8 lines
235 B
Text
8 lines
235 B
Text
Clear[Ackermann3]
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$RecursionLimit=Infinity;
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Ackermann3[0,n_]:=n+1;
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Ackermann3[1,n_]:=n+2;
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Ackermann3[2,n_]:=3+2n;
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Ackermann3[3,n_]:=5+8 (2^n-1);
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Ackermann3[m_,0]:=Ackermann3[m-1,1];
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Ackermann3[m_,n_]:=Ackermann3[m-1,Ackermann3[m,n-1]]
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