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 A056588 Coefficient triangle of certain polynomials. 18
 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS G.f. for column m: see column sequences: A000012, A000071, A056589-91, for m=0..4. The row polynomials p(n,x) := sum(a(n,m)*x^m) occur as numerators of the g.f. for the (n+1)-th power of Fibonacci numbers A000045. The corresponding denominator polynomials are the row polynomials q(n+2,x) = Sum_{m=0..n+2} A055870(n+2, m)*x^m (signed Fibonomial triangle). The row polynomials p(n,x) and the companion denominator polynomials q(n,x) can be deduced from Riordan's recursion result. The explicit formula is found from the recursion relation for powers of Fibonacci numbers (see Knuth's exercise with solution). - Roger L. Bagula, Apr 03 2010 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. FORMULA a(n, m)=0 if n=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 EXAMPLE 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) 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 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 *) PROG (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 CROSSREFS Cf. A055870, A000012, A000071, A056589-91, A056592 (row sums), A000045, A007598, A056570-4, A056585-7. Sequence in context: A203948 A296990 A156184 * A126770 A202979 A306326 Adjacent sequences:  A056585 A056586 A056587 * A056589 A056590 A056591 KEYWORD easy,sign,tabl AUTHOR Wolfdieter Lang, Jul 10 2000 STATUS approved

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Last modified September 22 03:21 EDT 2020. Contains 337289 sequences. (Running on oeis4.)