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<DIV> For A053735 Sum of digits of (n written in base 3), the given generating function,</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} x^(3^k)/(1+x^(3^k)))/(1-x),</DIV>
<DIV> </DIV>
<DIV>is wrong. The correct generating function is</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} (x^(3^k)+2*x^(2*3^k))/(1+x^(3^k)+x^(2*3^k)))/(1-x).</DIV>
<DIV> </DIV>
<DIV>In general, the sum of digits of (n written in base b) has generating function</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} (Sum_{0<=i<b} i*x^(i*b^k))/(Sum_{0<=i<b} x^(i*b^k)))/(1-x);</DIV>
<DIV> </DIV>
<DIV>in particular, for b=4, A053737 the generating function is</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} (x^(4^k)+2*x^(2*4^k)+3*x^(3*4^k))/(1+x^(4^k)+x^(2*4^k)+x^(3*4^k))/(1-x).</DIV>
<DIV> </DIV>
<DIV>(There is a typo in the example line of this sequence: it should end with 4, not 3.)</DIV>
<DIV> </DIV>
<DIV>Also, the generating function for the number of digits equal to d in the base b representation of n (0<d<b) is</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} x^(d*b^k)/(Sum_{0<=i<b} x^(i*b^k)))/(1-x);</DIV>
<DIV> </DIV>
<DIV>in particular, for d=1, b=3, A062756 has generating function</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} x^(3^k)/(1+x^(3^k)+x^(2*3^k)))/(1-x).</DIV>
<DIV> </DIV>
<DIV>For d=0, use the above formula with d=b:</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} x^(b^(k+1))/(Sum_{0<=i<b} x^(i*b^k)))/(1-x),</DIV>
<DIV> </DIV>
<DIV>adding 1 if you consider the representation of 0 to have one zero digit. Thus for A077267, the G.F. is</DIV>
<DIV> </DIV>
<DIV>(Sum_{k>=0} x^(3^(k+1))/(1+x^(3^k)+x^(2*3^k)))/(1-x),</DIV>
<DIV> </DIV>
<DIV>and for A081602 it is</DIV>
<DIV> </DIV>
<DIV>
<DIV>1+(Sum_{k>=0} x^(3^(k+1))/(1+x^(3^k)+x^(2*3^k)))/(1-x).</DIV>
<DIV> </DIV>
<DIV>(And these two sequences should have cross-references indicating that they are essentially the same.)</DIV>
<DIV> </DIV>
<DIV>Finally, the sequence that got me into this: A033095 (total number of 1's in all bases) has G.F.</DIV>
<DIV> </DIV>
<DIV>x+(Sum_{b>=2} (Sum_{k>=0} x^(b^k)/(Sum_{0<=i<b} x^(i*b^k)))/(1-x) - x).</DIV>
<DIV> </DIV>
<DIV>If the initial term of A033095 was 0, the initial "x+" would not be needed. This value is rather arbitrary; changing the "n+1" in the definition to "n" would make it 0.</DIV></DIV></DIV></DIV></DIV></DIV></DIV>
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