

A265227


Nonnegative m for which k*floor(m^2/9) = floor(k*m^2/9), with 2 < k < 9.


2



0, 1, 3, 6, 8, 9, 10, 12, 15, 17, 18, 19, 21, 24, 26, 27, 28, 30, 33, 35, 36, 37, 39, 42, 44, 45, 46, 48, 51, 53, 54, 55, 57, 60, 62, 63, 64, 66, 69, 71, 72, 73, 75, 78, 80, 81, 82, 84, 87, 89, 90, 91, 93, 96, 98, 99, 100, 102, 105, 107, 108, 109, 111, 114
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OFFSET

1,3


COMMENTS

Also, nonnegative m congruent to 0, 1, 3, 6 or 8 (mod 9). The product of any two terms belongs to the sequence and so also a(n)^2, a(n)^3, a(n)^4, etc.
Integers x >= 0 satisfying k*floor(x^2/9) = floor(k*x^2/9) with k >= 0:
k = 0, 1: x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... (A001477);
k = 2: x = 0, 1, 2, 3, 6, 7, 8, 9, 10, 11, 12, 15, ... (A060464);
k = 3..8: x = 0, 1, 3, 6, 8, 9, 10, 12, 15, 17, 18, ... (this sequence);
k > 8: x = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ... (A008585).
Primes in sequence: 3, 17, 19, 37, 53, 71, 73, 89, 107, 109, 127, ...


LINKS

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,1).


FORMULA

G.f.: x^2*(1 + 2*x + 3*x^2 + 2*x^3 + x^4)/((1  x)^2*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n1) + a(n5)  a(n6) for n>6.


MATHEMATICA

Select[Range[0, 120], 3 Floor[#^2/9] == Floor[3 #^2/9] &]
Select[Range[0, 120], MemberQ[{0, 1, 3, 6, 8}, Mod[#, 9]] &]
LinearRecurrence[{1, 0, 0, 0, 1, 1}, {0, 1, 3, 6, 8, 9}, 70]


PROG

(Sage) [n for n in (0..120) if 3*floor(n^2/9) == floor(3*n^2/9)]
(MAGMA) [n: n in [0..120]  3*Floor(n^2/9) eq Floor(3*n^2/9)]; /* or, by the definition: */ K:=[3..8]; [<k, [n: n in [0..30]  k*Floor(n^2/9) eq Floor(k*n^2/9)]>: k in K];


CROSSREFS

Cf. A001477, A008585, A060464.
Cf. similar sequences listed in A265188.
Sequence in context: A163463 A137386 A153307 * A201817 A299240 A004715
Adjacent sequences: A265224 A265225 A265226 * A265228 A265229 A265230


KEYWORD

nonn,easy


AUTHOR

Bruno Berselli, Dec 06 2015


STATUS

approved



