is this sequence interesting?

[Peter Pein]

Dear Seq-fans,

let t(k)=k*(k+1)/2 be the k-th triangular number and write n>=2 as difference
of two positive triangular numbers: n = t(k) - t(j).

Now define a(n) as the minimal k needed for the minuend:

a(n) := min(k > 0: there exists j>0 with n = t(k)-t(j) )


example:

  a(30)=8, because 30 = t(30) - t(29)
                      = t(11) - t( 8)
                      = t( 9) - t( 5)
                      = t( 8) - t( 3)

and t(8) is the smallest minuend.

The sequence starts (with offset 2,1):

     2,  3,  4,  3,  6,  4,  8,  4, 10,
 6,  5,  7,  5,  6, 16,  9,  6, 10,  6,
 8,  7, 12,  9,  7,  8,  7, 28, 15,  8,
16, 32,  8, 10,  8, 13, 19, 11,  9, 10,
21,  9, 22,  9, 10, 13, 24, 17, 10, 12,
11, 10, 27, 10, 13, 11, 12, 16, 30, 11,
31, 17, 11, 64, 11, 18, 34, 12, 14, 13,
36, 12, 37, 20, 12, 13, 12, 21, 40, 18,
13, 22, 42, 14, 13, 23, 17, 13, 45, 13,
16, 15, 18, 25, 14, 33, 49, 17, 14, 16

Mathematica:

triadiff[n_] := Block[{c1, c2, k2, d = Divisors[n]},
 c2 = d - 1;
 c1 = (Reverse[d] - 7 - 2*c2)/4;
 k2 = Pick[4 + 2*(c1 + c2), (#1 >= 0 && IntegerQ[#1] & ) /@ c1];
 c1 = (Reverse[d] - 5 - 2*c2)/4;
 k2 = {k2, Pick[3 + 2*(c1 + c2), (#1 >= 0 && IntegerQ[#1] & ) /@ c1]};
 c2 = Cases[(d - 1)/2, _Integer];
 c1 = (n/(1 + 2*c2) - 3 - c2)/2;
 k2 = {k2, Pick[3 + 2*(c1 + c2), (#1 >= 0 && IntegerQ[#1] & ) /@ c1]};
 c1 = (n/(1 + 2*c2) - 2 - c2)/2;
 k2 = {k2, Pick[2 + 2*(c1 + c2), (#1 >= 0 && IntegerQ[#1] & ) /@ c1]};
 Min[k2]
]

triadiff /@ Range[2, 30]
--> {2, 3, 4, 3, 6, 4, 8, 4, 10, 6, 5, 7, 5, 6, 16, 9, 6, 10, 6, 8, 7, 12, 9,
7, 8, 7, 28, 15, 8}


Superseeker did not find any similarity with existing sequences in the EIS.
If you think it is worth to be kept, I'll be glad to enter it into the
web-form during the next days.

 Please correct me if my language is once again unclear or wrong and/or if you
find any errors in the above!


Thank you!

  Peter