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RosettaCodeData/Task/Floyds-triangle/Mathematica/floyds-triangle.math

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f=Function[n,
Most/@(Range@@@Partition[FindSequenceFunction[{1,2,4,7,11}]/@Range[n+1],2,1])]
TableForm[f@5,TableAlignments->Right,TableSpacing->{1,1}]
TableForm[f@14,TableAlignments->Right,TableSpacing->{1,1}]
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