A276600 Values of m such that m^2 + 6 is a triangular number (A000217).
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
Examples
7 is in the sequence because 7^2 + 6 = 55, which is a triangular number.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-1).
Crossrefs
Programs
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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
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Mathematica
LinearRecurrence[{0,0,6,0,0,-1}, {0,2,3,7,15,20}, 41] (* G. C. Greubel, Sep 15 2021 *)
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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)))
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Sage
def A276600_list(prec): P.
= 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
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)
Comments