A033043 Sums of distinct powers of 6.
0, 1, 6, 7, 36, 37, 42, 43, 216, 217, 222, 223, 252, 253, 258, 259, 1296, 1297, 1302, 1303, 1332, 1333, 1338, 1339, 1512, 1513, 1518, 1519, 1548, 1549, 1554, 1555, 7776, 7777, 7782, 7783, 7812, 7813, 7818, 7819, 7992, 7993, 7998, 7999, 8028, 8029, 8034
Offset: 0
Links
- T. D. Noe, Table of n, a(n) for n = 0..1023
- Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 6.
Programs
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Julia
function a(n) m, r, b = n, 0, 1 while m > 0 m, q = divrem(m, 2) r += b * q b *= 6 end r end; [a(n) for n in 0:46] |> println # Peter Luschny, Jan 03 2021
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Maple
S:= {0,1}: for i from 1 to 6 do S:= S union (S +~ 6^i) od: sort(convert(S,list)); # Robert Israel, Apr 04 2025
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Mathematica
t = Table[FromDigits[RealDigits[n, 2], 6], {n, 0, 100}] (* Clark Kimberling, Aug 02 2012 *) FromDigits[#,6]&/@Tuples[{0,1},6] (* Harvey P. Dale, Mar 31 2016 *)
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PARI
A033043(n,b=6)=subst(Pol(binary(n)),'x,b) \\ M. F. Hasler, Feb 01 2016
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PARI
a(n)=fromdigits(binary(n), 6) \\ Charles R Greathouse IV, Jan 11 2017
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Python
def A033043(n): return int(bin(n)[2:],6) # Chai Wah Wu, Apr 04 2025
Formula
a(n) = Sum_{i=0..m} d(i)*6^i, where Sum_{i=0..m} d(i)*2^i is the base-2 representation of n.
a(n) = A097252(n)/5.
a(2n) = 6*a(n), a(2n+1) = a(2n)+1.
a(n) = Sum_{k>=0} A030308(n,k)*6^k. - Philippe Deléham, Oct 20 2011
G.f.: (1/(1 - x))*Sum_{k>=0} 6^k*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jun 04 2017
Extensions
Extended by Ray Chandler, Aug 03 2004
Comments