login
Inverse Moebius transform of powers of 10.
4

%I #12 May 25 2017 04:21:08

%S 1,10,110,1010,10110,100010,1001110,10000010,100010110,1000001010,

%T 10000100110,100000000010,1000001011110,10000000000010,

%U 100000010000110,1000000000101010,10000000100010110,100000000000000010,1000000001001001110,10000000000000000010,100000000010000110110

%N Inverse Moebius transform of powers of 10.

%C A055895 to base 2.

%C The triangle formed by stacking the reversals of a(n) is A113704.

%C Using decimal positional notation, write a 1 at position d for all divisors d of n. All other digits are zeros and leading zeros are not permitted. - _Michael De Vlieger_, May 24 2017

%H Michael De Vlieger, <a href="/A113705/b113705.txt">Table of n, a(n) for n = 0..999</a>

%F a(n) = Sum_{k=0..n} if(mod(n, k)=0, 10^k, 0).

%F G.f.: Sum_{k>=1} 10^k*x^k/(1 - x^k). - _Ilya Gutkovskiy_, May 23 2017

%e From _Michael De Vlieger_, May 24 2017: (Start)

%e First 20 terms of a(n), replacing zeros with "." to more clearly show positions of 1s in positions corresponding to terms in row n of A027750. This chart also pertains to terms of A055895 written in binary.

%e n a(n) A027750(n)

%e ---------------------------------------

%e 0: 1 {}

%e 1: 1. 1

%e 2: 11. 1,2

%e 3: 1.1. 1,3

%e 4: 1.11. 1,2,4

%e 5: 1...1. 1,5

%e 6: 1..111. 1,2,3,6

%e 7: 1.....1. 1,7

%e 8: 1...1.11. 1,2,4,8

%e 9: 1.....1.1. 1,3,9

%e 10: 1....1..11. 1,2,5,10

%e 11: 1.........1. 1,11

%e 12: 1.....1.1111. 1,2,3,4,6,12

%e 13: 1...........1. 1,13

%e 14: 1......1....11. 1,2,7,14

%e 15: 1.........1.1.1. 1,3,5,15

%e 16: 1.......1...1.11. 1,2,4,8,16

%e 17: 1...............1. 1,17

%e 18: 1........1..1..111. 1,2,3,6,9,18

%e 19: 1.................1. 1,19

%e 20: 1.........1....11.11. 1,2,4,5,10,20 (End)

%t Table[If[n == 0, 1, Total[10^Divisors[n]]], {n, 0, 20}] (* or *)

%t Table[If[n == 0, 1, Sum[If[Mod[n, k] == 0, 10^k, 0], {k, n}]], {n, 0,

%t 20}] (* or *)

%t Table[Boole[n == 0] + Total@ MapIndexed[Boole[Divisible[n, #1]]*10^First@ #2 &, Range@ n], {n, 0, 20}] (* or *)

%t Table[If[n == 0, 1, Function[d, FromDigits @Reverse@ ReplacePart[#, Map[# + 1 -> 1 &, d]] &@ ConstantArray[0, n + 1]]@ Divisors@ n], {n, 0, 20}]

%t (* _Michael De Vlieger_, May 24 2017 *)

%o (PARI) a(n) = if (n==0, 1, sum(k=1, n, if (! (n % k), 10^k))); \\ _Michel Marcus_, May 23 2017

%Y Cf. A027750, A055895, A113704.

%K easy,nonn

%O 0,2

%A _Paul Barry_, Nov 05 2005