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A251769
Numbers n such that the sum of the heptagonal numbers H(n), H(n+1) and H(n+2) is equal to the sum of three consecutive squares.
2
17, 239, 25529, 345647, 36813809, 498423743, 53085488057, 718726692767, 76549236965393, 1036403392547279, 110383946618609657, 1494492973326484559, 159173574474798161009, 2155057831133398187807, 229528184008712329566329, 3107591898001386860334143
OFFSET
1,1
COMMENTS
Also nonnegative integers x in the solutions to 15*x^2-6*y^2+21*x-12*y+6 = 0, the corresponding values of y being A251770.
FORMULA
a(n) = a(n-1)+1442*a(n-2)-1442*a(n-3)-a(n-4)+a(n-5).
G.f.: x*(x^4+6*x^3-776*x^2-222*x-17) / ((x-1)*(x^2-38*x+1)*(x^2+38*x+1)).
EXAMPLE
17 is in the sequence because H(17)+H(18)+H(19) = 697+783+874 = 2354 = 729+784+841 = 27^2+28^2+29^2.
PROG
(PARI) Vec(x*(x^4+6*x^3-776*x^2-222*x-17)/((x-1)*(x^2-38*x+1)*(x^2+38*x+1)) + O(x^100))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 08 2014
STATUS
approved