A252359 Numbers n such that the sum of the heptagonal numbers H(n) and H(n+1) is equal to the pentagonal number P(m) for some m.

158, 76252, 36753402, 17715063608, 8538623905750, 4115599007507988, 1983710182994944562, 956144192604555770992, 460859517125212886673678, 222133331110160006820941900, 107067804735579998074807322218, 51606459749218448912050308367272

Offset

1,1

Comments

Also positive integers x in the solutions to 10*x^2-3*y^2+4*x+y+2 = 0, the corresponding values of y being A252360.

Links

Colin Barker, Table of n, a(n) for n = 1..372

Index entries for linear recurrences with constant coefficients, signature (483,-483,1).

Formula

a(n) = 483*a(n-1)-483*a(n-2)+a(n-3).

G.f.: 2*x*(31*x-79) / ((x-1)*(x^2-482*x+1)).

a(n) = (-24+(12-5*sqrt(30))*(241+44*sqrt(30))^(-n)+(12+5*sqrt(30))*(241+44*sqrt(30))^n)/120. - Colin Barker, Apr 26 2016

Example

158 is in the sequence because H(158)+H(159) = 62173+62964 = 125137 = P(289).

Prog

(PARI) Vec(2*x*(31*x-79)/((x-1)*(x^2-482*x+1)) + O(x^100))

Crossrefs

Cf. A000326, A000566, A252360.

Sequence in context: A180096 A250996 A280483 * A159418 A159434 A287798

Adjacent sequences: A252356 A252357 A252358 * A252360 A252361 A252362

Keyword

nonn,easy

Author

Colin Barker, Dec 17 2014

Status

approved