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%I #24 Jun 27 2021 07:57:49
%S 9,25,49,50,98,121,169,242,289,338,361,484,529,578,676,722,841,961,
%T 1058,1156,1369,1444,1681,1682,1849,1922,2116,2209,2312,2738,2809,
%U 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
%N Numbers m with the property that the symmetric representation of sigma(m) has three parts of width one.
%C 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.
%C Since the sequence is defined by a two-parameter expression it can be written naturally as a triangle as shown in the Example section.
%C 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
%H Hartmut F. W. Hoft, <a href="/A247687/a247687.pdf">Three regions width one - triangle formula proof</a>
%F As an irregular triangle, T(n, k) = 2^k * prime(n)^2 where n >= 2 and 0 <= k <= floor(log_2(prime(n)) - 1).
%e We show portions of the first eight columns, powers of two 0 <= k <= 7, and 55 rows of the triangle through prime(56) = 263.
%e p/k 0 1 2 3 4 5 6 7
%e 3 9
%e 5 25 50
%e 7 49 98
%e 11 121 242 484
%e 13 169 338 676
%e 17 289 578 1156 2312
%e 19 361 722 1444 2888
%e 23 529 1058 2116 4232
%e 29 841 1682 3364 6728
%e 31 961 1922 3844 7688
%e 37 1369 2738 5476 10952 21904
%e 41 1681 3362 6724 13448 26896
%e 43 1849 3698 7396 14792 29584
%e 47 2209 4418 8836 17672 35344
%e 53 2809 5618 11236 22472 44944
%e 59 3481 6962 13924 27848 55696
%e 61 3721 7442 14884 29768 59536
%e 67 4489 8978 17956 35912 71824 143648
%e 71 5041 10082 20164 40328 80656 161312
%e . . . . . . .
%e . . . . . . .
%e 131 17161 34322 68644 137288 274567 549152 1098304
%e 137 18769 37538 75076 150152 300304 600608 1201216
%e . . . . . . . .
%e . . . . . . . .
%e 257 66049 132098 264196 528392 1056784 2113568 4227136 8454272
%e 263 69169 138338 276676 553352 1106704 2213408 4426816 8853632
%e 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).
%t (* path[n] and a237270[n] are defined in A237270 *)
%t atmostOneDiagonalsQ[n_] := SubsetQ[{0, 1}, Union[Flatten[Drop[Drop[path[n], 1], -1] - path[n-1], 1]]]
%t (* data *)
%t Select[Range[10000], atmostOneDiagonalsQ[#] && Length[a237270[#]]==3 &]
%t (* expression for the triangle in the Example section *)
%t 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]]]}]
%Y Cf. A000203, A237270, A237271, A237593, A241008, A241010, A246955, A250068, A250070, A250071.
%Y Cf. A070552, A263951.
%K nonn
%O 1,1
%A _Hartmut F. W. Hoft_, Sep 22 2014
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