A372718 Triangular numbers that are 2 mod 4, halved.

3, 5, 33, 39, 95, 105, 189, 203, 315, 333, 473, 495, 663, 689, 885, 915, 1139, 1173, 1425, 1463, 1743, 1785, 2093, 2139, 2475, 2525, 2889, 2943, 3335, 3393, 3813, 3875, 4323, 4389, 4865, 4935, 5439, 5513, 6045, 6123, 6683, 6765, 7353, 7439, 8055, 8145, 8789, 8883, 9555, 9653

Offset: 1

Author

Tanya Khovanova and the PRIMES STEP senior group, May 11 2024

Keywords

• nonn
• easy

Comments

The sum of the first 2*a(n) numbers of any Fibonacci-like sequence equals its (a(n)+2)-nd term times the a(n)-th Lucas number.

Examples

			10 is a triangular number that has a remainder of 2 when divided by 4. Therefore, its half, 5, is in this sequence. Moreover, the sum of the first 5*2 Fibonacci numbers is 143 (not counting zero). This sum is a product of 13, which is the (5+2 = 7)-th term of the Fibonacci sequence times 11, which is the fifth Lucas number.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A000032, A000045, A000217. A372048, A372049, A372050, A372051, A372070.

Programs

Mathematica

Select[Table[n (n + 1)/2, {n, 200}], Mod[#, 4] == 2 &]/2

Formula

a(n) = A000217(A047457(n))/2 = A372070(n)/2.