A055041 Numbers of the form 3^(2i+1)*(3*j+1).
3, 12, 21, 27, 30, 39, 48, 57, 66, 75, 84, 93, 102, 108, 111, 120, 129, 138, 147, 156, 165, 174, 183, 189, 192, 201, 210, 219, 228, 237, 243, 246, 255, 264, 270, 273, 282, 291, 300, 309, 318, 327, 336, 345, 351, 354, 363, 372, 381, 390, 399
Offset: 1
Keywords
Links
- Amiram Eldar, 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
f[p_, e_] := (p^Mod[e, 2]); sqfpart[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[400], Mod[sqfpart[#], 9] == 3 &] (* Amiram Eldar, Mar 08 2021 *)
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Python
from sympy import integer_log def A055041(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)*3 # Chai Wah Wu, Feb 14 2025
Formula
a(n) = A055047(n) * 3. - Peter Munn, May 17 2020
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