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A056573
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Sixth power of Fibonacci numbers A000045.
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8
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0, 1, 1, 64, 729, 15625, 262144, 4826809, 85766121, 1544804416, 27680640625, 496981290961, 8916100448256, 160005726539569, 2871098559212689, 51520374361000000, 924491486192068809, 16589354847268067929
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OFFSET
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0,4
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COMMENTS
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A divisibility sequence; that is, if n divides m, then a(n) divides a(m).
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REFERENCES
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D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 85, (exercise 1.2.8. Nr. 30) and p. 492 (solution).
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LINKS
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Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 38, 2012, pp. 1871-1876. Mathematical Reviews, MR2959001. Zentralblatt MATH, Zbl 1255.05003.
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums II, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 42, 2012, pp. 2053-2059. Mathematical Reviews, MR2980853. Zentralblatt MATH, Zbl 1255.05004.
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FORMULA
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a(n) = F(n)^6, where F(n) = A000045(n).
G.f.: x*p(6, x)/q(6, x) with p(6, x) := sum_{m=0..5} A056588(5, m)*x^m = (1-x)*(1 - 11*x - 64*x^2 - 11*x^3 + x^4) and q(6, x) := sum_{m=0..7} A055870(7, m)*x^m = (1+x)*(1 - 3*x + x^2)*(1 + 7*x + x^2)*(1 - 18*x + x^2) (denominator factorization deduced from Riordan result).
Recursion (cf. Knuth's exercise): sum_{m=0..7} A055870(7, m)*a(n-m) = 0, n >= 7; inputs: a(n), n=0..6. a(n) = 13*a(n-1) + 104*a(n-2) - 260*a(n-3) - 260*a(n-4) + 104*a(n-5) + 13*a(n-6) - a(n-7).
a(n) = (F(3*n)^2 - (-1)^n*6*F(n)*F(3*n) + 9*F(n)^2)/25.
a(n) = (10*F(n)^3*F(3*n) - F(3*n)^2 + 9*F(n)^2)/25. (End)
a(n+1) = 2*[2*F(n+1)^2-(-1)^n]^3+3*F(n)^2*F(n+1)^2*F(n+2)^2-[F(n)^6+F(n+2)^6] = {Sum(0 <= j <= [n/2]; binomial(n-j, j))}^6, for n (this is Theorem 2.2 (vi) of Azarian's second paper in the references for this sequence). - Mohammad K. Azarian, Jun 29 2015
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MAPLE
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MATHEMATICA
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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