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%I #33 Jul 06 2024 19:04:33
%S 0,0,2,0,1,6,0,2,4,14,0,1,2,7,30,0,2,4,6,16,62,0,1,4,3,14,25,126,0,2,
%T 2,6,8,14,52,254,0,1,4,5,4,13,30,79,510,0,2,4,6,8,10,28,62,160,1022,0,
%U 1,2,3,8,5,22,23,62,241,2046,0,2,4,6,8,10,12,34,48,126,484,4094
%N T(b,n) is the largest m such that there exists N such that none of S(N), S(N+1), ..., S(N+m-1) is divisible by n, where S(N) is the sum of digits of N in base b. Square array read by ascending antidiagonals.
%C Write n = (b-1)*s + t, 1 <= t <= b-1. The smallest N_0 such that none of S(N_0), S(N_0+1), ..., S(N_0+m-1) is divisible by n is given by N_0 = b^(u_0) - b^s*(t-gcd(t,b-1)+1) + 1, where u_0 is the smallest nonnegative solution to (b-1)*u == -gcd(t,b-1) (mod n). See my link below for more detailed information.
%H Jianing Song, <a href="/A331787/b331787.txt">Table of n, a(n) for n = 2..7627</a> (Note: T(b,n) occurs at the ((n+b-2)*(n+b-1)/2-b+3)-th place)
%H Jianing Song, <a href="/A331787/a331787.pdf">Proof of the formula, and the examples N_0 such that none of S(N_0), S(N_0+1), ..., S(N_0+m-1) is divisible by n</a>.
%F If n = (b-1)*s + t, 1 <= t <= b-1, then T(b,n) = b^s*(2*t-gcd(t,b-1)+1) - 2. See my link for a proof of the formula.
%F T(b,n) = T(b,n-1) + b*T(b,n-b+1) - b*T(b,n-b) for b >= 2, n >= b+1.
%F T(b,n) = O(b^(n/(b-1))).
%e Table begins
%e b\n 1 2 3 4 5 6 7 8 9 10
%e 2 0 2 6 14 30 62 126 254 510 1022
%e 3 0 1 4 7 16 25 52 79 160 241
%e 4 0 2 2 6 14 14 30 62 62 126
%e 5 0 1 4 3 8 13 28 23 48 73
%e 6 0 2 4 6 4 10 22 34 46 34
%e 7 0 1 2 5 8 5 12 19 26 47
%e 8 0 2 4 6 8 10 6 14 30 46
%e 9 0 1 4 3 8 9 12 7 16 25
%e 10 0 2 2 6 8 8 12 14 8 18
%o (PARI) T(b,n) = my(s=(n-1)\(b-1), t=(n-1)%(b-1)+1); b^s*(2*t-gcd(t,b-1)+1)-2
%Y Cf. A331789.
%Y Cf. A000918 (row 2), A164123 (row 3), A331786 (row 10).
%K nonn,base,easy,tabl
%O 2,3
%A _Jianing Song_, Jan 25 2020