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A054390
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Number of ways of writing n as a sum of powers of 3, each power being used at most three times.
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11
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1, 1, 1, 2, 1, 1, 2, 1, 1, 3, 2, 2, 3, 1, 1, 2, 1, 1, 3, 2, 2, 3, 1, 1, 2, 1, 1, 4, 3, 3, 5, 2, 2, 4, 2, 2, 5, 3, 3, 4, 1, 1, 2, 1, 1, 3, 2, 2, 3, 1, 1, 2, 1, 1, 4, 3, 3, 5, 2, 2, 4, 2, 2, 5, 3, 3, 4, 1, 1, 2, 1, 1, 3, 2, 2, 3, 1, 1, 2, 1, 1, 5, 4, 4, 7, 3, 3, 6, 3, 3, 8, 5, 5, 7, 2, 2, 4, 2, 2, 6, 4, 4, 6, 2, 2
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OFFSET
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0,4
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COMMENTS
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Let M be an infinite matrix with (1, 1, 1, 1, 0, 0, 0, ...) in each column shifted down thrice from the previous column (for k>0). Then A054390 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Apr 14 2010
Conjecture: Number of ways of partitioning n into distinct parts of A038754. - R. J. Mathar, Mar 01 2023
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LINKS
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FORMULA
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a(0)=1, a(1)=1, a(2)=1 and, for n>0, a(3n)=a(n)+a(n-1), a(3n+1)=a(n), a(3n+2)=a(n).
G.f.: Product_{j >= 0} (1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))). - Emeric Deutsch, Apr 02 2006
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3) * A(x^3). - Ilya Gutkovskiy, Jul 09 2019
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EXAMPLE
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a(33) = 4 because we have 33 = 27+3+3 = 27+3+1+1+1 = 9+9+9+3+3 = 9+9+9+3+1+1+1.
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MAPLE
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a[0]:=1: a[1]:=1: a[2]:=1: for n from 1 to 35 do a[3*n]:=a[n]+a[n-1]: a[3*n+1]:=a[n]: a[3*n+2]:=a[n] od: A:=[seq(a[n], n=0..104)]; # Emeric Deutsch, Apr 02 2006
g:=product((1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))), j=0..10): gser:=series(g, x=0, 125): seq(coeff(gser, x, n), n=0..104); # Emeric Deutsch, Apr 02 2006
# third Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
add(`if`(n-j*3^i<0, 0, b(n-j*3^i, i-1)), j=0..3)))
end:
a:= n-> b(n, ilog[3](n)):
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MATHEMATICA
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a[0]=1; a[1]=1; a[2]=1; For[n=1, n <= 35, n++, a[3*n] = a[n] + a[n-1]; a[3*n+1] = a[n]; a[3*n+2] = a[n]]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Dec 20 2016, after Emeric Deutsch *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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