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Triangular numbers n for which A240542(n) = A240542(n-1).
5

%I #17 Mar 31 2023 07:09:37

%S 3,10,21,55,78,105,136,171,253,351,406,465,595,666,741,820,903,1081,

%T 1275,1378,1711,1830,1953,2211,2485,2628,2775,2926,3081,3403,3741,

%U 3916,4465,4656,5050,5253,5671,5886,6105,6328,7021,7503,7750,8001,8515,9045,9316,9591

%N Triangular numbers n for which A240542(n) = A240542(n-1).

%C Number n is in this sequence exactly when two parts of the symmetric representation of sigma(n) meet at the diagonal.

%C Proof: If n = k*(2*k+1) is in this sequence then the length of row n in A240542 is 2*k and that of row n-1 is 2*k-1, i.e., the last leg of the Dyck path for n down to the diagonal is vertical and that for n-1 is horizontal to the same point on the diagonal. Therefore, one part of the symmetric representation of sigma(n) ends at the diagonal and so does its symmetric copy. Conversely, if two parts meet at the diagonal then the number of legs in the Dyck path to the diagonal for n, i.e., the length of row n in A240542, is one larger than that for n-1 and must be even, i.e., n has the form n = k*(2*k+1).

%C A156592 is a subsequence since for every number of the form n = p*(2*p+1) where both p and 2*p+1 are primes A240542(n) = A240542(n-1). For a proof let T(n,k) = ceiling((n+1)/k - (k+1)/2) for 1 <= k <= floor((sqrt(8*n+1) - 1)/2) = 2*p, see A235791; then T(n,k) = T(n-1,k) + 1 for k = 1, 2, p, 2*p, and T(n,k) = T(n-1,k) for all other k. Therefore, the two alternating sums defining A240542(n) and A240542(n-1) are equal, i.e., their Dyck paths meet at the diagonal.

%C Except for missing 10 the intersection of this sequence and A298855 equals A156592. Sequence A262259 is a subsequence of this sequence.

%C The five known members of A191363 belong to this sequence, and since their symmetric representation consists of two parts of width one (the respective rows of triangle A237048 have the form 1 0 ... 0 1) they also belong to A262259.

%C Subsequence of A014105. - _Omar E. Pol_, Jan 31 2018

%C Second hexagonal numbers without middle divisors. - _Omar E. Pol_, Mar 10 2023

%e 3, 10 and 21 are in the sequence as the illustration of Dyck paths in sequence A237593 shows.

%e The sequence contains triangular numbers n*(2n+1) where neither n nor 2n+1 are prime. Numbers 1275=25*51 and 2926=38*77 are examples, however, 36 = 4*9 does not belong to the sequence.

%e 78 is the first number in the sequence whose two parts of its symmetric representation contain pieces of width two.

%t (* Function path[] is defined in A237270 *)

%t meetAtDiagonalQ[n_] := Module[{diags=Transpose[{Drop[Drop[path[n], 1], -1], path[n-1]}]}, Length[Union[diags[[n]]]]==1 && First[diags[[n-1]]]!=Last[diags[[n-1]]]]

%t a298856[m_, n_] := Select[Map[#(2#+1)&, Range[m, n]], meetAtDiagonalQ]

%t a298856[1, 70] (* data *)

%Y Cf. A000217, A156592, A191363, A237048, A237270, A237593, A240542, A262259, A298855.

%Y Cf. A067742.

%K nonn

%O 1,1

%A _Hartmut F. W. Hoft_, Jan 27 2018