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.

A232731 Number of numbers that yield an n-digit number after Reverse and Add!.

Original entry on oeis.org

4, 50, 450, 4590, 45405, 454950, 4544550, 45454500, 454495500, 4545045000, 45449955000, 454500450000, 4544999550000, 45450004500000, 454499995500000, 4545000045000000, 45449999955000000, 454500000450000000, 4544999999550000000, 45450000004500000000
Offset: 1

Views

Author

Lars Blomberg, Nov 29 2013

Keywords

Comments

A190481 is the number of distinct n-digit numbers after Reverse and Add!.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{10,10,-100},{4,50,450,4590,45405},20] (* Harvey P. Dale, Sep 22 2017 *)
  • PARI
    Vec(x*(4 + 10*x - 90*x^2 - 10*x^3 + 5*x^4) / ((1 - 10*x)*(1 - 10*x^2)) + O(x^30)) \\ Colin Barker, Mar 20 2017

Formula

a(1) = A232729(1), a(n) = A232729(n) + A232730(n-1), n > 1.
From Colin Barker, Mar 20 2017: (Start)
G.f.: x*(4 + 10*x - 90*x^2 - 10*x^3 + 5*x^4) / ((1 - 10*x)*(1 - 10*x^2)).
a(n) = 10*a(n-1) + 10*a(n-2) - 100*a(n-3) for n>5.
(End)

A232729 Number of n-digit numbers that yield an n-digit number after Reverse and Add.

Original entry on oeis.org

4, 45, 405, 4095, 40500, 405450, 4050000, 40504500, 405000000, 4050045000, 40500000000, 405000450000, 4050000000000, 40500004500000, 405000000000000, 4050000045000000, 40500000000000000, 405000000450000000, 4050000000000000000, 40500000004500000000, 405000000000000000000
Offset: 1

Views

Author

Lars Blomberg, Nov 29 2013

Keywords

Examples

			There are 4 1-digit numbers (1,2,3,4) that yield a 1-digit number (2,4,6,8), so a(1)=4.
		

Crossrefs

Programs

  • PARI
    Vec(x*(4+5*x-85*x^2-5*x^3)/((1-10*x)*(1-10*x^2)) + O(x^50)) \\ Colin Barker, Apr 22 2016

Formula

a(n) + A232730(n) = 9*10^(n-1).
a(1) = 4, a(3) = 405, a(2k+1) = 100*a(2k-1), k > 1.
a(2) = 45, a(4) = 4095, a(2k) = 110*a(2k-2) - 1000*a(2k-4), k > 2.
From Colin Barker, Apr 21 2016: (Start)
a(n) = 10*a(n-1)+10*a(n-2)-100*a(n-3) for n>4.
G.f.: x*(4+5*x-85*x^2-5*x^3) / ((1-10*x)*(1-10*x^2)). (End)
E.g.f.: (81*cosh(10*x) + 90*cosh(sqrt(10)*x) + 81*sinh(10*x) - 10*x - 171)/200. - Stefano Spezia, Oct 27 2022
Showing 1-2 of 2 results.