A272525 -id:A272525 - cp's OEIS frontend

A277830 Number of digits '0' 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, ...).

1, 1, 2, 23, 344, 4665, 58986, 713307, 8367628, 96021949, 1083676272, 12071330614, 133058985146, 1454046641578, 15775034317010, 170096022182442, 1824417011947874, 19478738020713306, 207133059219478738, 2194787382318244170, 23182441724417009624, 244170096256515775267

Offset: 0

M. F. Hasler, Nov 01 2016

The first 10 terms are given by a simple explicit formula and linear recurrence, which does not hold for n > 9. Note that A007908 (concat(1..n)) differs from A014824 (a(n) = a(n-1)*10 + n) for n > 9. - M. F. Hasler, Nov 07 2020

M. F. Hasler, Table of n, a(n) for n = 0..100

Cf. A277831 - A277838, A277849, A277635, A272525, A083449, A014824.

print1(c=1);N=0;for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==0,digits(k))))) \\ For purpose of illustration.

apply( A277830(n)={A061217(A014824(n)+!n)+1}, [0..22]) \\ Thanks to Kevin Ryde's formula. - M. F. Hasler, Nov 07 2020

a(n) = A083449(n) + 1 for n <= 9.

a(n) = 1 + A061217(A014824(n)), taking A061217(0)=0. - Kevin Ryde, Nov 07 2020

Incorrect data, b-file, links, formulas and programs deleted by M. F. Hasler, following observations by Kevin Ryde, Nov 07 2020

A277849 Number of digits '9' 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, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713306, 8367627, 96021949, 1083676281, 12071330713, 133058986145, 1454046651577, 15775034417009, 170096023182441, 1824417021947873, 19478738120713305, 207133060219478737, 2194787392318244180, 23182441824417009723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n = 2 there is only one digit '9' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '9' in { 19, 29, ..., 89, 90, ..., 99, 109, 119 }, where 99 accounts for two '9's.

Lars Blomberg, Table of n, a(n) for n = 0..997

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

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==9,digits(k)))))

A014824(n)=(10^n-1)*(10/81)-n/9;
A102684(n)=my(pow,f,g,h);sum(j=1,#Str(n),pow=10^j;f=floor(n/pow);g=floor(n/pow+1/10);h=(4/5+g)*pow;g*(2*n+2-h)-f*(2*n+2-(1+f)*pow))/2;
A277849(n)=A102684(A014824(n));
vector(50,n,A277849(n-1)) \\ Lars Blomberg, Nov 11 2020

a(n) = A083449(n) = A277830(n) - 1 for 0 < n < 9.
a(n) = A277838(n) for n < 8, and a(8) = A277838(8) - 1.
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

Replaced incorrect b-file by Lars Blomberg, Nov 11 2020

A277635 Number of 7's appearing in the sequence of consecutive natural numbers from 1 to A007908(n), where A007908 = (1, 12, 123, 1234, ...).

Original entry on oeis.org

0, 1, 22, 343, 4664, 58985, 713307, 8367637, 96022049

Offset: 1

Alexander R. Povolotsky, Oct 24 2016

First 6 terms are the same as in A083449, also see A272525. [See the OEIS wiki page for more details. - M. F. Hasler, Dec 29 2020]
a(n) gives the number of times the digit 7 occurs in all terms of A000027 in the interval [A000027(1), A007908(n)]. - Felix Fröhlich, Oct 28 2016
The sequence was initially defined only up to n = 9 and then extended using A007908 = concat(1..n); see A277837 for the extension using A014824 (a(n) = 10 a(n-1) + n) leading to a smoother growth, in particular at powers of 10. - M. F. Hasler, Nov 01 2016, edited Dec 29 2020

			22 is the third term of the sequence because there are 22 occurrences of the digit '7' contained in numbers within the range of 1 to 123.
96022049 is the 9th term of the sequence because there are 96022049 occurrences of the digit '7' contained in numbers within the range of 1 to 123456789.

Puzzling Stack Exchange, How many sevens?
M. F. Hasler, Digits d in 0 through 123...n, OEIS Wiki, Nov. 2016.

Cf. A083449, A272525, A014824.

Cf. A277830 - A277838 and A277849: analog for digits 0 .. 9, but based on A014824 instead of A083449.

