A001744 Numbers n such that every digit contains a loop (version 2).

0, 4, 6, 8, 9, 40, 44, 46, 48, 49, 60, 64, 66, 68, 69, 80, 84, 86, 88, 89, 90, 94, 96, 98, 99, 400, 404, 406, 408, 409, 440, 444, 446, 448, 449, 460, 464, 466, 468, 469, 480, 484, 486, 488, 489, 490, 494, 496, 498, 499, 600, 604, 606, 608, 609, 640, 644, 646

Offset: 1

Author

N. J. A. Sloane

Keywords

• base
• nonn
• easy

Comments

See A001743 for the other version.

If n-1 is represented as a base-5 number (see A007091) according to n-1 = d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n)= Sum_{j=0..m} c(d(j))*10^j, where c(k)=0,4,6,8,9 for k=0..4. - Hieronymus Fischer, May 30 2012

Examples

			a(1000) = 46999.
a(10^4) = 809999.
a(10^5) = 44499999.
a(10^6) = 668999999.
		

Links

• Hieronymus Fischer, Table of n, a(n) for n = 1..10000
• Index entries for 10-automatic sequences.

Crossrefs

Cf. A061371, A029581, A007091, A046034, A084544, A084984, A017042, A001743, A014261, A014263, A202267, A202268.

Programs

Mathematica

FromDigits/@Tuples[{0,4,6,8,9},3] (* Harvey P. Dale, Aug 16 2018 *)

PARI

is(n) = #setintersect(vecsort(digits(n), , 8), [1, 2, 3, 5, 7])==0 \\ Felix Fröhlich, Sep 09 2019

Formula

From Hieronymus Fischer, May 30 2012: (Start)

a(n) = ((2*b_m(n)) mod 8 + 4 + floor(b_m(n)/4) - floor((b_m(n)+1)/4))*10^m + sum_{j=0..m-1} ((2*b_j(n))) mod 10 + 2*floor((b_j(n)+4)/5) - floor((b_j(n)+1)/5) -floor(b_j(n)/5)))*10^j, where n>1, b_j(n)) = floor((n-1-5^m)/5^j), m = floor(log_5(n-1)).

a(1*5^n+1) = 4*10^n.

a(2*5^n+1) = 6*10^n.

a(3*5^n+1) = 8*10^n.

a(4*5^n+1) = 9*10^n.

a(n) = 4*10^log_5(n-1) for n=5^k+1,

a(n) < 4*10^log_5(n-1), otherwise.

a(n) > 10^log_5(n-1) n>1.

a(n) = 4*A007091(n-1), iff the digits of A007091(n-1) are 0 or 1.

G.f.: g(x) = (x/(1-x))*sum_{j>=0} 10^j*x^5^j*(1-x^5^j)*(4 + 6x^5^j + 8(x^2)^5^j + 9(x^3)^5^j)/(1-x^5^(j+1)).

Also: g(x) = (x/(1-x))*(4*h_(5,1)(x) + 2*h_(5,2)(x) + 2*h_(5,3)(x) + h_(5,4)(x) - 9*h_(5,5)(x)), where h_(5,k)(x) = sum_{j>=0} 10^j*(x^5^j)^k/(1-(x^5^j)^5). (End)

Extensions

Ambiguous comment deleted by Zak Seidov, May 25 2010

Examples added by Hieronymus Fischer, May 30 2012