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

0,2

Comments

a(1000)=793775, a(10000)=79261054, a(100000)=7924941755, a(1000000)=792482542841.

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.

Formula

a(n) = A071521(A000244(n)); A003586(a(n)) = A000244(n). - Reinhard Zumkeller, May 09 2006

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

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 *)

Prog

(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
(PARI) a(n)=sum(k=0, n, logint(3^k, 2))+n+1 \\ Charles R Greathouse IV, Nov 22 2022

Crossrefs

Cf. A022331, A020914 (first differences).

Cf. A000244, A003586, A071521.

Sequence in context: A006317 A194147 A077043 * A303279 A024219 A371701

Adjacent sequences: A022327 A022328 A022329 * A022331 A022332 A022333

Keyword

nonn

Author

Clark Kimberling

Status

approved