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%I #35 Dec 31 2023 11:35:20
%S 1,166,18313,2014318,221556721,24369225046,2680393198393,
%T 294818882598238,32427396692607841,3566718817304264326,
%U 392306642506776468073,43150163956928107223758,4746125728619585018145361,522030679984197423888766006,57418628672533097042746115353,6315527123298656477278183922878
%N Numbers n such that there exists an integer k with (n+1)^3 - n^3 = 7*k^2.
%D E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185.
%D P.-F. Teilhet, Query 2228, L'Intermédiaire des Mathématiciens, 11 (1904), 44-45. - _N. J. A. Sloane_, Mar 08 2022
%H Colin Barker, <a href="/A144929/b144929.txt">Table of n, a(n) for n = 1..450</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (111,-111,1).
%F a(n+2) = 110*a(n+1)-a(n)+54.
%F a(n) = 111*a(n-1)-111*a(n-2)+a(n-3), with n>3. - _Harvey P. Dale_, Jun 11 2011
%F G.f.: x*(-1-55*x+2*x^2) / ((x-1)*(x^2-110*x+1)). - _Harvey P. Dale_, Jun 11 2011
%F a(n) = (A350967(n)-3)/6. - _N. J. A. Sloane_, Mar 03 2022
%e a(1) = 1 because 2^3-1^3 = 7 = 7*1^2.
%t RecurrenceTable[{a[1]==1,a[2]==166,a[n]==54+110a[n-1]-a[n-2]},a[n],{n,20}] (* or *) LinearRecurrence[{111,-111,1},{1,166,18313},20] (* _Harvey P. Dale_, Jun 11 2011 *)
%t Rest@ CoefficientList[Series[x (-1 - 55 x + 2 x^2)/((x - 1) (x^2 - 110 x + 1)), {x, 0, 16}], x] (* or *) Last /@ Table[n /. {ToRules[Reduce[n > 0 && k >= 0 && (n + 1)^3 - n^3 == 7 k^2, n, Integers] /. C[1] -> c]} // Simplify, {c, 1, 16}] // Union (* _Michael De Vlieger_, Jul 14 2016 *)
%o (PARI) Vec(x*(-1-55*x+2*x^2)/((x-1)*(x^2-110*x+1)) + O(x^20)) \\ _Colin Barker_, Jul 14 2016
%Y Cf. A144927, A144928, A144930, A350967, A350968.
%K nonn,easy
%O 1,2
%A _Richard Choulet_, Sep 25 2008
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