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A252003
Numbers n such that the sum of the octagonal numbers N(n) and N(n+1) is equal to the sum of the pentagonal numbers P(m) and P(m+1) for some m.
2
232, 267920, 309179640, 356793036832, 411738855324680, 475146282251644080, 548318397979541943832, 632758956122109151538240, 730203287046515981333185320, 842653960492723320349344321232, 972421940205315665167162013516600, 1122174076342973784879584614253835360
OFFSET
1,1
COMMENTS
Also nonnegative integers x in the solutions to 6*x^2-3*y^2+2*x-2*y = 0, the corresponding values of y being A252004.
FORMULA
a(n) = 1155*a(n-1)-1155*a(n-2)+a(n-3).
G.f.: 8*x*(5*x-29) / ((x-1)*(x^2-1154*x+1)).
a(n) = (-2 - (sqrt(2)-1)*(577+408*sqrt(2))^(-n) + (sqrt(2)+1)*(577+408*sqrt(2))^n) / 12. - Colin Barker, May 30 2017
EXAMPLE
232 is in the sequence because N(232)+N(233) = 161008+162401 = 323409 = 161212+162197 = P(328)+P(329).
MATHEMATICA
LinearRecurrence[{1155, -1155, 1}, {232, 267920, 309179640}, 20] (* Harvey P. Dale, Sep 14 2020 *)
PROG
(PARI) Vec(8*x*(5*x-29)/((x-1)*(x^2-1154*x+1)) + O(x^100))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 12 2014
STATUS
approved