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A252585
Numbers n such that the sum of the pentagonal numbers P(n) and P(n+1) is equal to the heptagonal number H(m) for some m.
2
3, 234, 1617, 112948, 779551, 54440862, 375742125, 26240382696, 181106924859, 12647810018770, 87293162040073, 6096218188664604, 42075122996390487, 2938364519126320518, 20280121991098174821, 1416285602000697825232, 9774976724586323873395
OFFSET
1,1
COMMENTS
Also positive integers x in the solutions to 6*x^2-5*y^2+4*x+3*y+2 = 0, the corresponding values of y being A252586.
FORMULA
a(n) = a(n-1)+482*a(n-2)-482*a(n-3)-a(n-4)+a(n-5).
G.f.: x*(11*x^3+63*x^2-231*x-3) / ((x-1)*(x^2-22*x+1)*(x^2+22*x+1)).
EXAMPLE
3 is in the sequence because P(3)+P(4) = 12+22 = 34 = H(4).
MATHEMATICA
LinearRecurrence[{1, 482, -482, -1, 1}, {3, 234, 1617, 112948, 779551}, 20] (* Jean-François Alcover, Nov 13 2017 *)
PROG
(PARI) Vec(x*(11*x^3+63*x^2-231*x-3)/((x-1)*(x^2-22*x+1)*(x^2+22*x+1)) + O(x^100))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Dec 18 2014
STATUS
approved