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

Author

Ron Knott, Aug 25 2006, corrected Aug 29 2006

Keywords

• nonn

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).

Crossrefs

Cf. A000045, A000071, A013583, A000119, A122194.

Programs

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

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);

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