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A052101
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One of the three sequences associated with the polynomial x^3 - 2.
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8
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1, 1, 1, 3, 9, 21, 45, 99, 225, 513, 1161, 2619, 5913, 13365, 30213, 68283, 154305, 348705, 788049, 1780947, 4024809, 9095733, 20555613, 46454067, 104982561, 237252321, 536171481, 1211705163, 2738358009, 6188472981, 13985460405
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OFFSET
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0,4
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COMMENTS
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If x^3 = 2 and n >= 0, then there are unique integers a, b, c such that (1 + x)^n = a + b*x + c*x^2. The coefficient a is a(n).
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REFERENCES
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R. Schoof, Catalan's Conjecture, Springer-Verlag, 2008, pp. 17-18.
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LINKS
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FORMULA
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a(n) = 3*a(n-1) - 3*a(n-2) + 3*a(n-3).
a(n)/a(n-1) tends to 2.259921049... = 1 + 2^(1/3) (a real root to (x - 1)^3 = 2 or x^3 - 3x^2 + 3x - 3 = 0). A 3 X 3 matrix corresponding to the latter polynomial is [0 1 0 / 0 0 1 / 3 -3 3]. Let the matrix = M. Then a(n) = the center term in M^n * [1, 1, 1]. M^[1, 1, 1] = [9, 21, 45], center term = a(4) - Gary W. Adamson, Mar 28 2004
a(n) = Sum_{0..floor(n/3)}, 2^k * binomial(n, 3*k). - Ralf Stephan, Aug 30 2004
Equals the first differences of A052102.
Equals the second differences of A052103.
Equals the binomial transform of A077959.
a(n) = 4*a(n-1) - 6*a(n-2) + 6*a(n-3) - 3*a(n-4).
A052103 is binomial transform of c(n)=0, 1, 1, 0, 2, 2, 0, 4, 4, 0, 8, 8, ... b(n+1) - 2*b(n) is essentially 3*b(n). (End)
G.f.: (1 - x)^2 / (1 - 3*x + 3*x^2 - 3*x^3).
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EXAMPLE
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From the Schoof reference, pp. 17, 18: Set pi = 1 + sqrt[3]{2}. For every integer k >= 0, there are unique a_k,b_k,c_k in Q such that pi^k = a_k + b_k sqrt[3]{2} + c_k sqrt[3]{4}. The coefficients a_k,b_k,c_k are actually in Z:
Coefficients a_k, b_k, c_k:
k 0 1 2 3 4 5 6
----------------------------------------------
a_k 1 1 1 3 9 21 45
b_k 0 1 2 3 6 15 36
c_k 0 0 1 3 6 12 27
----------------------------------------------
G.f. = 1 + x + x^2 + 3*x^3 + 9*x^4 + 21*x^5 + 45*x^6 + 99*x^7 + 225*x^8 + ...
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MAPLE
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A052101 := n -> add(2^j*binomial(n, 3*j), j = 0..floor(n/3));
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MATHEMATICA
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LinearRecurrence[{3, -3, 3}, {1, 1, 1}, 31] (* Ray Chandler, Sep 23 2015 *)
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PROG
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(PARI) {a(n) = polcoeff( lift( Mod(1 + x, x^3 - 2)^n ), 0)} /* Michael Somos, Aug 05 2009 */
(PARI) {a(n) = sum(k=0, n\3, 2^k * binomial(n, 3*k))} /* Michael Somos, Aug 05 2009 */
(PARI) {a(n) = if( n<0, 0, polcoeff( (1 - x)^2 / (1 - 3*x + 3*x^2 - 3*x^3) + x * O(x^n), n))} /* Michael Somos, Aug 05 2009 */
(Magma) [n le 3 select 1 else 3*(Self(n-1) -Self(n-2) +Self(n-3)): n in [1..41]]; // G. C. Greubel, Apr 15 2021
(Sage) [sum(2^j*binomial(n, 3*j) for j in (0..n//3)) for n in (0..40)] # G. C. Greubel, Apr 15 2021
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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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Ashok K. Gupta and Ashok K. Mittal (akgjkiapt(AT)hotmail.com), Jan 20 2000
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STATUS
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approved
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