A122195 Numbers that are the sum of exactly 3 sets of Fibonacci numbers.
8, 11, 13, 14, 18, 19, 22, 23, 30, 31, 36, 38, 49, 51, 59, 62, 80, 83, 96, 101, 130, 135, 156, 164, 211, 219, 253, 266, 342, 355, 410, 431, 554, 575, 664, 698, 897, 931, 1075, 1130, 1452, 1507, 1740, 1829, 2350, 2439, 2816, 2960, 3803, 3947, 4557, 4790, 6154
Offset: 1
Keywords
Examples
8 is the sum of only 3 sets of Fibonacci numbers: {8}, {3,5} and {1,2,5};
11 is the sum of only {3,8}, {1,2,8}, {1,2,3,5}.
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
- M. Bicknell-Johnson & D. C. Fielder, The number of Representations of N Using Distinct Fibonacci Numbers, Counted by Recursive Formulas, Fibonacci Quart. 37.1 (1999) pp. 47 ff.
- Ron Knott, Sumthing about Fibonacci Numbers
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1,0,0,1,-1).
Programs
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GAP
a:=[11,13,14,18,19,22,23,30];; for n in [9..60] do a[n]:=a[n-4]+a[n-8]+1; od; Concatenation([8], a); # G. C. Greubel, Jul 13 2019
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Magma
R
:=PowerSeriesRing(Integers(), 60); Coefficients(R!( (8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1 - x-x^4+x^5-x^8+x^9) )); // G. C. Greubel, Jul 13 2019 -
Maple
# first N terms: series((8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(x^9-x^8+x^5-x^4-x+1),x,N+1);
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Mathematica
CoefficientList[Series[(8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1 - x-x^4+x^5-x^8+x^9), {x, 0, 60}], x] (* G. C. Greubel, Jul 13 2019 *) -
PARI
my(x='x+O('x^60)); Vec((8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8 -3*x^9)/(1-x-x^4+x^5-x^8+x^9)) \\ G. C. Greubel, Jul 13 2019 -
Sage
((8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1 - x-x^4+x^5-x^8+x^9)).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Jul 13 2019
Formula
G.f.: (8+3*x+2*x^2+x^3-4*x^4-2*x^5+x^6-5*x^8-3*x^9)/(1-x-x^4+x^5-x^8+x^9).
a(n) = a(n-4) + a(n-8) + 1.
a(0)=8, a(1)=11, a(2)=13, a(3)=18, then: a(4n) = A022318(n+3) = 2*A000045(n+5) + A000045(n+3) - 1, a(4n+1) = A022406(n+2) = 4*A000045(n+4) - 1, a(4n+2) = A022308(n+4) = 2*A000045(n+4) + A000045(n+6) - 1, a(4n+3) = 3*A000045(n+4) - 1, for n>=1.
a(n) = a(n-1) +a(n-4) -a(n-5) +a(n-8) -a(n-9). - G. C. Greubel, Jul 13 2019