A275019 2-adic valuation of tetrahedral numbers C(n+2,3) = n(n+1)(n+2)/6 = A000292(n).
0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 4, 3, 4, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 5, 4, 5, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 4, 3, 4, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 6, 5, 6, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 4, 3, 4, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 5, 4, 5, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 4
Offset: 1
Keywords
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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Magma
[Valuation(n*(n+1)*(n+2)/6, 2): n in [1..100]]; // Vincenzo Librandi, Dec 04 2016
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Maple
seq(padic:-ordp(n*(n+1)*(n+2)/6,2),n=1..100); # Robert Israel, Dec 04 2016
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Mathematica
a[n_] := IntegerExponent[n*(n+1)*(n+2)/6, 2]; Array[a, 100] (* Amiram Eldar, Sep 13 2024 *)
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PARI
a(n)=valuation(n*(n+1)*(n+2)/6,2)
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Python
def A275019(n): return (~(m:=n*(n+1)*(n+2)//6)& m-1).bit_length() # Chai Wah Wu, Jul 07 2022
Formula
From Robert Israel, Dec 04 2016: (Start)
G.f.: (1+x+x^2)*Sum_{k>=1} x^(2^k-2)/(1-x^(2^k)) - 1/(1-x). (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 2. - Amiram Eldar, Sep 13 2024
Comments