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a(n) = binomial(n,2) + (2-adic valuation of n).
1

%I #76 Jul 01 2022 15:17:38

%S 0,2,3,8,10,16,21,31,36,46,55,68,78,92,105,124,136,154,171,192,210,

%T 232,253,279,300,326,351,380,406,436,465,501,528,562,595,632,666,704,

%U 741,783,820,862,903,948,990,1036,1081,1132,1176,1226,1275,1328,1378

%N a(n) = binomial(n,2) + (2-adic valuation of n).

%C 2^a(n) is the smallest integer m >= n such that binomial(m,n) is divisible by 2^binomial(n,2).

%C 2^a(n) is conjectured to be the order of the smallest n-symmetric graph.

%H Sebastian Jeon, Tanya Khovanova, <a href="https://arxiv.org/abs/2003.03870">3-Symmetric Graphs</a>, arXiv:2003.03870 [math.CO], 2020.

%F a(n) = A007814(n) + A161680(n).

%e Binomial(4,2) is 6. In addition, the 2-adic value of 4 is 2, so a(4) = 8.

%t a[n_] := Binomial[n, 2] + IntegerExponent[n, 2]; Array[a, 60] (* _Giovanni Resta_, Dec 03 2019 *)

%o (Python)

%o for i in range(1, 70):

%o j = i

%o res = i*(i-1)//2

%o while j%2 == 0:

%o res = res + 1

%o j = j // 2

%o print(str(res), end = ', ')

%o (Python)

%o def A326714(n): return (n*(n-1)>>1)+(~n & n-1).bit_length() # _Chai Wah Wu_, Jul 01 2022

%Y Cf. A007814, A161680, A329952.

%K nonn

%O 1,2

%A Sebastian Jeon and _Tanya Khovanova_, Dec 02 2019