A247687 Numbers m with the property that the symmetric representation of sigma(m) has three parts of width one.

9, 25, 49, 50, 98, 121, 169, 242, 289, 338, 361, 484, 529, 578, 676, 722, 841, 961, 1058, 1156, 1369, 1444, 1681, 1682, 1849, 1922, 2116, 2209, 2312, 2738, 2809, 2888, 3362, 3364, 3481, 3698, 3721, 3844, 4232, 4418, 4489, 5041, 5329, 5476, 5618, 6241, 6724, 6728, 6889, 6962, 7396, 7442, 7688, 7921, 8836, 8978, 9409

Offset: 1

Author

Hartmut F. W. Hoft, Sep 22 2014

Keywords

• nonn

Comments

The symmetric representation of sigma(m) has 3 regions of width 1 where the two extremal regions each have 2^k - 1 legs and the central region starts with the p-th leg of the associated Dyck path for sigma(m) precisely when m = 2^(k - 1) * p^2 where 2^k < p <= row(m), k >= 1, p >= 3 is prime and row(m) = floor((sqrt(8*m + 1) - 1)/2). Furthermore, the areas of the two outer regions are (2^k - 1)*(p^2 + 1)/2 each so that the area of the central region is (2^k - 1)*p; for a proof see the link.

Since the sequence is defined by a two-parameter expression it can be written naturally as a triangle as shown in the Example section.

A263951 is a subsequence of this sequence containing the squares of all those primes p for which the areas of the 3 regions in the symmetric representation of p^2 (p once and (p^2 + 1)/2 twice) are primes; i.e., p^2 and p^2 + 1 are semiprimes (see A070552). - Hartmut F. W. Hoft, Aug 06 2020

Examples

			We show portions of the first eight columns, powers of two 0 <= k <= 7, and 55 rows of the triangle through prime(56) = 263.
p/k     0       1       2       3       4       5       6       7
3       9
5       25      50
7       49      98
11      121     242     484
13      169     338     676
17      289     578     1156    2312
19      361     722     1444    2888
23      529     1058    2116    4232
29      841     1682    3364    6728
31      961     1922    3844    7688
37      1369    2738    5476    10952   21904
41      1681    3362    6724    13448   26896
43      1849    3698    7396    14792   29584
47      2209    4418    8836    17672   35344
53      2809    5618    11236   22472   44944
59      3481    6962    13924   27848   55696
61      3721    7442    14884   29768   59536
67      4489    8978    17956   35912   71824   143648
71      5041    10082   20164   40328   80656   161312
.       .       .       .       .       .       .
.       .       .       .       .       .       .
131     17161   34322   68644   137288  274567  549152  1098304
137     18769   37538   75076   150152  300304  600608  1201216
.       .       .       .       .       .       .       .
.       .       .       .       .       .       .       .
257     66049   132098  264196  528392  1056784 2113568 4227136 8454272
263     69169   138338  276676  553352  1106704 2213408 4426816 8853632
Number 4 is not in this sequence since the symmetric representation of sigma(4) consists of a single region. Column k=0 contains the squares of primes (A001248(n), n>=2), column k=1 contains double the squares of primes (A079704(n), n>=2) and column k=2 contains four times the squares of primes (A069262(n), n>=5).
		

Links

• Hartmut F. W. Hoft, Three regions width one - triangle formula proof

Crossrefs

Cf. A000203, A237270, A237271, A237593, A241008, A241010, A246955, A250068, A250070, A250071.

Cf. A070552, A263951.

Programs

Mathematica

(* path[n] and a237270[n] are defined in A237270 *)
atmostOneDiagonalsQ[n_] := SubsetQ[{0, 1}, Union[Flatten[Drop[Drop[path[n], 1], -1] - path[n-1], 1]]]
(* data *)
Select[Range[10000], atmostOneDiagonalsQ[#] && Length[a237270[#]]==3 &]
(* expression for the triangle in the Example section *)
TableForm[Table[2^k Prime[n]^2, {n, 2, 57}, {k, 0, Floor[Log[2, Prime[n]] - 1]}], TableDepth -> 2, TableHeadings -> {Map[Prime, Range[2, 57]], Range[0, Floor[Log[2, Prime[57] - 1]]]}]

Formula

As an irregular triangle, T(n, k) = 2^k * prime(n)^2 where n >= 2 and 0 <= k <= floor(log_2(prime(n)) - 1).