A123490 Triangle whose k-th column satisfies a(n) = (k+3)*a(n-1)-(k+2)*a(n-2).
1, 2, 1, 4, 2, 1, 8, 5, 2, 1, 16, 14, 6, 2, 1, 32, 41, 22, 7, 2, 1, 64, 122, 86, 32, 8, 2, 1, 128, 365, 342, 157, 44, 9, 2, 1, 256, 1094, 1366, 782, 260, 58, 10, 2, 1, 512, 3281, 5462, 3907, 1556, 401, 74, 11, 2, 1, 1024, 9842, 21846, 19532, 9332, 2802, 586, 92, 12, 2, 1
Offset: 0
Examples
Triangle begins 1; 2, 1; 4, 2, 1; 8, 5, 2, 1; 16, 14, 6, 2, 1; 32, 41, 22, 7, 2, 1; 64, 122, 86, 32, 8, 2, 1; 128, 365, 342, 157, 44, 9, 2, 1; 256, 1094, 1366, 782, 260, 58, 10, 2, 1; 512, 3281, 5462, 3907, 1556, 401, 74, 11, 2, 1; 1024, 9842, 21846, 19532, 9332, 2802, 586, 92, 12, 2, 1;
Links
- G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Crossrefs
Programs
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Magma
[((k+2)^(n-k) +k)/(k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, Jun 15 2021
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Mathematica
Table[((k+2)^(n-k) +k)/(k+1), {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 14 2017 *)
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PARI
for(n=0, 10, for(k=0,n, print1(((k+2)^(n-k)+k)/(k+1), ", "))) \\ G. C. Greubel, Oct 14 2017
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Sage
flatten([[((k+2)^(n-k) +k)/(k+1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 15 2021