A103737 Define a(1)=0, a(2)=0, a(3)=3, a(4)=7 such that from i=1 to 4: 30*a(i)^2 + 30*a(i) + 1 = j(i)^2, j(1)=1, j(2)=1, j(3)=19, j(4)=41 Then a(n) = a(n-4) + 4*sqrt(30*(a(n-2)^2) + 30*a(n-2) + 1).

0, 0, 3, 7, 76, 164, 1679, 3611, 36872, 79288, 809515, 1740735, 17772468, 38216892, 390184791, 839030899, 8566292944, 18420462896, 188068259987, 404411152823, 4128935426780, 8878624899220, 90648511129183, 194925336630027, 1990138309415256, 4279478780961384, 43692394296006459

Offset

1,3

Comments

By construction and recurrence, 30*a(n)^2 + 30*a(n) + 1 = j(n)^2.

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,22,-22,-1,1).

Formula

G.f.: x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)). - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009

Mathematica

Rest[CoefficientList[Series[x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)), {x, 0, 50}], x]] (* G. C. Greubel, Jul 15 2018 *)
LinearRecurrence[{1, 22, -22, -1, 1}, {0, 0, 3, 7, 76}, 40] (* Harvey P. Dale, Mar 05 2025 *)

Prog

(PARI) x='x+O('x^30); concat([0, 0], Vec(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))) \\ G. C. Greubel, Jul 15 2018
(Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); [0, 0] cat Coefficients(R!(x^3*(3*x^2+4*x+3)/((1-x)*(x^4-22*x^2+1)))); // G. C. Greubel, Jul 15 2018

Crossrefs

Cf. A053141, A103200.

Sequence in context: A172995 A325476 A234536 * A354480 A108537 A364660

Adjacent sequences: A103734 A103735 A103736 * A103738 A103739 A103740

Keyword

nonn,changed

Author

Pierre CAMI, Mar 27 2005

Extensions

Terms a(19) onward added by G. C. Greubel, Jul 15 2018

Status

approved