A389111 - OEIS

%I #27 Nov 16 2025 16:31:16

%S 11,13,15,17,19,21,23,25,27,56,58,60,62,64,66,68,70,72,74,76,78,80,82,

%T 84,86,88,90,137,139,141,143,145,147,149,151,153,155,157,159,161,163,

%U 165,167,169,171,173,175,177,179,181,183,185,187,189

%N 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).

%H Chinese BBS, <a href="https://bbs.emath.ac.cn/thread-50045-1-1.html">Triangular graph of triangular numbers</a>.

%F a(n) = 2*n + 9*round(sqrt(2*(n + 1))/3)^2.

%F 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

%e The spiral begins:

%e             15

%e             / \

%e           16  14

%e           /     \

%e         17   3  13

%e         /   / \   \

%e       18   4   2  12

%e       /   /     \   \

%e     19   5   0---1  11

%e     /   /             \

%e   20   6---7---8---9--10

%e   /

%e a(1) = 11, since 1 sits at the midpoint between 0 and 11.

%e a(2) = 13, since 2 sits at the midpoint between 0 and 13.

%e a(3) = 15, since 3 sits at the midpoint between 0 and 15.

%e a(4) = 17, since 4 sits at the midpoint between 0 and 17.

%t Table[2 n + 9 Round[Sqrt[2 (n + 1)]/3]^2, {n, 60}]

%Y Cf. A062708.

%K nonn,easy

%O 1,1

%A _Zhining Yang_, Sep 23 2025