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 1-2 of 2 results.

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

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 1-2 of 2 results.

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