Table[a[n] = Count[Flatten@ Map[IntegerDigits, Range@ FromDigits@ Range@ n], k_ /; k == 8]; Print@ a@ n; an = a[n]; an, {n, 0, 9}] (* Michael De Vlieger, Oct 30 2016 *)

print1(c=0);N=1;for(n=2,8,print1(","c+=sum(k=N+1,N=eval(Str(N,n)),#select(d->d==7,digits(k))))) \\ For illustration; more efficient code below. - M. F. Hasler, Oct 31 2016

A277635(n, m=7)=if(n>m,A277635(n, m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ Valid only for n <= 9. - M. F. Hasler, Nov 02 2016

A277838 Number of '8' 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, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713306, 8367628, 96021959, 1083676380, 12071331701, 133058996022, 1454046750343, 15775035404664, 170096033058985, 1824417120713306, 19478739108367627, 207133070096021958, 2194787491083676380, 23182442812071331701

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '8' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '8' in { 18, 28, ..., 78, 80, ..., 89, 98, 108, 118 }, where 88 accounts for two '8's.

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

Cf. A277830 - A277837, A277849, A277635, A272525, A083449, A014824.

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==8,digits(k)))))

A277838(n,m=8)=if(n>m,A277838(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 8, a(8) = A277849(8) + 1 = A277837(8) - 9.

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

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

A277837 Number of '7' 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, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4664, 58985, 713307, 8367637, 96022049, 1083677281, 12071340713, 133059086145, 1454047651577, 15775044417009, 170096123182441, 1824418021947873, 19478748120713314, 207133160219478837, 2194788392318245180, 23182451824417019723

Offset: 0

M. F. Hasler, Nov 01 2016

This sequence is very similar (actually equal, for 1 <= n <= 9) to A277635, which was the original motivation for considering the family A277830 - A277838 and A277849. The main difference is that A277635 is based on A007908 (where 123456789 is followed by 12345678910) while this family is based on A014824, starts as the latter at offset 0, and therefore has a strongly different growth for n > 9.

			For n=2 there is only one digit '7' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '7' in { 17, 27, ..., 67, 70, ..., 79, 87, 97, 107, 117 }, where 77 accounts for two '7's.

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

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

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==7,digits(k)))))

A277837(n,m=7)=if(n>16,error("n>16 not yet implemented"), n>m,A277837(n,m+1)+(m+2)*10^(n-m-1),(9*n-11)*(10^n+1)\729+2-(m>n)) \\ Edited by M. F. Hasler, Dec 29 2020

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 7,
a(n) = A277836(n) - 8*10^(n-7) [for n >= 7] = A277838(n) + 9*10^(n-8) [for n >= 8].
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

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

A277831 Number of '1' 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, ...).

Original entry on oeis.org

0, 1, 5, 57, 689, 8121, 93553, 1058985, 11824417, 130589849, 1429355281, 15528120716, 167626886179, 1799725651922, 19231824420465, 204663923217008, 2170096022293551, 22935528124170094, 241700960254046637, 2540466392663923180, 26639231827873799724

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 are counted the same '1' as for n=1, plus the 4 additional digits '1' in 10, 11 and 12.

David A. Corneth, Table of n, a(n) for n = 0..998
M. F. Hasler, "Digits d in 0 through 123...n", OEIS Wiki, Nov. 2016.

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

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==1,digits(k)))))

A277831(n)=if(n<2,n, n<11, A277832(n)+3*10^(n-2), error("n > 10 not yet implemented")) \\ M. F. Hasler, Nov 02 2016, edited Dec 28 2020

a(n) = A277832(n) + 3*10^(n-2), for 2 <= n <= 10.

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

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

A277832 Number of '2' 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, ...).

Original entry on oeis.org

0, 0, 2, 27, 389, 5121, 63553, 758985, 8824417, 100589849, 1129355281, 12528120713, 137626886149, 1499725651622, 16231824417465, 174663923187008, 1870096021993551, 19935528121170094, 211700960224046637, 2240466392363923180, 23639231824873799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 are counted the two '2's in { 2, 12 }.

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

Cf. A102685, A277830 - A277838, A277849, A277635, A272525, A083449, A014824.

