A332382 If n = Sum (2^e_k) then a(n) = Product (prime(e_k + 2)).
1, 3, 5, 15, 7, 21, 35, 105, 11, 33, 55, 165, 77, 231, 385, 1155, 13, 39, 65, 195, 91, 273, 455, 1365, 143, 429, 715, 2145, 1001, 3003, 5005, 15015, 17, 51, 85, 255, 119, 357, 595, 1785, 187, 561, 935, 2805, 1309, 3927, 6545, 19635, 221, 663, 1105, 3315, 1547, 4641, 7735, 23205
Offset: 0
Keywords
Examples
21 = 2^0 + 2^2 + 2^4 so a(21) = prime(2) * prime(4) * prime(6) = 3 * 7 * 13 = 273.
Crossrefs
Programs
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Maple
a:= n-> (l-> mul(ithprime(i+1)^l[i], i=1..nops(l)))(convert(n, base, 2)): seq(a(n), n=0..55); # Alois P. Heinz, Feb 10 2020
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Mathematica
nmax = 55; CoefficientList[Series[Product[(1 + Prime[k + 2] x^(2^k)), {k, 0, Floor[Log[2, nmax]]}], {x, 0, nmax}], x] a[0] = 1; a[n_] := Prime[Floor[Log[2, n]] + 2] a[n - 2^Floor[Log[2, n]]]; Table[a[n], {n, 0, 55}]
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PARI
a(n) = my(b=Vecrev(binary(n))); prod(k=1, #b, if (b[k], prime(k+1), 1)); \\ Michel Marcus, Feb 10 2020
Formula
G.f.: Product_{k>=0} (1 + prime(k+2) * x^(2^k)).
a(0) = 1; a(n) = prime(floor(log_2(n)) + 2) * a(n - 2^floor(log_2(n))).
a(2^(k-1)-1) = A002110(k)/2 for k > 0.
From Peter Munn, May 02 2020: (Start)
a(2n) = A003961(a(n)).
a(2n+1) = 3 * a(2n).
a(n) = A225546(4^n).
a(n+k) = A331590(a(n), a(k)).
A048675(a(n)) = 2n.
(End)
a(n+1) = A334748(a(n)). - Peter Munn, Mar 04 2022
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