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A252360
Numbers n such that the pentagonal number P(n) is equal to the sum of the heptagonal numbers H(m) and H(m+1) for some m.
2
289, 139217, 67102225, 32343133153, 15589323077441, 7514021380193329, 3621742715930107057, 1745672475056931408065, 841410511234725008580193, 405558120742662397204244881, 195478172787452040727437452369, 94220073725431140968227647796897
OFFSET
1,1
COMMENTS
Also positive integers y in the solutions to 10*x^2-3*y^2+4*x+y+2 = 0, the corresponding values of x being A252359.
FORMULA
a(n) = 483*a(n-1)-483*a(n-2)+a(n-3).
G.f.: -x*(x^2-370*x+289) / ((x-1)*(x^2-482*x+1)).
EXAMPLE
289 is in the sequence because P(289) = 125137 = 62173+62964 = H(158)+H(159).
MATHEMATICA
LinearRecurrence[{483, -483, 1}, {289, 139217, 67102225}, 20] (* Harvey P. Dale, May 29 2016 *)
PROG
(PARI) Vec(-x*(x^2-370*x+289)/((x-1)*(x^2-482*x+1)) + O(x^100))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 17 2014
STATUS
approved