A055048 Numbers of the form 9^i*(3*j+2).
2, 5, 8, 11, 14, 17, 18, 20, 23, 26, 29, 32, 35, 38, 41, 44, 45, 47, 50, 53, 56, 59, 62, 65, 68, 71, 72, 74, 77, 80, 83, 86, 89, 92, 95, 98, 99, 101, 104, 107, 110, 113, 116, 119, 122, 125, 126, 128, 131, 134, 137, 140, 143, 146, 149, 152, 153, 155
Offset: 1
Keywords
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
- L. J. Mordell, A new Waring's problem with squares of linear forms, Quart. J. Math., 1 (1930), 276-288 (see p. 283).
Crossrefs
Programs
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Haskell
a055048 n = a055048_list !! (n-1) a055048_list = filter (s 0) [1..] where s t u | m > 0 = even t && m == 2 | m == 0 = s (t + 1) u' where (u',m) = divMod u 3 -- Reinhard Zumkeller, Apr 07 2012
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Mathematica
max = 200; Select[ Union[ Flatten[ Table[ 9^i*(3*j + 2), {i, 0, Ceiling[Log[max]/Log[9]]}, {j, 0, Ceiling[( max/9^i - 2)/3]}]]], # <= max &] (* Jean-François Alcover, Oct 13 2011 *)
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PARI
is(n)=n/=9^valuation(n, 9); n%3==2 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013
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Python
from sympy import integer_log def A055048(n): def bisection(f,kmin=0,kmax=1): while f(kmax) > kmax: kmax <<= 1 kmin = kmax >> 1 while kmax-kmin > 1: kmid = kmax+kmin>>1 if f(kmid) <= kmid: kmax = kmid else: kmin = kmid return kmax def f(x): return n+x-sum((x//9**i-2)//3+1 for i in range(integer_log(x,9)[0]+1)) return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025
Formula
a(n) = A055040(n)/3. - Peter Munn, May 17 2020
Comments