A055047 Numbers of the form 9^i*(3*j+1).
1, 4, 7, 9, 10, 13, 16, 19, 22, 25, 28, 31, 34, 36, 37, 40, 43, 46, 49, 52, 55, 58, 61, 63, 64, 67, 70, 73, 76, 79, 81, 82, 85, 88, 90, 91, 94, 97, 100, 103, 106, 109, 112, 115, 117, 118, 121, 124, 127, 130, 133, 136, 139, 142, 144, 145, 148, 151, 154, 157, 160, 163
Offset: 1
Keywords
Links
- Paolo Xausa, 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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Mathematica
A055047Q[k_] := Mod[k/9^IntegerExponent[k, 9], 3] == 1; Select[Range[300], A055047Q] (* Paolo Xausa, Mar 20 2025 *)
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PARI
is(n)=n/=9^valuation(n,9); n%3==1 \\ Charles R Greathouse IV and V. Raman, Dec 19 2013
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Python
from sympy import integer_log def A055047(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-1)//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) = 8n/3 + O(log n). - Charles R Greathouse IV, Dec 19 2013
a(n) = A055041(n)/3. - Peter Munn, May 17 2020
Comments