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a(n) = position of 2*n^3 in A003325.
1

%I #12 Sep 20 2024 03:04:47

%S 1,3,6,11,18,25,33,44,57,68,81,99,116,134,152,177,200,223,246,276,304,

%T 331,360,397,433,465,501,541,579,617,662,707,749,793,845,895,944,995,

%U 1051,1105,1161,1214,1279,1337,1397,1456,1528,1591,1657,1722,1799,1870

%N a(n) = position of 2*n^3 in A003325.

%H Robert Israel, <a href="/A024667/b024667.txt">Table of n, a(n) for n = 1..2000</a>

%F It appears that a(n) ~ 2^(2/3) * Pi^2 * n^2/(9 * Gamma(2/3)^3). - _Robert Israel_, Sep 19 2024

%p M:= 200: # to get a(1) .. a(M)

%p N:= 2*N^3:

%p A:=sort(convert({seq(seq(x^3 + y^3, y = 1 .. floor((N-x^3)^(1/3))),x=1..floor(N^(1/3)))},list)):

%p filter:= proc(n) local F;

%p if n::odd then return false fi;

%p F:= ifactors(n/2)[2][..,2] mod 3;

%p andmap(`=`,F,0)

%p end proc:

%p select(t -> filter(A[t]), [$1..nops(A)]); # _Robert Israel_, Sep 19 2024

%Y Cf. A003325.

%K nonn

%O 1,2

%A _Clark Kimberling_