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

1,1

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.

Links

Table of n, a(n) for n=1..50.

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

Formula

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

From Elmo R. Oliveira, Apr 10 2026: (Start)

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 5.

a(n) = [x^n] x*(3 + 2*x + 22*x^2 + 2*x^3 + 3*x^4)/((1-x)^3*(1+x)^2). (End)

Example

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.

Mathematica

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

Crossrefs

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

Sequence in context: A199774 A235267 A284379 * A103010 A225866 A332704

Adjacent sequences: A372715 A372716 A372717 * A372719 A372720 A372721

Keyword

nonn,easy

Author

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

Status

approved