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.

A072543 Numbers whose largest decimal digit is also the initial digit.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 21, 22, 30, 31, 32, 33, 40, 41, 42, 43, 44, 50, 51, 52, 53, 54, 55, 60, 61, 62, 63, 64, 65, 66, 70, 71, 72, 73, 74, 75, 76, 77, 80, 81, 82, 83, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 110, 111, 200, 201
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 04 2002

Keywords

Comments

A054055(a(n)) = A000030(a(n));
the sequence differs from A009996, A032873 and A032907: a(66)=101 is not in A009996, a(67)=110 is not in A032873 and a(65)=100 is not in A032907.

Examples

			a(10^ 1) = 9
a(10^ 2) = 411
a(10^ 3) = 6216
a(10^ 4) = 73474
a(10^ 5) = 813826
a(10^ 6) = 8512170
a(10^ 7) = 88368780
a(10^ 8) = 911960211
a(10^ 9) = 9237655227
a(10^10) = 93323313303
		

Crossrefs

Cf. A072544.

Programs

  • Haskell
    a072543 n = a072543_list !! (n-1)
    a072543_list = [x | x <- [0..], a054055 x == a000030 x]
    -- Reinhard Zumkeller, Apr 25 2012
    
  • Maple
    for i from 1 to 10 do A[i]:= i-1 od:
    count:= 10:
    for i from 1 to 9 do P[i]:= [seq([j],j=0..i)]; od:
    for d from 2 to 4 do
      for x from 1 to 9 do
        for p in P[x] do
          count:= count+1;
          A[count]:= add(p[k]*10^(k-1),k=1..d-1) + x*10^(d-1);
        od:
        P[x]:= [seq(seq([op(v),t], v=P[x]),t=0..x)];
      od
    od:
    seq(A[i],i=1..count); # Robert Israel, Feb 01 2015
  • Mathematica
    Select[Range[0,250],Max[IntegerDigits[#]]==First[IntegerDigits[#]]&] (* Harvey P. Dale, Apr 28 2016 *)
  • PARI
    is(n)=n=digits(n); !#n || n[1]==vecmax(n) \\ Charles R Greathouse IV, Jan 02 2014
    
  • PARI
    a(n)={d = 0; r = 1; s = 0; i = 0; if(n == 1, 0, n-=2; while(n > sum(i=0, 9,(i+1)^d), n-=sum(i=0, 9, (i+1)^d); n++; d++); while(n >= (r+1)^d, n -= (r+1)^d; r++);s = r * 10^d; while(n, s += 10^i*(n%(r+1)); n \= (r+1); i++));s } \\ David A. Corneth, Jan 31 2015

Extensions

Offset corrected by Reinhard Zumkeller, Apr 25 2012