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05AB1E, 16 10 bytes LDmTªPγθg< -6 bytes thanks to @lyxal. Actually calculates \$H(n)\$ and counts the amount of trailing 0s, so this is a much slower approach. Still valid though, since the new 05AB1 …

Java 10, 44 bytes f->b->{int r=1;for(;b>0;b-=f)r*=b;return r;} Takes the factorial as first input, base as second. Try it online. … integer return-type int r=1; // Result-integer, starting at 1 for(;b>0; // Loop as long as the base is still larger than 0 b-=f) // After every iteration: decrease the base by the factorial …

91 bytes [S S S T N Push_1][S N S _Duplicate_1][S N S _Duplicate_1][T N T T _Read_STDIN_as_integer_(base)][T T T _Retrieve_base][S S S N _Push_0][T N T T _Read_STDIN_as_integer_(factorial … base = base - factorial Go to next iteration of LOOP function PRINT_RESULT: Print result as integer to STDOUT …

# Take the factorial of it (strings remain the same) s # Swap to get the duplicate value again > # Increase it by 1 (strings remain the same) η … # Reverse it J # And join them together to a single string M # And then push the largest value on the stack # (which is either the factorial-integer …

05AB1E (legacy), 5 bytes FL}˜P Port of @Lynn's Jelly answer, so make sure to upvote him/her as well! Try it online or verify the first \$[0,7]\$ test cases. Explanation: F # Loop the (implicit) i …

05AB1E, 10 8 7 bytes ݦRIιнP Input as two separated inputs: first input being base; second input being factorial. … factorial=8: [[10,2],[9,1],[8],[7],[6],[5],[4],[3]] # i.e. base=420, factorial=30: [[420,390,360,...,90,60,30],[419,389,359,...],...] …

Java 8, 141 136 134 bytes s->{float q=s.split("!",-1).length,n=new Float(s.split("!")[0]),i=n,r=n;for(;--i>0;r=q<3?i*r:q<4?i+r:q<5?i-r:i/r);return n<1&q<3?1:r;} -5 bytes (141 → 136) thanks to @Carlo …

# Get the factorial (aka gamma(n+1)) of the (implicit) input-integer # (output the entire stack joined together implicitly as result) …

05AB1E, 3 bytes L.¿ Pretty similar as most golfing languages. Try it online or verify all test cases. Explanation: L # Push a list in the range [1, (implicit) input] .¿ # Pop and push the LCM ( …

return-type int s=0; // Sum for(String i:(n+"").split("")) // Loop over the digits of `n` s+=f(new Long(i)); // And add the factorial … } // End of separate method (1) long f(long x){ // Separate method (2) with long parameter and long return-type (calculates the factorial …

# Get the factorial of the current integer IÅ? # And check if it starts with the input # (after which the result is output implicitly) …

# Take the factorial on the second s # Swap so the other is at the top m # Exponent the two together }P # After the map: take the product of the values I # Push the … input .E # Evaluate it as Elixir code, which calculates the value for factorial # inputs and will keep integer inputs unchanged Q # Check if they're the same }í # After …