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A triangle sequence of holes in a triangle sequence numbering of n^3+m^3 (A003325) by comparing the n level to the 2*n level.
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%I #8 Mar 20 2013 04:06:25

%S 28,35,65,72,91,126,126,133,152,189,217,224,243,250,217,224,243,280,

%T 341,344,351,370,407,432,468,513,344,351,370,407,468,513,520,539,559,

%U 576,637,686,728,730,737,756,793,513,520,539,576,637,728,730,737,756,793

%N A triangle sequence of holes in a triangle sequence numbering of n^3+m^3 (A003325) by comparing the n level to the 2*n level.

%C Row sums are:

%C {0, 63, 354, 1534, 4190, 9714, 21632, 40635, 72215, 123741}.

%C The estimate of coverage is:

%C Table[N[Length[c[[n]]]/Length[a[2*n]]], {n, 1, 10}]

%C {0., 0.2, 0.190476, 0.222222, 0.218182, 0.220779, 0.230769, 0.231343, 0.230769, 0.235577};

%C that is only 25% aren't in the a(n) compared to a(2*n).

%F a(n) =triangle_sequenceto_level(n); b(n)=lower_half_of_a(2*n); c(n)=elements in b(n) but not in a(n).

%e {28, 35},

%e {65, 72, 91, 126},

%e {126, 133, 152, 189, 217, 224, 243, 250},

%e {217, 224, 243, 280, 341, 344, 351, 370, 407, 432, 468, 513},

%e {344, 351, 370, 407, 468, 513, 520, 539, 559, 576, 637, 686, 728, 730, 737, 756, 793},

%e {513, 520, 539, 576, 637, 728, 730, 737, 756, 793, 854, 855, 945, 1001, 1008, 1024, 1027, 1064, 1072, 1125, 1216, 1241, 1332, 1339},

%e {730, 737, 756, 793, 854, 945, 1001, 1008, 1027, 1064, 1072, 1125, 1216, 1241, 1332, 1339, 1343, 1358, 1395, 1456, 1458, 1512, 1547, 1674, 1729, 1736, 1755, 1792, 1843, 1853, 1944},

%e {1001, 1008, 1027, 1064, 1125, 1216, 1332, 1339, 1343, 1358, 1395, 1456, 1512, 1547, 1674, 1729, 1736, 1755, 1792, 1843, 1853, 1944, 2000, 2060, 2071, 2198, 2205, 2224, 2240, 2261, 2322, 2331, 2413, 2457, 2540, 2662, 2709, 2728, 2745}, {1332, 1339, 1358, 1395, 1456, 1547, 1674, 1736, 1755, 1792, 1843, 1853, 1944, 2060, 2071, 2198, 2205, 2224, 2240, 2261, 2322, 2331, 2413, 2457, 2540, 2662, 2709, 2728, 2745, 2752, 2771, 2808, 2869, 2926, 2960, 3059, 3087, 3197, 3256, 3376, 3383, 3402, 3439, 3456, 3473, 3500, 3528, 3591, 3718}

%t Clear[a, b, c, n, m, k]; t[n_, m_] = n^3 + m^3; a[k_] := Union[Flatten[Table[Table[t[n, m], {m, 1, n}], {n, 1, k}]]]; b[n_] := Table[a[2*n][[k]], {k, 1, Floor[Length[a[2*n]]/2]}]; c = Table[Complement[b[n], a[n]], {n, 1, 10}]; Flatten[c]

%Y Cf. A003325.

%K nonn,tabf,less,uned

%O 1,1

%A _Roger L. Bagula_, Oct 17 2008