cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

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, ...).

Original entry on oeis.org

1, 1, 2, 23, 344, 4665, 58986, 713307, 8367628, 96021949, 1083676272, 12071330614, 133058985146, 1454046641578, 15775034317010, 170096022182442, 1824417011947874, 19478738020713306, 207133059219478738, 2194787382318244170, 23182441724417009624, 244170096256515775267
Offset: 0

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Comments

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

Links

Crossrefs

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

Programs

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

Formula

a(n) = A083449(n) + 1 for n <= 9.
a(n) = 1 + A061217(A014824(n)), taking A061217(0)=0. - Kevin Ryde, Nov 07 2020

Extensions

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

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.

Extensions

More terms from Lars Blomberg, Nov 05 2016
Replaced incorrect b-file by Lars Blomberg, Nov 11 2020

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

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.

Extensions

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Comments

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.

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

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.

Extensions

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

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

Links

Crossrefs

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

Programs

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

Formula

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.

Extensions

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

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

Links

Crossrefs

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

Programs

  • Mathematica
    Array[Total@ DigitCount[Range[Sum[10^i - 1, {i, #}]/9], 10, 2] &, 7] (* Michael De Vlieger, Dec 31 2020 *)
  • PARI
    print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==2,digits(k)))))
  • PARI
    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
  • PARI
    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

Formula

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.

Extensions

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

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.

Extensions

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

			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.

Links

Crossrefs

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

Programs

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

Formula

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.

Extensions

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

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

			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.

Links

Crossrefs

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

Programs

  • 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 *)
  • 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

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.

Extensions

More terms from Lars Blomberg, Nov 05 2016
Removed incorrect b-file. - David A. Corneth, Dec 31 2020

A277834 Number of '4' 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, 344, 4671, 59053, 713985, 8374417, 96089849, 1084355281, 12078120713, 133126886145, 1454725651577, 15781824417015, 170163923182508, 1825096021948551, 19485528120720094, 207200960219546637, 2195466392318923180, 23189231824423799723
Offset: 0

Views

Author

M. F. Hasler, Nov 01 2016

Keywords

Examples

			For n=2 there is only one digit '4' in the sequence 0, 1, 2, ..., 12.
For n=3 there are 11 + 10 = 21 more digits '4' in { 14, 24, 34, 40, ..., 49, 54, ..., 114 }, where 44 accounts for two '4's.

Links

Crossrefs

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

Programs

  • PARI
    print1(c=N=0);for(n=1,8,print1(","c+=sum(k=N+1,N=N*10+n,#select(d->d==4,digits(k)))))
  • PARI
    A277834(n,m=4)=if(n>m,A277833(n,m+1)+(m+2)*10^(n-m-1),A277830(n)-(m>n))

Formula

a(n) = A277849(n) = A083449(n) = A277830(n) - 1 for n < 4,
a(n) = A277833(n) - 5*10^(n-4) for n >= 4, a(n) = A277835(n) + 6*10^(n-5) for n >= 5.
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.

Extensions

More terms from Lars Blomberg, Nov 05 2016
Removed incorrect b-file. - David A. Corneth, Dec 31 2020
Showing 1-10 of 10 results.

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