Array[Total@ DigitCount[Range[Sum[10^i - 1, {i, #}]/9], 10, 2] &, 7] (* Michael De Vlieger, Dec 31 2020 *)

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==2,digits(k)))))

A277832(n)=if(n<3,(n==2)*2, n<13,A277833(n)+4*10^(n-3), error("n > 12 not yet implemented")) \\ M. F. Hasler, Nov 02 2016, edited Dec 28 2020

a(n) = {if(n == 0, return(0)); n = (10^(n+1)\9-n)\9; f(n, 2) }
f(n, {c = 2}) = { my(d = digits(n), res = 0); for(i = 1, #d - 1, res += d[i] * (#d - i)*10^(#d - i - 1); if(d[i]==c, res+=(n % (10^(#d - i)) + 1); ); if(d[i] > c, res+=(10^(#d - i)) ); ); if(d[#d] >= c, res++); res } \\ David A. Corneth, Dec 31 2020

a(n) = A277831(n) - 3*10^(n-2) [for n >= 2] = A277833(n) + 4*10^(n-3) [n >= 3].

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

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

A277833 Number of '3' 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, ...).

Original entry on oeis.org

0, 0, 1, 23, 349, 4721, 59553, 718985, 8424417, 96589849, 1089355281, 12128120713, 133626886145, 1459725651582, 15831824417065, 170663923183008, 1830096021953551, 19535528120770094, 207700960220046637, 2200466392323923180, 23239231824473799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n=2 there is only one digit '3' in the sequence 0, 1, 2, *3*, 4, ..., 12.
For n=3 there are 12 + 10 = 22 more digits '3' in { 13, 23, 30, ..., 39, 43, 53, ..., 123 }, where 33 accounts for two '3's.

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

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

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==3,digits(k)))))

A277833(n,m=3)=if(n>12, error("not yet implemented"), n>m, A277833(n,m+1)+(m+2)*10^(n-m-1), (9*n-11)*(10^n+1)\729+2-(m>n)) \\ M. F. Hasler, Nov 02 2016, edited Dec 29 2020

a(n) = A277832(n) - 4*10^(n-3) [for n >= 3] = A277834(n) + 5*10^(n-4) [for n >= 4].

More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

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

A277835 Number of '5' 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, ...).

Original entry on oeis.org

0, 0, 1, 22, 343, 4665, 58993, 713385, 8368417, 96029849, 1083755281, 12072120713, 133066886145, 1454125651577, 15775824417009, 170103923182448, 1824496021947951, 19479528120714094, 207140960219486637, 2194866392318323180, 23183231824417799723

Offset: 0

M. F. Hasler, Nov 01 2016

			For n = 2 there is only one digit '5' in the sequence 0, 1, 2, ..., 12.
For n = 3 there are 11 + 10 = 21 more digits '5' in { 15, 25, ..., 45, 50, ..., 59, 65, ..., 115 }, where 55 accounts for two '5's.

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

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

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==5,digits(k)))))

A277835(n,m=5)=if(n>m,A277835(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n)) \\ M. F. Hasler, Nov 02 2016

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 5,
a(5) = A277836(5) + 1, a(n) = A277836(n) + 7*10^(n-6) for n >= 6.
More generally, for m = 0, ..., 9, let a[m] denote A277830, ..., A277838 and A277849, respectively. Then a[0](n) = a[n](n) = a[m](n) + 1 for all m > n >= 0, and a[m-1](n) = a[m](n) + (m+1)*10^(n-m) for all n >= m > 1.

More terms from Lars Blomberg, Nov 05 2016

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

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, ...).

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Nov 01 2016

			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.

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

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

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 *)

print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==6,digits(k)))))

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

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.

More terms from Lars Blomberg, Nov 05 2016

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

cp's OEIS Frontend

A277830 Number of digits '0' 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, ...).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A277849 Number of digits '9' 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, ...).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A277635 Number of 7's appearing in the sequence of consecutive natural numbers from 1 to A007908(n), where A007908 = (1, 12, 123, 1234, ...).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

A277838 Number of '8' 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, ...).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A277837 Number of '7' 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, ...).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A277831 Number of '1' 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, ...).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

Extensions

A277832 Number of '2' 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, ...).

Views

Author

Keywords

Examples

Links