|
%I #49 Mar 08 2022 03:02:47
%S 0,1,3,6,10,55,66,666
%N Triangular numbers k such that every permutation of the digits of k is a triangular number.
%C I assume this sequence is finite.
%C From _Jon E. Schoenfield_, Mar 05 2022: (Start)
%C If k is a term of this sequence and is not a repdigit, then at least one of its permutations (a triangular number, call it k1) will have a digit d0 in its ones place and a digit d1 != d0 in its tens place, and the number that results from reversing those last two digits will be a triangular number k2 = k1 + 9*(d0 - d1), so the two distinct triangular numbers k2 and k1 will differ by no more than 9*(9-0) = 81. But the j-th triangular number T(j) differs from the nearest other triangular number by T(j) - T(j-1) = j, so if two distinct triangular numbers T(k1) > T(k2) differ by no more than 81, then k1 <= 81. No triangular numbers <= T(81) = 3321 are terms that exceed 666, so if there exists any term > 666, it is a repdigit triangular number.
%C However, Ballew and Weger proved (see A045914) that the only repdigit triangular numbers are 0, 1, 3, 6, 55, 66, and 666. Thus, 666 is the last term of this sequence. (End)
%e 10 = 4*5/2, 01 = 1*2/2.
%t triQ[n_] := IntegerQ @ Sqrt[8*n + 1]; Select[Range[0, 1000], AllTrue[Permutations[ IntegerDigits[#] ], triQ[FromDigits[#1]] &] &] (* _Amiram Eldar_, Apr 30 2021 *)
%t pdtQ[n_]:=AllTrue[FromDigits/@Permutations[IntegerDigits[n]],OddQ[ Sqrt[ 8#+1]]&]; Select[Accumulate[Range[0,5000]],pdtQ] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Jul 04 2021 *)
%Y Cf. A000217, A003459, A045914, A077671.
%K nonn,base,fini,full
%O 1,3
%A _Ctibor O. Zizka_, Apr 30 2021
|
