(1)
The sequence 1, 3, 17, 99, 577, …, which is A001541 in the Online Encyclopedia of Integer Sequences, has some very interesting properties, some of them described in the Encyclopedia itself. Its connection with triangular and square numbers is also known, see https://www.cut-the-knot.org/do_you_know/triSquare.shtml.
In the forum, Ramsey2879 also used the sequence in the topic "Recursive Series Problem" just recently.

The conjecture is the following:

Let
a(0) = 1, a(1) = 3, and a(n) = 6a(n-1) – a(n-2) ,
b(0) = 0, b(1) = 2, and b(n) = 6b(n-1) – b(n-2),
A(1) = 3, A(n) = A(n-1) + a(n),
B(1) = 2, B(n) = B(n-1) + b(n), and
T(n) = n(n+1)/2 < triangular numbers >.

Then T(n) = 2T(m) if and only if n = A(i) and m = B(i).

For instance,
i=1, A(1) = 3, B(1) = 2, T(3) = 6, and T(2) = 3, that is T(3) = 2T(2),
i=2, A(2) = 20, B(2) = 14, T(20) = 210, T(14) = 105, that is T(20) = 2T(14), or
i=3, A(3) = 119, B(3) = 84, T(119) = 7140, T(84) = 3570, that is T(119) = 2T(84).

I would also note that B(n) / A(n) seems to converge on a number, which, with four decimal places, is 0.7071. Does anyone know what this number is?

(2)
2^n can not be obtained by adding up (more than one) consecutive nonnegative integers.

(3)
The sum of a sequence of consecutive nonnegative integers starting with n is never a square for any n, if and only if the number of the terms in the sequence can be expressed as
( 2m – 1 ) * 2^( 2i ), i and m being any nonnegative integers.

For instance, if i = 1, and m = 1, the hypothesis says that the sum of 4 consecutive integers, ie
n + n + 1 + n + 2 + n + 3 = 4n + 6 is never a square, for any n.

A few numbers generated by ( 2m – 1 ) * 2^( 2i ):
4, 12, 16, 20, 28, 36, 44, 48, 52, 60, 64, 68, 80, 112, …

This is almost true.

Suppose K and N are nonnegative integers, N > K, and SUM_i=K_to_N = 2^n for some n.

Then, 2^n = N(N+1)/2 - (K-1)K/2 so that

2^(n+1) = N(N+1) - (K-1)K = (N-K+1)(N+K)

Checking parity, we see that these last two terms cannot both be even, and therefore one of them is odd.

Since their product is 2^(n+1), the odd term must be equal to 1.

However, N-K+1 > 1 (since N > K). So we must have N+K = 1, yielding the trivial sum 0+1 = 2^0 as the only solution.

If the statement is reworded as "2^n cannot be obtained as the sum of (more than one) consecutive positive integers", then the proof above shows that there is no solution.

This should be amended to change "nonnegative" to positive since if i = 0 and m = 25 there are in fact solutions to m consecutive positive integers summing to a square. For instance the sum of 1,2,3,...49 is a square.