

A306388


a(n) is a decimal number k having a length n binary expansion which encodes, from left to right at digit j, the coprimality (0) or noncoprimality (1) of j to n, for 1 < j <= n, except for the first digit, which is always 1.


1



1, 3, 5, 13, 17, 61, 65, 213, 329, 885, 1025, 3933, 4097, 13781, 22121, 54613, 65537, 251741, 262145, 906613, 1364681, 3497301, 4194305, 16111453, 17859617, 55932245, 86282825, 225793493, 268435457, 1064687485, 1073741825, 3579139413, 5526297161, 14316688725
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OFFSET

1,2


COMMENTS

Let Sum* be a special summation procedure carried out on the binary expansions of each of the decimal values produced by the following expression for all distinct prime factors of n. That is, when 'adding' the various binary expansions of said decimal results for each p dividing n, p prime, allow that 1 + q + r + ... + s = 1, and 0 + 0 + ... + 0 = 0. Then, Sum*_{pn} 2^(p1) * ((2^p+1) * 2^(np)  2)/(2^p  1) + 1, when reverted to decimal, gives a(n).
a(n) in binary, and recorded as a triangle gives a 'Totient map' for the naturals.
1 1
2 11
3 101
4 1101
5 10001
6 111101
7 1000001
8 11010101
9 101001001
10 1101110101
11 10000000001
12 111101011101
13 1000000000001
14 11010111010101
15 101011001101001
16 1101010101010101
...


LINKS



EXAMPLE

a(p), p prime, are always 2^(p1)+1, a result of ((2^p+1)*2^(np)2)/(2^p1) the main parenthetical term in Sum* being equal to 1.
a(c), c composite, is computable as follows:
a(6) = 61 because 6 has the distinct prime factors 2 and 3. So, the special summation of 2^(21) * ((2^2 + 1) * 2^(62)  2)/(2^2  1) + 1 = 53, a decimal number which has a length 6 binary expansion (110101), and 2^(31) * ((2^3 + 1) * 2^(63)  2)/(2^3  1) + 1 = 41, another decimal number which has a length 6 binary expansion (101001), gives Sum* =
110101
+ 101001
_______
111101, which, when reverted to decimal, gives a(6).
a(12) = 3933 because 12 has the distinct prime factors 2 and 3. So, the special summation of 2^(21) * ((2^2 + 1) * 2^(122)  2)/(2^2  1) + 1 = 3413, a decimal number which has a length 12 binary expansion (110101010101), and 2^(31) * ((2^3 + 1) * 2^(123)  2)/(2^3  1) + 1 = 2633, another decimal number which has a length 12 binary expansion (101001001001), gives Sum* =
110101010101
+ 101001001001
______________
111101011101, which, when reverted to decimal, gives a(12).
Likewise, a(30) = 1064687485 because 30 has the distinct prime factors 2, 3, and 5. So, the special summation of 2^(21) * ((2^2 + 1) * 2^(302)  2)/(2^2  1) + 1 = 894784853 = 110101010101010101010101010101 (length 30), and 2^(31) *((2^3 + 1) * 2^(303)  2)/(2^3  1) + 1 = 690262601 = 101001001001001001001001001001, and 2^(51) * ((2^5 + 1) * 2^(305)  2)/(2^5  1) + 1 = 571507745 = 100010000100001000010000100001, gives Sum* =
110101010101010101010101010101
101001001001001001001001001001
+ 100010000100001000010000100001
______________________________
111111011101011101011101111101, which, when reverted to decimal, gives a(30).


MATHEMATICA

a[n_] := FromDigits[Boole@(#==1  GCD[#, n] > 1) &/@ Range[n], 2]; Array[a, 30] (* Amiram Eldar, Mar 26 2019 *)


PROG

(PARI) a(n) = my(v=vector(n, k, if (k==1, 1, gcd(k, n) != 1))); fromdigits(v, 2); \\ Michel Marcus, Mar 28 2019


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



