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A332589
Let t_k denote the triangular number k*(k+1)/2. Suppose 0 < x < y < z are integers satisfying t_x + t_y = t_p, t_y + t_z = t_q, t_x + t_z = t_r, for integers p,q,r. Sort the triples [x,y,z] first by x, then by y. Sequence gives the values of y.
2
13, 51, 50, 30, 124, 65, 123, 119, 245, 69, 143, 244, 99, 135, 426, 195, 459, 90, 125, 125, 425, 284, 679, 678, 678, 170, 363, 551, 429, 1016, 1547, 174, 245, 154, 1015, 602, 1449, 260, 1859, 398, 579, 469, 1448, 1007, 305, 1275, 1990, 1989, 1044, 1575, 390, 429
OFFSET
1,1
COMMENTS
Ulas gives a table assuming 0 < x < y < z < 1000. Because of the assumption z < 1000, only the entries with x < 46 can be relied upon (above this it is possible that there are gaps in the table).
LINKS
Ulas Maciej, A note on Sierpinski's problem related to triangular numbers, arXiv:0810.0222 [math.NT], 2008. See Table 1.
Ulas Maciej, A note on Sierpinski's problem related to triangular numbers, Colloq. Math. 117 (2009), no. 2, 165-173. See MR2550124. See Table 1.
EXAMPLE
The initial values of x, y, z, p, q, r are:
x y z p q r
-- --- --- --- ---- ---
9 13 44 16 46 45
14 51 104 53 116 105
20 50 209 54 215 210
23 30 90 38 95 93
27 124 377 127 397 378
35 65 86 74 108 93
35 123 629 128 641 630
41 119 285 126 309 288
44 245 989 249 1019 990
...
MATHEMATICA
L = {}; t[n_] := n (n + 1)/2; Do[ syp = Solve[t[x] + t[y] == t[p] && p > 0 && y > x , {p, y}, Integers]; If[syp =!= {}, Do[{y1, p1} = {y, p} /. e; s = Solve[ t[y1] + t[z] == t[q] && t[x] + t[z] == t[r] && q > 0 && z > y1 && r > 0, {z, q, r}, Integers]; If[s =!= {}, L = Join[L, {x, y1, z, p1, q, r} /. s]], {e, syp}]], {x, 54}]; Sort[L][[All, 2]] (* Giovanni Resta, Mar 02 2020 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Feb 29 2020
EXTENSIONS
Terms a(10) and beyond from Giovanni Resta, Mar 02 2020
STATUS
approved