A331787 -id:A331787 - cp's OEIS frontend

A331786 a(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.

0, 2, 2, 6, 8, 8, 12, 14, 8, 18, 38, 38, 78, 98, 98, 138, 158, 98, 198, 398, 398, 798, 998, 998, 1398, 1598, 998, 1998, 3998, 3998, 7998, 9998, 9998, 13998, 15998, 9998, 19998, 39998, 39998, 79998, 99998, 99998, 139998, 159998, 99998, 199998, 399998, 399998, 799998

Offset: 1

Jianing Song, Jan 25 2020

Write n = 9*s + t, 1 <= t <= 9. 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 = 10^(u_0) - 10^s*(t-gcd(t,9)+1) + 1, where u_0 is the smallest nonnegative solution to 9*u == -gcd(t,9) (mod n). See A331787 for more detailed information.
From Bernard Schott, Mar 25 2022: (Start)
Equivalently, a(n) is the largest number of consecutive integers whose sum of digits (A007953) is never divisible by n (this is the answer to problem of Diophante link).
a(n) ends with 8 when n = 5, 6 and n >= 9 (see formula). (End)

			The following list gives the smallest example for each 2 <= n <= 27:
   2: 9..10 (2)
   3: 1..2 (2)
   4: 997..1002 (6)
   5: 6..13 (8)
   6: 7..14 (8)
   7: 994..1005 (12)
   8: 9999993..10000006 (14)
   9: 1..8 (8)
  10: 1..18 (18)
  11: 999981..1000018 (38)
  12: 1..38 (38)
  13: 9999999961..10000000038 (78)
  14: 951..1048 (98)
  15: 961..1058 (98)
  16: 9999931..10000068 (138)
  17: 999999999999921..1000000000000078 (158)
  18: 1..98 (98)
  19: 1..198 (198)
  20: 99999999801..100000000198 (398)
  21: 1..398 (398)
  22: 99999999999999601..100000000000000398 (798)
  23: 99501..100498 (998)
  24: 99601..100598 (998)
  25: 99999999301..100000000698 (1398)
  26: 99999999999999999999201..100000000000000000000798 (1598)
  27: 1..998 (998)

Jianing Song, Table of n, a(n) for n = 1..1000
Diophante, A389 - Les décaXphobes (in French).
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,10,-10).

Cf. A007953 (S(N)), A051885, A331788.

Row 10 of A331787.

a(n) = my(s=(n-1)\9, t=(n-1)%9+1); 10^s*(2*t-gcd(t,9)+1)-2

If n = 9*s + t, 1 <= t <= 9, then a(n) = 10^s*(2*t-gcd(t,9)+1) - 2. See A331787 for a proof of the formula in base b.
Conjectures from Colin Barker, Jan 26 2020: (Start)
G.f.: 2*x^2*(1 + 2*x^2 + x^3 + 2*x^5 + x^6 - 3*x^7 + 5*x^8) / ((1 - x)*(1 - 10*x^9)).
a(n) = a(n-1) + 10*a(n-9) - 10*a(n-10) for n>10.
(End) [This conjecture is correct.]
a(n) = O(10^(n/9)).

A331788 a(n) is the smallest m such that for any N, at least one of S(N), S(N+1), ..., S(N+m-1) is divisible by n, where S(N) is the sum of digits of N.

Original entry on oeis.org

1, 3, 3, 7, 9, 9, 13, 15, 9, 19, 39, 39, 79, 99, 99, 139, 159, 99, 199, 399, 399, 799, 999, 999, 1399, 1599, 999, 1999, 3999, 3999, 7999, 9999, 9999, 13999, 15999, 9999, 19999, 39999, 39999, 79999, 99999, 99999, 139999, 159999, 99999, 199999, 399999, 399999, 799999

Offset: 1

Jianing Song, Jan 25 2020

The main sequence is A331786; this is added because some people may search for this.

			See A331786.

Jianing Song, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,10,-10).

Cf. A007953 (S(N)), A051885, A331786.

Row 10 of A331789.

a(n) = my(s=(n-1)\9, t=(n-1)%9+1); 10^s*(2*t-gcd(t,9)+1)-1

If n = 9*s + t, 1 <= t <= 9, then a(n) = 10^s*(2*t-gcd(t,9)+1) - 1. See A331787 for a proof of the formula in base b.
a(n) = A331786(n) + 1.
Conjectures from Colin Barker, Jan 26 2020: (Start)
G.f.: x*(1 + 2*x + 4*x^3 + 2*x^4 + 4*x^6 + 2*x^7 - 6*x^8) / ((1 - x)*(1 - 10*x^9)).
a(n) = a(n-1) + 10*a(n-9) - 10*a(n-10) for n>10.
(End) [This conjecture is correct.]
a(n) = O(10^(n/9)).

A331789 T(b,n) is the smallest m such that for any N, at least one 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.

Original entry on oeis.org

1, 1, 3, 1, 2, 7, 1, 3, 5, 15, 1, 2, 3, 8, 31, 1, 3, 5, 7, 17, 63, 1, 2, 5, 4, 15, 26, 127, 1, 3, 3, 7, 9, 15, 53, 255, 1, 2, 5, 6, 5, 14, 31, 80, 511, 1, 3, 5, 7, 9, 11, 29, 63, 161, 1023, 1, 2, 3, 4, 9, 6, 23, 24, 63, 242, 2047, 1, 3, 5, 7, 9, 11, 13, 35, 49, 127, 485, 4095

Offset: 2

Jianing Song, Jan 25 2020

The main sequence is A331787; this is added because some people may search for this.

			Table begins
  b\n  1  2  3   4   5   6    7    8    9    10
   2   1  3  7  15  31  63  127  255  511  1023
   3   1  2  5   8  17  26   53   80  161   242
   4   1  3  3   7  15  15   31   63   63   127
   5   1  2  5   4   9  14   29   24   49    74
   6   1  3  5   7   5  11   23   35   47    35
   7   1  2  3   6   9   6   13   20   27    48
   8   1  3  5   7   9  11    7   15   31    47
   9   1  2  5   4   9  10   13    8   17    26
  10   1  3  3   7   9   9   13   15    9    19

Jianing Song, Table of n, a(n) for n = 2..7627 (Note: T(b,n) occurs at the ((n+b-2)*(n+b-1)/2-b+3)-th place)

Cf. A331787.

Cf. A000225 (row 2), A062318 (row 3 with an offset shift), A331788 (row 10).

T(b,n) = my(s=(n-1)\(b-1), t=(n-1)%(b-1)+1); b^s*(2*t-gcd(t,b-1)+1)-1

If n = (b-1)*s + t, 1 <= t <= b-1, then T(b,n) = b^s*(2*t-gcd(t,b-1)+1) - 1. See A331787 for a proof of the formula in base b.
T(b,k) = A331787(b,k) + 1.
T(b,n) = T(b,n-1) + b*T(b,n-b+1) - b*T(b,n-b) for b >= 2, n >= b+1.
T(b,n) = O(b^(n/(b-1))).

cp's OEIS Frontend

A331786 a(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.

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A331788 a(n) is the smallest m such that for any N, at least one of S(N), S(N+1), ..., S(N+m-1) is divisible by n, where S(N) is the sum of digits of N.

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A331789 T(b,n) is the smallest m such that for any N, at least one 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.

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