A056588 Coefficient triangle of certain polynomials.
1, 1, -1, 1, -2, -1, 1, -4, -4, 1, 1, -7, -16, 7, 1, 1, -12, -53, 53, 12, -1, 1, -20, -166, 318, 166, -20, -1, 1, -33, -492, 1784, 1784, -492, -33, 1, 1, -54, -1413, 9288, 17840, -9288, -1413, 54, 1, 1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1
Offset: 0
Examples
Row polynomial for n=4: p(4,x) = 1 - 7*x - 16*x^2 + 7*x^3 + x^4. x*p(4,x) is the numerator of the g.f. for A056572(n), n >= 0 (fifth power of Fibonacci numbers) {0,1,1,32,243,...}. The denominator polynomial is Sum_{m=0..6} A055870(6,m)*x^m (n=6 row polynomial of signed fibonomial triangle). From _Roger L. Bagula_, Apr 03 2010: (Start) 1; 1, -1; 1, -2, -1; 1, -4, -4, 1; 1, -7, -16, 7, 1; 1, -12, -53, 53, 12, -1; 1, -20, -166, 318, 166, -20, -1; 1, -33, -492, 1784, 1784, -492, -33, 1; 1, -54, -1413, 9288, 17840, -9288, -1413, 54, 1; 1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1; (End)
References
- D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 84, (exercise 1.2.8. Nr. 30) and p. 492 (solution).
Links
- A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
- S. Falcon, On The Generating Functions of the Powers of the K-Fibonacci Numbers, Scholars Journal of Engineering and Technology (SJET), 2014; 2 (4C):669-675.
- J. Riordan, Generating functions for powers of Fibonacci numbers, Duke. Math. J. 29 (1962) 5-12.
Crossrefs
Programs
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Maple
A056588 := proc(n,k) if k = 0 then 1; elif k >n then 0; else combinat[fibonacci](k+1)^(n+1)+add( A055870(n+2, j)*(combinat[fibonacci](k+1-j)^(n+1)), j=1..k) ; end if; end proc: # R. J. Mathar, Jun 14 2015
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Mathematica
p[x_, n_] = Sum[(((1 + Sqrt[5])^k - (1 - Sqrt[5])^k)/(2^k*Sqrt[5]))^n*x^k, {k, 0, Infinity}]; a = Table[CoefficientList[FullSimplify[Numerator[p[ x, n]]/x], x]/2^(1 + Floor[n/2]), {n, 1, 10}]; Table[a[[n]]/a[[n]][[1]], {n, 1, 10}]; Flatten[%] (* Roger L. Bagula, Apr 03 2010 *)
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PARI
S(n, k) = (-1)^floor((k+1)/2)*(prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j))); T(n, k) = sum(j=0, k, fibonacci(k+1-j)^(n+1) * S(n+2, j)); tabl(m) = for (n=0, m, for (k=0, n, print1(T(n, k), ", ")); print); tabl(9); \\ Tony Foster III, Aug 20 2018
Formula
a(n, m)=0 if nA000045(n) (Fibonacci) and sfibonomial(n, m) := A055870(n, m).
From Roger L. Bagula, Apr 03 2010: (Start)
p(x,n) = Sum_{k>=0} (((1 + sqrt(5))^k - (1 - sqrt(5))^k)/(2^k*sqrt(5)))^n*x^k;
t(n,m) = Numerator_coefficients(p(x,n)/x)/2^(1 + floor(n/2));
out(n,m) = t(n,m)/t(n,1). (End)
T(n, k) = Sum_{j=0..k} Fibonacci(k+1-j)^(n+1) * A055870(n+2, j). - Tony Foster III, Aug 20 2018
Sum_{j=0..n-1} a(n-1, n-1-j)*A010048(k+j, n) = Fibonacci(k)^n. - Tony Foster III, Jul 24 2018
Comments