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.

1, 9, 25, 49, 81, 121, 169, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1369, 1521, 1681, 1849, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 4225, 4489, 4761, 5041, 5329, 6241, 6561, 6889, 7225, 7569, 7921, 8649, 9025, 9409, 10201, 10609, 11449, 11881, 12321, 12769, 13225, 14161, 14641, 15129, 15625

Offset: 1

Author

Hartmut F. W. Hoft, Nov 03 2024

Keywords

• nonn

Comments

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.

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.

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.

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.

Conjectures (true for all a(n) <= 10^8):

(1) The central part of SRS(a(n)) is the minimum of all parts of SRS(a(n)), 1 <= n.

(2) The terms in this sequence are the squares of the terms in A244579.

Examples

			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.
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.
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.
		

Crossrefs

Cf. A001248, A030514, A030516, A030629, A030631, A179645 (even prime power sequences).

Cf. A003056, A016754, A237591, A237593, A244579, A247687, A249223, A319529, A341969, A357581.

Programs

Mathematica

(* t237591 and partsSRS compute rows in A237270 and A237591, respectively *)
(* t249223 and widthPattern are also defined in A376829 *)
row[n_] := Floor[(Sqrt[8 n+1]-1)/2]
t237591[n_] := Map[Ceiling[(n+1)/#-(#+1)/2]-Ceiling[(n+1)/(#+1)-(#+2)/2]&, Range[row[n]]]
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]
t249223[n_] := FoldList[#1+(-1)^(#2+1)KroneckerDelta[Mod[n-#2 (#2+1)/2, #2]]&, 1, Range[2, row[n]]]
widthPattern[n_] := Map[First, Split[Join[t249223[n], Reverse[t249223[n]]]]]
centerQ[n_] := Module[{pS=partsSRS[n]}, Sqrt[n]==pS[[(Length[pS]+1)/2]]]/;OddQ[n]
widthQ[n_] := Module[{wP=SplitBy[widthPattern[n], #!=0&]}, wP[[(Length[wP]+1)/2]]]=={1}/;OddQ[n]
a377654[m_, n_] := Select[Map[#^2&, Range[m, n, 2]], centerQ[#]&&widthQ[#]&]/;OddQ[m]
a377654[1, 125]