A277836 Number of '6' digits in the set of all numbers from 0 to A014824(n) = Sum_{i=1..n} i*10^(n-i) = (0, 1, 12, 123, 1234, 12345, ...).

0, 0, 1, 22, 343, 4664, 58986, 713315, 8367717, 96022849, 1083685281, 12071420713, 133059886145, 1454055651577, 15775124417009, 170096923182441, 1824426021947881, 19478828120713394, 207133960219479637, 2194796392318253180, 23182531824417099723

Offset

0,4

Links

David A. Corneth, Table of n, a(n) for n = 0..998

Formula

a(n) = A277839(n) = A083449(n) = A277830(n) - 1 for n < 6,

a(n) = A277835(n) - 7*10^(n-6) for n >= 6,

a(n) = A277837(n) + 8*10^(n-7) for n >= 7.

Example

For n=2 there is only one digit '6' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '6' in { 16, 26, ..., 56, 60, ..., 69, 76, 86, ..., 116 }, where 66 accounts for two '6's.

Mathematica

T[int_Integer, {bndsLow_Integer, bndsUpp_Integer}] := Table[
   Count[
    Flatten[Table[
      IntegerDigits[m],
      {m, 1, Sum[
         10^i - 1,
         {i, n}
         ]/9
       }
      ]],
    int
    ],
   {n, bndsLow, bndsUpp}
   ];
T[6, {0, 7}](* Robert P. P. McKone, Jan 01 2021 *)

Prog

(PARI) print1(c=N=0); for(n=1, 8, print1(", "c+=sum(k=N+1, N=N*10+n, #select(d->d==6, digits(k)))))
(PARI) A277836(n, m=6)=if(n>m, A277836(n, m+1)+(m+2)*10^(n-m-1), A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

Crossrefs

Cf. A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

Sequence in context: A277838 A277635 A277837 * A277835 A018055 A277834

Adjacent sequences: A277833 A277834 A277835 * A277837 A277838 A277839

Keyword

nonn,base

Author

M. F. Hasler, Nov 01 2016

Extensions

More terms from Lars Blomberg, Nov 05 2016

Removed incorrect b-file. - David A. Corneth, Dec 31 2020

Status

approved