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A377654 Numbers m^2 for which the center part (containing the diagonal) of its symmetric representation of sigma, SRS(m^2), has width 1 and area m.

%I A377654 #5 Nov 17 2024 07:39:37
%S A377654 1,9,25,49,81,121,169,289,361,441,529,625,729,841,961,1089,1369,1521,
%T A377654 1681,1849,2209,2401,2601,2809,3025,3249,3481,3721,4225,4489,4761,
%U A377654 5041,5329,6241,6561,6889,7225,7569,7921,8649,9025,9409,10201,10609,11449,11881,12321,12769,13225,14161,14641,15129,15625
%N A377654 Numbers m^2 for which the center part (containing the diagonal) of its symmetric representation of sigma, SRS(m^2), has width 1 and area m.
%C A377654 Since for numbers m^2 in the sequence the width at the diagonal of SRS(m^2) is 1, the area m of its center part is odd so that this sequence is a proper subsequence of A016754 and since SRS(m^2) has an odd number of parts it is a proper subsequence of A319529. The smallest odd square not in this sequence is 225 = 15^2.  SRS(225) is {113, 177, 113}, its center part has maximum width 2, its width at the diagonal is 1.
%C A377654 The k+1 parts of SRS(p^(2k)), p an odd prime and k >= 0, through the diagonal including the center part have areas (p^(2k-i) + p^i)/2 for 0 <= i <= k. They form a strictly decreasing sequence. Since p^(2k) has 2k+1 divisors and SRS(p^(2k)) has 2k+1 parts, all of width 1 (A357581), the even powers of odd primes form a proper subsequence of A244579. For the subsequence of squares of odd primes p, SRS(p^2) consists of the 3 parts { (p^2 + 1)/2, p, (p^2 + 1)/2 } see A001248, A247687 and A357581.
%C A377654 The areas of the parts of SRS(m^2) need not be in descending order through the diagonal as a(112) = 275^2 = 75625 with SRS(75625) = (37813, 7565, 3443, 1525, 715, 738, 275, 738, 715, 1525, 3443, 7565, 37813) demonstrates.
%C A377654 An equivalent description of the sequence is: The center part of SRS(m^2) has width 1, m is odd, and A249223(m^2, m-1) = 0.
%C A377654 Conjectures (true for all a(n) <= 10^8):
%C A377654 (1) The central part of SRS(a(n)) is the minimum of all parts of SRS(a(n)), 1 <= n.
%C A377654 (2) The terms in this sequence are the squares of the terms in A244579.
%e A377654 The center part of SRS(a(3)) = SRS(25) has area 5, all 3 parts have width 1, and 25 with 3 divisors also belongs to A244579.
%e A377654 The center part of SRS(a(7)) = SRS(169) has area 13, all 3 parts have width 1, and 169 with 3 divisors also belongs to A244579.
%e A377654 The center part of SRS(a(10)) = SRS(441) has area 21 and width 1, but the maximum width of SRS(441) is 2. Number 441 has 9 divisors and SRS(441) has 7 parts while 21 has 4 divisors and SRS(21) has 4 parts so that 21 is in A244579 while 441 is not.
%t A377654 (* t237591 and partsSRS compute rows in A237270 and A237591, respectively *)
%t A377654 (* t249223 and widthPattern are also defined in A376829 *)
%t A377654 row[n_] := Floor[(Sqrt[8 n+1]-1)/2]
%t A377654 t237591[n_] := Map[Ceiling[(n+1)/#-(#+1)/2]-Ceiling[(n+1)/(#+1)-(#+2)/2]&, Range[row[n]]]
%t A377654 partsSRS[n_] := Module[{widths=t249223[n], legs=t237591[n], parts, srs}, parts=widths legs; srs=Map[Apply[Plus, #]&, Select[SplitBy[Join[parts, Reverse[parts]], #!=0&], First[#]!=0&]]; srs[[Ceiling[Length[srs]/2]]]-=Last[widths]; srs]
%t A377654 t249223[n_] := FoldList[#1+(-1)^(#2+1)KroneckerDelta[Mod[n-#2 (#2+1)/2, #2]]&, 1, Range[2, row[n]]]
%t A377654 widthPattern[n_] := Map[First, Split[Join[t249223[n], Reverse[t249223[n]]]]]
%t A377654 centerQ[n_] := Module[{pS=partsSRS[n]}, Sqrt[n]==pS[[(Length[pS]+1)/2]]]/;OddQ[n]
%t A377654 widthQ[n_] := Module[{wP=SplitBy[widthPattern[n], #!=0&]}, wP[[(Length[wP]+1)/2]]]=={1}/;OddQ[n]
%t A377654 a377654[m_, n_] := Select[Map[#^2&, Range[m, n, 2]], centerQ[#]&&widthQ[#]&]/;OddQ[m]
%t A377654 a377654[1, 125]
%Y A377654 Cf. A001248, A030514, A030516, A030629, A030631, A179645 (even prime power sequences).
%Y A377654 Cf. A003056, A016754, A237591, A237593, A244579, A247687, A249223, A319529, A341969, A357581.
%K A377654 nonn
%O A377654 1,2
%A A377654 _Hartmut F. W. Hoft_, Nov 03 2024