proposed
approved
proposed
approved
editing
proposed
1, 3, 9, 14, 23, 31, 43, 55, 70, 85, 103, 122, 143, 165, 189, 214, 41, 241, 269, 299, 331, 364, 399, 435, 473, 512, 553, 596, 640, 686, 733, 782, 832, 884, 937, 992, 1048, 1106, 1166, 1227, 1290, 1354, 1420, 1487, 1556, 1626, 1698, 1771, 1846, 1923, 2001, 2081
Clark Kimberling, <a href="/A294480/b294480.txt">Table of n, a(n) for n = 0..1000</a>
allocated for Clark KimberlingSolution of the complementary equation a(n) = a(n-2) + b(n-1) + 2n, where a(0) = 1, a(1) = 3, b(0) = 2.
1, 3, 9, 14, 23, 31, 43, 55, 70, 85, 103, 122, 143, 165, 189, 214, 41, 269, 299, 331, 364, 399, 435, 473, 512, 553, 596, 640, 686, 733, 782, 832, 884, 937, 992, 1048, 1106, 1166, 1227, 1290, 1354, 1420, 1487, 1556, 1626, 1698, 1771, 1846, 1923, 2001, 2081
0,2
The complementary sequences a() and b() are uniquely determined by the titular equation and initial values. See A294476 for a guide to related sequences.
Clark Kimberling, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Kimberling/kimberling26.html">Complementary equations</a>, J. Int. Seq. 19 (2007), 1-13.
a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4, so that
a(2) = a(0) + b(1) = 9
Complement: (b(n)) = (2, 4, 5, 6, 7, 8, 10, 12, 13, 15, ...)
mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 3; b[0] = 2;
a[n_] := a[n] = a[n - 2] + b[n - 1] + 2n;
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 40}] (* A294480 *)
Table[b[n], {n, 0, 10}]
allocated
nonn,easy
Clark Kimberling, Nov 01 2017
approved
editing
allocated for Clark Kimberling
allocated
approved