cp's OEIS Frontend

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)

The first open case is 11. Is a(11) = -1? See A329493.

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 -7 in A328882 is at term 2^103 - 7.

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.)

a(n) differs from A113019(n+9) starting at n = 19. - M. F. Hasler, Nov 17 2019