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a(n) = 2*(2^(35*3^n) + 1).
0

%I #31 Sep 08 2022 08:46:23

%S 68719476738,81129638414606681695789005144066,

%T 133499189745056880149688856635597007162669032647290798121690100488888732861290034376435130433538

%N a(n) = 2*(2^(35*3^n) + 1).

%C Terms are a practical numbers.

%C a(3) has 285 digits.

%H Li-Yuan Wang and Zhi-Wei Sun, <a href="https://arxiv.org/abs/1809.01532">On practical numbers of some special forms</a>, arXiv preprint arXiv:1809.01532 [math.NT], 2018.

%F a(n) = A005843(A000079(35*A000244(n)) + 1).

%p a := n -> 2*(2^(35*3^n)+1):

%p seq(a(n), n = 0 .. 5);

%t a[n_]:=2*(2^(35*3^n)+1); Array[a, 5, 0]

%o (GAP) Flat(List([0..2], n->List([0..n], k->2*(2^(35*3^k)+1))));

%o (Magma) [2*(2^(35*3^n)+1): n in [0..5]];

%o (Maxima) a(n):=2*(2^(35*3^n)+1)$ makelist(a(n), n, 0, 5);

%o (PARI) a(n)=2*(2^(35*3^n)+1);

%o tabl(nn) = for(n=0, nn, print1(a(n), ", "));

%o tabl(5)

%o (Python) for n in range(5): print(2 * (2**(35 * 3**n) + 1), end=', ')

%Y Cf. A005843, A000079, A000244, A005153 (practical number).

%Y Subsequence of A321308.

%K nonn

%O 0,1

%A _Stefano Spezia_, Nov 02 2018