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A251914
Numbers n such that the sum of the triangular numbers T(n) and T(n+1) is equal to the pentagonal number P(m) for some m.
2
0, 98, 9700, 950598, 93149000, 9127651498, 894416697900, 87643708742798, 8588189040096400, 841554882220704498, 82463790268588944500, 8080609891439495856598, 791817305570802005002200, 77590015336047156994359098, 7603029685627050583442189500
OFFSET
1,2
COMMENTS
Also nonnegative integers x in the solutions to 2*x^2-3*y^2+4*x+y+2 = 0, the corresponding values of y being A046172.
FORMULA
a(n) = A046173(n)-1.
a(n) = 99*a(n-1)-99*a(n-2)+a(n-3).
G.f.: 2*x^2*(x-49) / ((x-1)*(x^2-98*x+1)).
a(n) = 2*(-1/2+1/48*(49+20*sqrt(6))^(-n)*(-12-5*sqrt(6)+(-12+5*sqrt(6))*(49+20*sqrt(6))^(2*n))). - Colin Barker, Mar 03 2016
EXAMPLE
98 is in the sequence because T(98)+T(99) = 4851+4950 = 9801 = P(81).
MATHEMATICA
LinearRecurrence[{99, -99, 1}, {0, 98, 9700}, 20] (* Vincenzo Librandi, Mar 03 2016 *)
PROG
(PARI) concat(0, Vec(2*x^2*(x-49) / ((x-1)*(x^2-98*x+1)) + O(x^100)))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 11 2014
STATUS
approved