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%I #36 Dec 20 2019 22:42:31
%S 1,30,66,204,819,70032,3000000000096
%N Let f(n) be equal to n + S(n) + S(S(n)) ... + S(S(S..(n))), where the last term is less than 10 and S(n) is the sum of digits. This is the sequence of numbers k such that the equation f(x) = k has a record number of solutions.
%C Conjecture from _Daniel Starodubtsev_ and _Dmitry Petukhov_, Nov 19 2019: (Start)
%C a(8) = 20000000000000046;
%C a(9) = 8900000000000000000000127. (End)
%e a(4) = 204, because 204 = f(179) = f(185) = f(191) = f(201), which has more solutions than any smaller number.
%t T = 0*Range[10^5]; f[n_] := Block[{x=n, s=n}, While[x >= 10, x = Plus@@ IntegerDigits[x]; s += x]; s]; Do[v = f[i]; If[v <= 10^5, T[[v]]++], {i, 10^5}]; Flatten[Position[T, #, 1, 1] & /@ Range[6]] (* _Giovanni Resta_, Jul 30 2019 *)
%o (PARI) f(n) = {s=n;m=n;while(sumdigits(s)>9,s=sumdigits(s);m+=s);if(n<10,m=0);m+sumdigits(s);}
%o g(n) = sum(k=1,n,f(k)==n);
%o lista(NN) = {x=1;print1(1);for(n=2,NN,if(g(n)>x,x=g(n);print1(", ",n)))} \\ _Jinyuan Wang_, Jul 31 2019
%Y Cf. A007953, A006064.
%K nonn,base,more
%O 1,2
%A _Daniel Starodubtsev_, Jul 28 2019
%E a(7) from _Bert Dobbelaere_, Aug 15 2019