 + 3*e3 = n, ordered with increasing value of e2, for n >= 2. The trivial solution [0, 0] for n = 0 is not recorded. There is no solution for n = 1.
%e A321201 The triangle T(n, k) begins (pairs are separated by commas):
%e A321201   n\k  0  1   2  3   4  5   6  7 ...
%e A321201   2:   1  0
%e A321201   3:   0  1
%e A321201   4:   2  0
%e A321201   5:   1  1
%e A321201   6:   0  2,  3  0
%e A321201   7:   2  1
%e A321201   8:   1  2,  4  0
%e A321201   9:   0  3,  3  1
%e A321201   10:  2  2,  5  0
%e A321201   11:  1  3,  4  1
%e A321201   12:  0  4,  3  2,  6  0
%e A321201   13:  2  3,  5  1,
%e A321201   14:  1  4,  4  2,  7  0
%e A321201   15:  0  5,  3  3,  6  1
%e A321201   16:  2  4,  5  2,  8  0
%e A321201   17:  1  5,  4  3,  7  1
%e A321201   18:  0  6,  3  4,  6  2,  9  0
%e A321201   19:  2  5,  5  3,  8  1
%e A321201   20:  1  6,  4  4,  7  2, 10  0
%e A321201   ...
%e A321201 n=8: the two solutions of 2*e2 + 3*e3 = 8 are [e2, e3] = [1, 2] and = [4, 0], and 1 < 4, therefore row 8 is 1  2  4  0, with a comma after the first pair.
%t A321201 row[n_] := Reap[Do[If[2 e2 + 3 e3 == n, Sow[{e2, e3}]], {e2, 0, n/2}, {e3, 0, n/3}]][[2, 1]];
%t A321201 Table[row[n], {n, 2, 20}] // Flatten (* _Jean-François Alcover_, Nov 23 2018 *)
%Y A321201 Cf. A008615, A321202, A321203.
%K A321201 nonn,tabf
%O A321201 2,5
%A A321201 _Wolfdieter Lang_, Nov 05 2018
%E A321201 Missing row 2 inserted by _Jean-François Alcover_, Nov 23 2018