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A144929
Numbers n such that there exists an integer k with (n+1)^3 - n^3 = 7*k^2.
7
1, 166, 18313, 2014318, 221556721, 24369225046, 2680393198393, 294818882598238, 32427396692607841, 3566718817304264326, 392306642506776468073, 43150163956928107223758, 4746125728619585018145361, 522030679984197423888766006, 57418628672533097042746115353, 6315527123298656477278183922878
OFFSET
1,2
REFERENCES
E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185.
P.-F. Teilhet, Query 2228, L'Intermédiaire des Mathématiciens, 11 (1904), 44-45. - N. J. A. Sloane, Mar 08 2022
FORMULA
a(n+2) = 110*a(n+1)-a(n)+54.
a(n) = 111*a(n-1)-111*a(n-2)+a(n-3), with n>3. - Harvey P. Dale, Jun 11 2011
G.f.: x*(-1-55*x+2*x^2) / ((x-1)*(x^2-110*x+1)). - Harvey P. Dale, Jun 11 2011
a(n) = (A350967(n)-3)/6. - N. J. A. Sloane, Mar 03 2022
EXAMPLE
a(1) = 1 because 2^3-1^3 = 7 = 7*1^2.
MATHEMATICA
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 *)
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 *)
PROG
(PARI) Vec(x*(-1-55*x+2*x^2)/((x-1)*(x^2-110*x+1)) + O(x^20)) \\ Colin Barker, Jul 14 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Richard Choulet, Sep 25 2008
STATUS
approved