This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A261699 #86 Aug 08 2017 22:18:55
%S A261699 1,1,1,3,1,0,1,5,1,0,3,1,7,0,1,0,0,1,9,3,1,0,0,5,1,11,0,0,1,0,3,0,1,
%T A261699 13,0,0,1,0,0,7,1,15,3,0,5,1,0,0,0,0,1,17,0,0,0,1,0,3,9,0,1,19,0,0,0,
%U A261699 1,0,0,0,5,1,21,3,0,0,7,1,0,0,11,0,0,1,23,0,0,0,0,1,0,3,0,0,0,1,25,0,0,5,0,1,0,0,13,0,0
%N A261699 Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists positive terms interleaved with k-1 zeros, starting in row k(k+1)/2. If k is odd the positive terms of column k are k's, otherwise if k is even the positive terms of column k are the odd numbers greater than k in increasing order.
%C A261699 Conjecture: the positive terms in row n are the odd divisors of n.
%C A261699 Note that the elements appear with an unusual ordering, for example; row 45 is 1, 45, 3, 0, 5, 15, 0, 0, 9.
%C A261699 The positive terms give A261697.
%C A261699 Row n has length A003056(n) hence column k starts in row A000217(k).
%C A261699 The number of positive terms in row n is A001227(n).
%C A261699 The sum of row n is A000593(n).
%C A261699 The connection with the symmetric representation of sigma is as follows: A237048 --> A235791 --> A237591 --> A237593.
%C A261699 Proof of the conjecture: let n = 2^m*s*t with s and t odd. The property stated in A237048 verifies the conjecture with odd divisor k <= A003056(n) of n in position k and odd divisor t > A003056(n) in position 2^(m+1)*s. Therefore reading in row n the nonzero odd positions from left to right and then the nonzero even positions from right to left gives all odd divisors of n in increasing order. - _Hartmut F. W. Hoft_, Oct 25 2015
%C A261699 A237048 gives the signum function (A057427) of this sequence. - _Omar E. Pol_, Nov 14 2016
%C A261699 From _Peter Munn_, Jul 30 2017: (Start)
%C A261699 Each odd divisor d of n corresponds to n written as a sum of consecutive integers (n/d - (d-1)/2) .. (n/d + (d-1)/2). After canceling any corresponding negative and positive terms and deleting any zero term, the lower bound becomes abs(n/d - d/2) + 1/2, leaving k terms where k = n/d + d/2 - abs(n/d - d/2). It can be shown T(n,k) = d.
%C A261699 This sequence thereby defines a one to one relationship between odd divisors of n and partitions of n into k consecutive parts.
%C A261699 The relationship is expressed below using 4 sequences (with matching row lengths), starting with this one:
%C A261699 A261699(n,k) = d, the odd divisor.
%C A261699 A211343(n,k) = abs(n/d - d/2) + 1/2, smallest part.
%C A261699 A285914(n,k) = k, number of parts.
%C A261699 A286013(n,k) = n/d + (d-1)/2, largest part.
%C A261699 If no partition of n into k consecutive parts exists, the corresponding sequence terms are 0.
%C A261699 (End)
%F A261699 From _Hartmut F. W. Hoft_, Oct 25 2015: (Start)
%F A261699 T(n, k) = 2n/k, if A237048(n, k) = 1 and k even,
%F A261699 and in accordance with the definition:
%F A261699 T(n, k) = k, if A237048(n, k) = 1 and k odd,
%F A261699 T(n, k) = 0 otherwise; for k <= A003056(n).
