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A276600
Values of m such that m^2 + 6 is a triangular number (A000217).
8
0, 2, 3, 7, 15, 20, 42, 88, 117, 245, 513, 682, 1428, 2990, 3975, 8323, 17427, 23168, 48510, 101572, 135033, 282737, 592005, 787030, 1647912, 3450458, 4587147, 9604735, 20110743, 26735852, 55980498, 117214000, 155827965, 326278253, 683173257, 908231938
OFFSET
1,2
COMMENTS
2*a(n+2) gives the y members of all positive solutions (x(n), y(n)), proper and improper, of the Pell equation x^2 - 2*y^2 = 7^2, n >= 0. The corresponding x members are x(n) = A106525(n). - Wolfdieter Lang, Sep 29 2016
FORMULA
a(n) = 6*a(n-3) - a(n-6) for n>6.
G.f.: x^2*(2 + 3*x + 7*x^2 + 3*x^3 + 2*x^4)/(1 - 6*x^3 + x^6).
From Wolfdieter Lang, Sep 29 2016: (Start)
Trisection:
a(2+3*n) = 15*S(n-1,6) - 2*S(n-2,6) = A275794(n),
a(3+3*n) = 20*S(n-1,6) - 3*S(n-2,6) = A275796(n),
a(4+3*n) = 7*(6*S(n-1,6) - S(n-2,6)) = 7*A001109(n+1) for n >= 0, with the Chebyshev polynomials S(n, 6) = A001109(n+1), n >= -1, with S(-2, 6) = -1.
(End)
EXAMPLE
7 is in the sequence because 7^2 + 6 = 55, which is a triangular number.
MATHEMATICA
LinearRecurrence[(0, 0, 6, 0, 0, -1}, {0, 2, 3, 7, 15, 20}, 41] (* G. C. Greubel, Sep 15 2021 *)
PROG
(PARI) concat(0, Vec(x^2*(2+3*x+7*x^2+3*x^3+2*x^4)/(1-6*x^3+x^6) + O(x^40)))
(Magma) I:=[0, 2, 3, 7, 15, 20]; [n le 6 select I[n] else 6*Self(n-3) - Self(n-6): n in [1..41]]; // G. C. Greubel, Sep 15 2021
(Sage)
def A276600_list(prec):
P.<x> = PowerSeriesRing(ZZ, prec)
return P( x^2*(2+3*x+7*x^2+3*x^3+2*x^4)/(1-6*x^3+x^6) ).list()
a=A276600_list(41); a[1:] # G. C. Greubel, Sep 15 2021
CROSSREFS
Cf. A001109 (k=0), A106328 (k=1), A077241 (k=2), A276598 (k=3), A276599 (k=5), A276601 (k=9), A276602 (k=10), where k is the value added to n^2.
Sequence in context: A375828 A077161 A068080 * A276047 A209658 A098763
KEYWORD
nonn,easy
AUTHOR
Colin Barker, Sep 07 2016
STATUS
approved