A251990 Numbers n such that the sum of the hexagonal numbers H(n) and H(n+1) is equal to the sum of the pentagonal numbers P(m) and P(m+1) for some m.

52, 10136, 1966380, 381467632, 74002754276, 14356152861960, 2785019652466012, 540279456425544416, 104811429526903150740, 20332877048762785699192, 3944473336030453522492556, 765207494312859220577856720, 148446309423358658338581711172

Offset

1,1

Comments

Also nonnegative integers x in the solutions to 4*x^2-3*y^2+2*x-2*y = 0, the corresponding values of y being A251991.

Links

Colin Barker, Table of n, a(n) for n = 1..437

Index entries for linear recurrences with constant coefficients, signature (195,-195,1).

Formula

a(n) = 195*a(n-1)-195*a(n-2)+a(n-3).

G.f.: 4*x*(x-13) / ((x-1)*(x^2-194*x+1)).

a(n) = (-6+(3-2*sqrt(3))*(97+56*sqrt(3))^(-n)+(3+2*sqrt(3))*(97+56*sqrt(3))^n)/24. - Colin Barker, Mar 02 2016

a(n) = 194*a(n-1)-a(n-2)+48. - Vincenzo Librandi, Mar 03 2016

Example

52 is in the sequence because H(52)+H(53) = 5356+5565 = 10921 = 5370+5551 = P(60)+P(61).

Mathematica

LinearRecurrence[{195, -195, 1}, {52, 10136, 1966380}, 30] (* Vincenzo Librandi, Mar 03 2016 *)

Prog

(PARI) Vec(4*x*(x-13)/((x-1)*(x^2-194*x+1)) + O(x^100))
(Magma) I:=[52, 10136]; [n le 2 select I[n] else 194*Self(n-1)- Self(n-2)+48: n in [1..20]]; // Vincenzo Librandi, Mar 03 2016

Crossrefs

Cf. A000326, A000384, A251991.

Sequence in context: A294507 A209500 A093252 * A209741 A209462 A209524

Adjacent sequences: A251987 A251988 A251989 * A251991 A251992 A251993

Keyword

nonn,easy

Author

Colin Barker, Dec 12 2014

Status

approved