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0, 1, 7, 42, 246, 1435, 8365, 48756, 284172, 1656277, 9653491, 56264670, 327934530, 1911342511, 11140120537, 64929380712, 378436163736, 2205687601705, 12855689446495, 74928449077266, 436715005017102, 2545361581025347, 14835454481134981, 86467365305784540
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OFFSET
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0,3
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COMMENTS
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Partial sums of A001109. - Barry Williams, May 03 2000.
Number m such that 16*m*(2*m+1)+1 is a square. - Bruno Berselli, Oct 19 2012
Consecutive terms (a(n-1),a(n))=(u,v) give all points on the hyperbola u^2-u+v^2-v-6*u*v=0 in quadrant 1 with both coordinates an integer.
Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/4.
(End)
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REFERENCES
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R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., 58:2 (2020), 140-142.
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LINKS
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FORMULA
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a(n) = 6*a(n-1)-a(n-2)+1, a(0)=0, a(1)=1.
G.f.: x/((1-x)*(1-6*x+x^2)).
a(n+1) = Sum_{k=0..n} S(k, 6) = Sum_{k=0..n} U(n, 3), Chebyshev polynomials of 2nd kind, A049310.
a(n+1) = (sqrt(2)-1)^(2*n)(5/8-7*sqrt(2)/16)+(sqrt(2)+1)^(2*n)*(7*sqrt(2)/16 + 5/8)-1/4. (End)
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).
a(n) = -(1/4) + (1-sqrt(2))/(-8*sqrt(2))*(3-2*sqrt(2))^n + (1+sqrt(2))/(8*sqrt(2))*(3+2*sqrt(2))^n. (End)
E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/8. - Stefano Spezia, Mar 16 2024
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MATHEMATICA
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LinearRecurrence[{7, -7, 1}, {0, 1, 7}, 30] (* G. C. Greubel, Jul 15 2018 *)
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PROG
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(PARI) {a=1+sqrt(2); b=1-sqrt(2); P(n) = (a^n - b^n)/(a-b)};
for(n=0, 30, print1(round((P(2*n+1) - 1)/4), ", ")) \\ G. C. Greubel, Jul 15 2018
(PARI) x='x+O('x^30); Vec(x/((1-x)*(1-6*x+x^2))) \\ G. C. Greubel, Jul 15 2018
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018
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CROSSREFS
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Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).
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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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