%F A261699 (End)
%F A261699 For m >= 1, d >= 1 and odd, T(m*d, m + d/2 - abs(m - d/2)) = d. - _Peter Munn_, Jul 24 2017
%e A261699 Triangle begins:
%e A261699 1;
%e A261699 1;
%e A261699 1, 3;
%e A261699 1, 0;
%e A261699 1, 5;
%e A261699 1, 0, 3;
%e A261699 1, 7, 0;
%e A261699 1, 0, 0;
%e A261699 1, 9, 3;
%e A261699 1, 0, 0, 5;
%e A261699 1, 11, 0, 0;
%e A261699 1, 0, 3, 0;
%e A261699 1, 13, 0, 0;
%e A261699 1, 0, 0, 7;
%e A261699 1, 15, 3, 0, 5;
%e A261699 1, 0, 0, 0, 0;
%e A261699 1, 17, 0, 0, 0;
%e A261699 1, 0, 3, 9, 0;
%e A261699 1, 19, 0, 0, 0;
%e A261699 1, 0, 0, 0, 5;
%e A261699 1, 21, 3, 0, 0, 7;
%e A261699 1, 0, 0, 11, 0, 0;
%e A261699 1, 23, 0, 0, 0, 0;
%e A261699 1, 0, 3, 0, 0, 0;
%e A261699 1, 25, 0, 0, 5, 0;
%e A261699 1, 0, 0, 13, 0, 0;
%e A261699 1, 27, 3, 0, 0, 9;
%e A261699 1, 0, 0, 0, 0, 0, 7;
%e A261699 ...
%e A261699 From _Omar E. Pol_, Dec 19 2016: (Start)
%e A261699 Illustration of initial terms in a right triangle whose structure is the same as the structure of A237591:
%e A261699 Row _
%e A261699 1 _|1|
%e A261699 2 _|1 _|
%e A261699 3 _|1 |3|
%e A261699 4 _|1 _|0|
%e A261699 5 _|1 |5 _|
%e A261699 6 _|1 _|0|3|
%e A261699 7 _|1 |7 |0|
%e A261699 8 _|1 _|0 _|0|
%e A261699 9 _|1 |9 |3 _|
%e A261699 10 _|1 _|0 |0|5|
%e A261699 11 _|1 |11 _|0|0|
%e A261699 12 _|1 _|0 |3 |0|
%e A261699 13 _|1 |13 |0 _|0|
%e A261699 14 _|1 _|0 _|0|7 _|
%e A261699 15 _|1 |15 |3 |0|5|
%e A261699 16 _|1 _|0 |0 |0|0|
%e A261699 17 _|1 |17 _|0 _|0|0|
%e A261699 18 _|1 _|0 |3 |9 |0|
%e A261699 19 _|1 |19 |0 |0 _|0|
%e A261699 20 _|1 _|0 _|0 |0|5 _|
%e A261699 21 _|1 |21 |3 _|0|0|7|
%e A261699 22 _|1 _|0 |0 |11 |0|0|
%e A261699 23 _|1 |23 _|0 |0 |0|0|
%e A261699 24 _|1 _|0 |3 |0 _|0|0|
%e A261699 25 _|1 |25 |0 _|0|5 |0|
%e A261699 26 _|1 _|0 _|0 |13 |0 _|0|
%e A261699 27 _|1 |27 |3 |0 |0|9 _|
%e A261699 28 |1 |0 |0 |0 |0|0|7|
%e A261699 ... (End)
%t A261699 T[n_, k_?OddQ] /; n == k (k + 1)/2 := k; T[n_, k_?OddQ] /; Mod[n - k (k + 1)/2, k] == 0 := k; T[n_, k_?EvenQ] /; n == k (k + 1)/2 := k + 1; T[n_, k_?EvenQ] /; Mod[n - k (k + 1)/2, k] == 0 := T[n - k, k] + 2; T[_, _] = 0; Table[T[n, k], {n, 1, 26}, {k, 1, Floor[(Sqrt[1 + 8 n] - 1)/2]}] // Flatten (* _Jean-François Alcover_, Sep 21 2015 *)
%t A261699 (* alternate definition using function a237048 *)
%t A261699 T[n_, k_] := If[a237048[n, k] == 1, If[OddQ[k], k, 2n/k], 0] (* _Hartmut F. W. Hoft_, Oct 25 2015 *)
%Y A261699 Cf. A000217, A000593, A001227, A003056, A005408, A027750, A057427, A182469, A196020, A211343, A236104, A235791, A236112, A237048, A237591, A237593, A261350, A261697, A261698, A285914, A286013.
%K A261699 nonn,tabf
%O A261699 1,4
%A A261699 _Omar E. Pol_, Sep 20 2015