GP code: gg(n) yields 1 if n is in A036301, 0 otherwise: gg(n) = yy=0;dd=digits(n);s1=sum(i=1,length(dd),dd[i]*(dd[i]%2));s2=sum(i=1,length(dd),dd[i]*(1-(dd[i]%2)));if(s1==s2,yy=1);return(yy) Counts in residue classes mod 9 up to 1.2*10^6: 9 [1400, 6240, 6480, 1565, 145, 4210, 7055, 3795, 485] Very few congruent to 4 mod 9. Why? Let n be in A036301. Let s1 be the sum of the odd digits of n, and s2 the sum of the even digits of n. Then s1=s2 and s1+s2 is congruent to n modulo 9. Suppose n is congruent to 4 mod 9. Then the sum of the digits of n must be one of 4, 13, 22, 31, 40, etc. Since s1=s2, the sum must be even, so the sum must be 4, 22, 40, 58, etc. But, s2 is even, so the sum of the digits cannot be twice an odd number, and so must be a multiple of 4. So the sum must be 4, 40, 76, 112, 148, etc. To get 76, we would need at least 76/9=8.444... digits. To get 40, we would need at least 7 digits, since we need at least 3 digits for a sum of 20, and at least 4 digits for a sum of 20 with odd summands (if we used 3 the sum would be odd.) So, for small n, the only possibility is that the sum of the digits is 4, and so n's digits must consist of two 1s, one 2, and otherwise zeros (like 110002). Since there are not so many such numbers, not many small (<10^6) terms in A036301 are congruent to 4 mod 9. The smallest n that has other digits is 1148899.