A372718 - OEIS

%I #14 May 25 2024 15:23:54

%S 3,5,33,39,95,105,189,203,315,333,473,495,663,689,885,915,1139,1173,

%T 1425,1463,1743,1785,2093,2139,2475,2525,2889,2943,3335,3393,3813,

%U 3875,4323,4389,4865,4935,5439,5513,6045,6123,6683,6765,7353,7439,8055,8145,8789,8883,9555,9653

%N Triangular numbers that are 2 mod 4, halved.

%C 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.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

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

%e 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.

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

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

%K nonn,easy

%O 1,1

%A _Tanya Khovanova_ and the PRIMES STEP senior group, May 11 2024