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 A000119 Number of representations of n as a sum of distinct Fibonacci numbers. (Formerly M0101 N0037) 63
 1, 1, 1, 2, 1, 2, 2, 1, 3, 2, 2, 3, 1, 3, 3, 2, 4, 2, 3, 3, 1, 4, 3, 3, 5, 2, 4, 4, 2, 5, 3, 3, 4, 1, 4, 4, 3, 6, 3, 5, 5, 2, 6, 4, 4, 6, 2, 5, 5, 3, 6, 3, 4, 4, 1, 5, 4, 4, 7, 3, 6, 6, 3, 8, 5, 5, 7, 2, 6, 6, 4, 8, 4, 6, 6, 2, 7, 5, 5, 8, 3, 6, 6, 3, 7, 4, 4, 5, 1, 5, 5, 4, 8, 4, 7, 7, 3, 9, 6, 6, 9, 3, 8, 8, 5 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Number of partitions into distinct Fibonacci parts (1 counted as single Fibonacci number). Inverse Euler transform of sequence has generating function sum_{n>1} x^F(n)-x^{2F(n)} where F() are the Fibonacci numbers. A065033(n) = a(A000045(n)). a(n) = 1 if and only if n+1 is a Fibonacci number. The lengths of such quasi-periods (from Fib(i)-1 to Fib(i+1)-1, inclusive) is a Fibonacci number + 1. The maximum value of a(n) within each subsequent quasi-period increases by a Fibonacci number. For example, from n = 143 to n = 232, the maximum is 13. From 232 to 376, the maximum is 16, an increase of 3. From 376 to 609, 21, an increase of 5. From 609 to 986, 26, increasing by 5 again. Each two subsequent maxima seem to increase by the same increment, the next Fibonacci number. - Kerry Mitchell, Nov 14 2009 a(A000071(n)) = 1. - Reinhard Zumkeller, Dec 28 2012 The maxima of the quasi-periods are in A096748. - Max Barrentine, Sep 13 2015 Stockmeyer proves that a(n) <= sqrt(n+1) with equality iff n = fibonacci(m)^2 - 1 for some m>=2 (cf. A080097). - Michel Marcus, Mar 02 2016 REFERENCES M. Bicknell-Johnson, pp. 53-60 in 'Applications of Fibonacci Numbers', volume 8, ed: F T Howard, Kluwer (1999); see Theorem 3. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS T. D. Noe, Table of n, a(n) for n = 0..6765 Jean Berstel, Home Page J. Berstel, An Exercise on Fibonacci Representations, RAIRO/Informatique Theorique, Vol. 35, No 6, 2001, pp. 491-498, in the issue dedicated to Aldo De Luca on the occasion of his 60th anniversary. Pierre Bonardo, Anna E. Frid, Number of valid decompositions of Fibonacci prefixes, arXiv:1806.09534 [math.CO], 2018. Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 54. D. A. Klarner, Representations of N as a sum of distinct elements from special sequences, part 1, part 2, Fib. Quart., 4 (1966), 289-306 and 322. Ron Knott Sumthing about Fibonacci Numbers J. Shallit, Number theory and formal languages, in D. A. Hejhal, J. Friedman, M. C. Gutzwiller and A. M. Odlyzko, eds., Emerging Applications of Number Theory, IMA Volumes in Mathematics and Its Applications, V. 109, Springer-Verlag, 1999, pp. 547-570. (Eq. 9.2.) Paul K. Stockmeyer, A Smooth Tight Upper Bound for the Fibonacci Representation Function R(N), Fibonacci Quarterly, Volume 46/47, Number 2, May 2009. Scott V. Tezlaf, On ordinal dynamics and the multiplicity of transfinite cardinality, arXiv:1806.00331 [math.NT], 2018. See p. 42. FORMULA a(n) = (1/n)*Sum_{k=1..n} b(k)*a(n-k), b(k) = Sum_{f} (-1)^(k/f+1)*f, where the last sum is taken over all Fibonacci numbers f dividing k. - Vladeta Jovovic, Aug 28 2002 a(n) = 1, if n <= 2; a(n) = a(fib(i-2)+k)+a(k) if n>2 and 0<=k<=fib(i-3); a(n)= 2*a(k) if n>2 and fib(i-3)<=k<=fib(i-2); a(n) = a(fib(i+1)-2-k) otherwise where fib(i) is largest Fibonacci number (A000045) <= n and k=n-fib(i). [Bicknell-Johnson] - Ron Knott, Dec 06 2004 a(n) = f(n,1,1) with f(x,y,z) = if x=1, 1 + q^F(n+1) ) = 1 + sum(n>=1, q^F(n+1) * prod(k=1..n-1, 1+ q^F(k+1) ) ). - Joerg Arndt, Oct 20 2012 MAPLE with(combinat): p := product((1+x^fibonacci(i)), i=2..25): s := series(p, x, 1000): for k from 0 to 250 do printf(`%d, `, coeff(s, x, k)) od: # James A. Sellers, May 29 2000 MATHEMATICA CoefficientList[ Normal@Series[ Product[ 1+z^Fibonacci[ k ], {k, 2, 13} ], {z, 0, 233} ], z ] PROG (PARI) a(n)=local(A, m, f); if(n<0, 0, A=1+x*O(x^n); m=2; while((f=fibonacci(m))<=n, A*=1+x^f; m++); polcoeff(A, n)) (PARI) f(x, y, z)=if(x

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