cp's OEIS Frontend

This sequence is a map from the set of nonnegative integers into the set of all integers. It is clearly not one-to-one. It is not known if it is onto.

Comments from N. J. A. Sloane, Nov 17 2019: (Start)

For m >= 0, A329002 gives an expression for the first time that m appears in this sequence (if it does appear), and A329492 plays a similar role for negative m.

In all these sequences it is safer to say "sum of digits" (which is A007953) rather than "digital sum" (which is also A007953), because many people confuse the latter term with the "digital root" (A010888). (End)

The first open case is 31. If a(31) is not -1, then it is at least 20000.

If a(i) = m > 0 then A328882 takes the value i for the first time at term 2^m+i. For example, the first appearance of 2 in A328882 is at term 2^103 + 2.

From Yusuf Gurtas, Dec 27 2019: (Start)

a(n)=1 if and only if n+2 is a positive integer whose sum of digits is 1. Since the only such numbers are 10^k the only solutions to a(n)=1 are n=10^k-2 for k=1,2,... In other words, a(n)=1 if and only if n = 8, 98, 998, 9998, 99998, 999998, .... In particular, a(n)=1 has infinitely many solutions.

Using the same idea, a(n)=2 can be solved. a(n)=2 if and only if n + 2^2 = n + 4 is a positive integer whose sum of digits is 2. Since the only such numbers are 2*10^k or 10^j*(10^k+1) for j=0,1,2,..., k=1,2,3,..., the only solutions to a(n)=2 are n = 10^j*(10^k+1) - 4 for j=0,1,2,..., k=1,2,3,.... The first 30 solutions are n = 7, 16, 97, 106, 196, 997, 1006, 1096, 1996, 9997, 10006, 10096, 10996, 19996, 99997, 100006, 100096, 100996, 109996, 199996, 999997, 1000006, 1000096, 1000996, 1009996, 1099996, 1999996, 9999997, 10000006, 10000096.

(End)

Is this ever zero? If not, this would prove that A329492(11) = -1, and that A328882 is never -11. (-11 is the first negative open case.)