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A389111
Write 0,1,2,3,4,... in a triangular spiral, a(n) is such that n lies geometrically half way between 0 and a(n).
0
11, 13, 15, 17, 19, 21, 23, 25, 27, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 137, 139, 141, 143, 145, 147, 149, 151, 153, 155, 157, 159, 161, 163, 165, 167, 169, 171, 173, 175, 177, 179, 181, 183, 185, 187, 189
OFFSET
1,1
FORMULA
a(n) = 2*n + 9*round(sqrt(2*(n + 1))/3)^2.
G.f. 2/(1-x)^2 + 9*F(x^9)/(1-x), where F(x) = Sum_{n>=0} (2*n+1) * x^(n*(n+1)/2) is the g.f. of A198954. - Paul D. Hanna, Nov 09 2025
EXAMPLE
The spiral begins:
15
/ \
16 14
/ \
17 3 13
/ / \ \
18 4 2 12
/ / \ \
19 5 0---1 11
/ / \
20 6---7---8---9--10
/
a(1) = 11, since 1 sits at the midpoint between 0 and 11.
a(2) = 13, since 2 sits at the midpoint between 0 and 13.
a(3) = 15, since 3 sits at the midpoint between 0 and 15.
a(4) = 17, since 4 sits at the midpoint between 0 and 17.
MATHEMATICA
Table[2 n + 9 Round[Sqrt[2 (n + 1)]/3]^2, {n, 60}]
CROSSREFS
Cf. A062708.
Sequence in context: A292513 A171491 A277268 * A337254 A382937 A152200
KEYWORD
nonn,easy
AUTHOR
Zhining Yang, Sep 23 2025
STATUS
approved