A262259 Numbers k such that the symmetric representation of sigma(k) has only two parts and they meet at the center of the Dyck path.

3, 10, 78, 136, 666, 820, 1830, 2628, 4656, 5886, 6328, 16290, 18528, 28920, 32896, 39340, 48828, 56616, 62128, 78606, 80200, 83436, 88410, 93528, 100576, 104196, 135460, 146070, 166176, 180300, 187578, 190036

Offset: 1

Author

Hartmut F. W. Hoft, Sep 16 2015

Keywords

• nonn
• more

Comments

For a proof of the formula see the link and also the links in A239929 and A071561. This formula allows for a fast computation of numbers in the sequence that does not require computations of Dyck paths.

Subsequence of A239929.

A191363 is a subsequence.

All terms are triangular numbers.

More precisely, all terms are second hexagonal numbers (A014105). There are no terms with middle divisors. - Omar E. Pol, Oct 31 2018

Numbers k such that the concatenation of the widths of the symmetric representation of sigma(k) is a cyclops numbers (A134808). - Omar E. Pol, Aug 29 2021

Examples

			q = 128 = 2^7 is the 15th term in A174973 for which 2*n+1 = 2^8 + 1 is prime so that a(15) = 2^7 * (2^8 + 1) = 32896. The two parts in the symmetric representation of sigma of a(15) have width 1 and sigma(a(15)) = 2 * a(15) - 2.
q = 308 is the 32nd term in A174973 for which 2*n+1 is prime so that a(32) = 308 * 617 = 190036. The maximum width of the two regions is 2 and sigma(a(32)) = 415296.
For n = 21, the symmetric representation of sigma(21) has two parts that meet at the center of the Dyck path, but 21 is not in the sequence because the symmetric representation of sigma(21) has more than two parts. - _Omar E. Pol_, Sep 18 2015
From _Omar E. Pol_, Oct 05 2015: (Start)
Illustration of initial terms (n = 1, 2):
. y
.  |
.  |_ _ _ _ _ _
.  |_ _ _ _ _  |
.  |         | |_
.  |         |_ _|_
.  |             | |_ _
.  |             |_ _  |
.  |                 | |
.  |_ _              | |
.  |_ _|_            | |
.  |   | |           | |
.  |_ _|_|_ _ _ _ _ _|_|_ _ x
.       3             10
.
The symmetric representation of sigma(3) = 2 + 2 = 4 has two parts and they meet at the point (2, 2), so a(1) = 3.
The symmetric representation of sigma(10) = 9 + 9 = 18 has two parts and they meet at the point (7, 7), so a(2) = 10.
(End)
Also 10 is in the sequence because the concatenation of the widths of the symmetric representation of sigma(10) is 1111111110111111111 and it is a cyclops number (A134808). - _Omar E. Pol_, Aug 29 2021
		

Links

• Hartmut F. W. Hoft, proof of formula

Crossrefs

Cf. A000203, A000217, A014105, A071561, A134808, A174973, A191363, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A239929, A245092, A249351 (widths), A262045, A262048, A262626.

Programs

Mathematica

(* test for membership in A174973 *)
a174973Q[n_]:=Module[{d=Divisors[n]}, Select[Rest[d] - 2 Most[d], #>0&]=={}]
a174973[n_]:=Select[Range[n], a174973Q]
(* compute numbers in the sequence *)
a262259[n_]:=Map[#(2#+1)&, Select[a174973[1, n], PrimeQ[2#+1]&]]
a262259[308] (* data *)

Formula

Terms are equal to q*(2*q + 1) where q is in A174973 and 2*q + 1 is prime.