A022330 Index of 3^n within sequence of numbers of form 2^i*3^j (A003586).
1, 3, 7, 12, 19, 27, 37, 49, 62, 77, 93, 111, 131, 152, 175, 199, 225, 252, 281, 312, 344, 378, 413, 450, 489, 529, 571, 614, 659, 705, 753, 803, 854, 907, 961, 1017, 1075, 1134, 1195, 1257, 1321, 1386, 1453, 1522, 1592, 1664, 1737, 1812, 1889, 1967, 2047, 2128
Offset: 0
Keywords
Links
- Zak Seidov, Table of n, a(n) for n = 0..10000 (terms for n = 0..1000 from Charles R Greathouse IV).
- N. Carey, Lambda Words: A Class of Rich Words Defined Over an Infinite Alphabet, arXiv preprint arXiv:1303.0888 [math.CO], 2013.
- N. Carey, Lambda Words: A Class of Rich Words Defined Over an Infinite Alphabet, J. Int. Seq. 16 (2013) #13.3.4.
Programs
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Mathematica
c[0] = 1; c[n_] := 1 + Sum[Ceiling[j*Log[2, 3]], {j, n}]; Table[c[i], {i, 0, 51}] (* Norman Carey, Jun 13 2012 *) -
PARI
listsm(lim)=my(v=List(),N); for(n=0,log(lim)\log(3),N=3^n; while(N<=lim,listput(v,N);N<<=1)); v=Vec(v); vecsort(v) list(lim)=my(v=listsm(3^floor(lim)));vector(floor(lim+1),i,setsearch(v,3^(i-1))) \\ Charles R Greathouse IV, Aug 19 2011
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PARI
a(n)=sum(k=0,n, logint(3^k,2))+n+1 \\ Charles R Greathouse IV, Nov 22 2022
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Python
def A022330(n): return sum((3**i).bit_length() for i in range(n+1)) # Chai Wah Wu, Sep 16 2024
Formula
a(n) ~ kn^2 with k = log(3)/log(4) = 0.792.... More exact asymptotics? - Zak Seidov, Dec 22 2011
a(n+1) = a(n) + A020914(n+1). - Ruud H.G. van Tol, Nov 25 2022
kn^2 + kn + 1 <= a(n) <= kn^2 + (k+1)n + 1, so a(n) = kn^2 + O(n) with k = log(3)/log(4). The law of the iterated logarithm suggests that a better error term might be possible. - Charles R Greathouse IV, Nov 28 2022
Comments