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.

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A085890 7-smooth numbers (A002473) using digits in descending order. Zero is followed by a 9.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 21, 32, 54, 98, 210, 432
Offset: 1

Views

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jul 10 2003

Keywords

Comments

No more terms < 10^1000. Probably no more terms. - David Wasserman, Feb 10 2005

Examples

			432 is a member as 432= 2^4*3^3.
		

Crossrefs

Intersection of A002473 and A158699.
Cf. A085889.

Programs

  • PARI
    hcn(n) = while (!(n%2), n \=2); while (!(n%3), n \=3); while (!(n%5), n \=5); while (!(n%7), n \=7); n == 1;
    for (i = 0, 1000, for (j = 1, 9, my(n = sum(k = j - i, j, (k%10)*10^(i - j + k))); if (hcn(n), print1(n, ", ")))); /* David Wasserman, Feb 10 2005 */

Extensions

More terms from David Wasserman, Feb 10 2005

A247107 a(0) = 0, a(n) = previous term + repunit of length of previous term for n > 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 21, 32, 43, 54, 65, 76, 87, 98, 109, 220, 331, 442, 553, 664, 775, 886, 997, 1108, 2219, 3330, 4441, 5552, 6663, 7774, 8885, 9996, 11107, 22218, 33329, 44440, 55551, 66662, 77773, 88884, 99995, 111106, 222217, 333328, 444439
Offset: 0

Views

Author

Dhilan Lahoti, Nov 30 2014

Keywords

Examples

			98 = 9*10 + 8 -> 10*10 + 9 = 109.
109 = 1*100 + 0*10 + 9*1 -> 2*100 + 1*10 + 10*1 = 220.
a(42) = 44440 + (10^(floor(log_10(44440))+1)-1) / 9 = 44440 + (10^(4+1)-1) / 9 = 44440 + 99999/9 = 44440 + 11111 = 55551.
		

Crossrefs

Similar to A158699, but with simpler rules.

Programs

  • Mathematica
    a[0]=0; a[n_]:=FromDigits[IntegerDigits[a[n-1]]+1]; Array[a,50,0] (* Stefano Spezia, Sep 19 2024 *)

Formula

a(0) = 0, a(n) = a(n-1) + A002275(A055642(a(n-1))) for n>0.
From Jon E. Schoenfield, Nov 30 2014: (Start)
For n > 1, a(n) = a(n-1) + (10^(floor(log_10(a(n-1))) + 1) - 1) / 9.
For n > 0, a(n) = ((n-1) mod 9 + 1) * (10^D - 1) / 9 + 1 - D, where D = floor((n-1)/9) + 1. (There are exactly D digits in a(n).) (End)
G.f.: -(10*x^10-10*x^9+1)*x/((10*x^9-1)*(x-1)^2). - Alois P. Heinz, Nov 30 2014
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