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 A082291 Expansion of x(2 + 5x - x^2)/((1 - x)(1 - 6x + x^2)). 2
 2, 19, 118, 695, 4058, 23659, 137902, 803759, 4684658, 27304195, 159140518, 927538919, 5406093002, 31509019099, 183648021598, 1070379110495, 6238626641378, 36361380737779, 211929657785302, 1235216565974039 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Integers n such that (n+1)^2 + (n+2)^2 is a square. From Paul Weisenhorn, Aug 03 2010: (Start) Place a(n) red and 2 blue balls in an urn; draw b(n) = A001109(n) balls without replacement; 2*probability(b(n) red balls) = probability(b(n) balls); this is equivalent to the Pell equation A(n)^2 - 8*B(n)^2 = 1 with a(n) = (A(n) + 4*B(n) - 3)/2; b(n) = B(n); and the solution (3;1) for the unit form. (End) LINKS Table of n, a(n) for n=1..20. Index entries for two-way infinite sequences Index entries for linear recurrences with constant coefficients, signature (7, -7, 1). FORMULA G.f.: x(2+5x-x^2)/((1-x)(1-6x+x^2)). a(n) = 6a(n-1) - a(n-2) + 6. a(-1-n) = -3 - a(n). A001109(n+1) + A001109(n) = 2a(n)+3, a(n+1) = 7a(n) - 4*A001109(n) + 9. - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de) From Paul Weisenhorn, Aug 03 2010: (Start) a(n) = round(((3 + sqrt(8))^n*(1 + sqrt(2)) - 6)/4); b(n) = round((sqrt(2)*(3+sqrt(8))^n)/8)=A001109(n); a(n+3) = 7*(a(n+2) - a(n+1)) + a(n); (End) a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) with a(1)=2, a(2)=19, and a(3)=118. - Harvey P. Dale, Jul 19 2012 EXAMPLE For n=3: a(3)=118; b(3)=35; 2*binomial(118,35) = binomial(120,35). - Paul Weisenhorn, Aug 03 2010 2*x + 19*x^2 + 118*x^3 + 695*x^4 + 4058*x^5 + 23659*x^6 + ... MAPLE for n from 1 to 20 do a[n]:=round((3+sqrt(8))^n*(1+sqrt(2))-6)/4): end do: # Paul Weisenhorn, Aug 03 2010 MATHEMATICA Rest[CoefficientList[Series[x (2+5x-x^2)/((1-x)(1-6x+x^2)), {x, 0, 20}], x]] (* or *) LinearRecurrence[{7, -7, 1}, {2, 19, 118}, 20] (* Harvey P. Dale, Jul 19 2012 *) PROG (PARI) {a(n) = if( n<0, -3 - a(-1 - n), if( n==0, -1, polcoeff( x * (2 + 5*x -x^2) / ((1 - x) * (1 - 6*x + x^2)) + x*O(x^n), n)))} (PARI) {a(n) = subst( poltchebi(n+1) - poltchebi(n) - 6, x, 3)/4} CROSSREFS Cf. A001652(n) = a(n) + 1. b(n) = A001109(n). - Paul Weisenhorn, Aug 03 2010 Sequence in context: A278405 A285791 A101253 * A055518 A289230 A350568 Adjacent sequences: A082288 A082289 A082290 * A082292 A082293 A082294 KEYWORD nonn,easy AUTHOR Michael Somos, Apr 07 2003 STATUS approved

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Last modified September 7 13:22 EDT 2024. Contains 375730 sequences. (Running on oeis4.)