cp's OEIS Frontend

Every row of this triangle represents a number which is normal in base b >= 2. Since there are infinitely many bases, every row represents a set of infinitely many numbers, each of which is normal in at least one base.

Treating A(n,k) as equal to k when k is greater than A084558(n) (A084558(n) is the length of the n-th row minus one), for any base b >= 2, the concatenation of the base-b expansion of the n-th row is normal in base b. In other words, for any n >= 0 and b >= 2, C(n,b) is normal in base b.

C(n,b) = Sum_{k >= 0} A(n,k)/(b^(Sum_{m=0..k} ceiling(log_b(A(n,m)+1))))

(The equation for Champernowne's constants using the n-th row of this triangle rather than 0,1,2,...)

C(0,b) is Champernowne's constant for base b (C_b).

Even though for any b and m >= 2, b != m, C(n,b) != C(n,m), it is possible for C(n,b) = C(m,p) where n != m and b != p. In such a case, C(n,b) is normal in two different bases. m will likely be significantly larger than n and p will likely be a power of b.

The base-n Champernowne constant (C_n) is normal in base n. A(n,k) is the (k+1)-th decimal digit of the fractional part of C_n.

This sequence uses the centered version of mod. The residue system modulo prime(n) is {-1*floor(prime(n)/2)..floor(prime(n)/2)}. This is so that this sequence will encode information about the numbers around 2^A007013(4). If a(n) = k and prime(n) < 2^A007013(4) - k, then 2^A007013(4) - k is not prime (prime(n) is a factor of 2^A007013(4) - k). For example, a(22) = -3, so prime(22) = 79 is a factor of 2^A007013(4) + 3.

The length of this sequence is the lowest value of n such that A014664(n) = A007013(4). This is because for any power of 2, 2^p, if p == 0 (mod A014664(n)), then 2^p == 1 (mod prime(n)) (prime(n) is a factor of A000225(p)). Since A007013(4) is prime, we can apply this to get: If A014664(n) = A007013(4) and prime(n) < A007013(5), then A007013(5) is not prime (prime(n) is a nontrivial factor).

For any n such that prime(n) < 5*(10^51 + 5*10^9), a(n) != 1.

Although the most significant digits of powers of 2 in base n are generally not periodic (the exception being when n is a power of 2), the least significant digits are. For example, 2 to an even power is congruent to 1 (mod 3) and 2 to an odd power is congruent to -1 (mod 3). This means that one can determine one of the prime factors of a Mersenne number, A000225, using the exponent. If n == 0 (mod 2), then A000225(n) == 0 (mod 3) (is a multiple of 3); if n == 0 (mod 4), then A000225(n) == 0 (mod 5); if n == 0 (mod 3), then A000225(n) == 0 (mod 7), and so on.

This general fact gives a reason for why certain Mersenne numbers are not prime (even with prime exponents). If p is congruent to 0 mod A014664(n) (the length of an n-th row) and prime(n) is less than the A000225(p), then prime(n) is a nontrivial factor of A000225(p).

The n-th row of this triangle is the cycle of the (n-th)-to-last digit of powers of 2.

The period of the last n digits of powers of 2 where the exponent is greater than or equal to n is A005054(n). As a result, this triangle can be used to get the (n-th)-to-last digit of a large power of 2; if p == k + A123384(n-1) (mod A005054(n)), then the (n-th)-to-last digit (base 10) of 2^p is T(n,k). For example, for n = 1, if p == 1 (mod 4), then 2^p == 2 (mod 10) and if p == 3 (mod 4), then 2^p == 8 (mod 10). For n = 2, if p == 4 (mod 20), then the second-to-last digit of 2^p (base 10) is 1 and if p == 7 (mod 20), then the second-to-last digit of 2^p (base 10) is 2.

Although the sequences for the most significant digits of Mersenne numbers, A000225, are not cyclic (the most significant is not cyclic, the second most is not, etc.), the sequences for the least significant digits are. For example, if p == 3 (mod 4), then A000225(p) == 7 (mod 10). Since A007013(n + 1) = A000225(A007013(n)) and A007013(1) == 3 (mod 4), all subsequent values will be congruent to 7 (mod 10). Similarly, if p == 7 (mod 20), A000225(p) == 27 (mod 100). In general, if p == x (mod A005054(n)), then A000225(p) == A000225(x) (mod 10^n).

There are many primes of the form Sum_{i=1..n} a(i)*10^(i - 1). The largest known is for n = 7032 (it is 7032 digits long).

A(n,k) is the least m such that m contains the base-n expansions of the active bits (plus 1) in k as substrings.

The Shortest Superstring Problem is, given a set of strings, S, to find the shortest string which contains each element of S as a substring. All possible solutions to this problem are contained in this array. The values of k represent the set of strings (where the active bits represent the strings in base n). The value of k for non-numeral strings (or numeral strings with an initial 0) is generated by mapping each character to a unique value 1 through n, converting from base n+1, subtracting 1 from each, raising 2 to the power of each and then summing the result. A(n+1,k) in base n+1 is the shortest superstring. The value of k for numeral strings in base n (without initial 0's) is generated by just raising 2 to the power of the value of each and then summing the result. A(n,k) in base n is the shortest superstring.

Complex binary (base -1 + i) has the ability to express all positive or negative, real or complex, integers with only 2 numerical symbols ('0' and '1') as integers, without the need for a sign marking the integers as such.

Converting a real number, n, to complex binary requires one to convert it to base -4 ((n + N) xor N, N = floor(4/5*16^(ceiling(log_4(abs(n))) + 1))), then adding 10 to every digit greater than 1, then treating it as a number in base 16 and converting that to binary. (E.g., -5 => [2,3] => [12,13] => 205 => 11001101.)

Converting a complex number, n + k*i, requires one to convert X = n + k into complex binary and then convert k into the same but shift it one digit to the left. After this, one must add them together. This functions much the same way as binary addition, but the carry is '110' rather than '1' and 11 + 111 = 0.

Pi is normal in base n >= 2 if and only if in every row N, such that N is a power of n, -1 does not appear. Pi is absolutely normal if and only if -1 never appears.

Conjecture: Pi is absolutely normal, meaning that -1 will never appear.

This triangle is an instance of the more general f(n,k,r), where f(n,k,r) is the smallest m >= 0 such that floor(r*n^m) == k (mod n) (-1 if one does not exist) and r is irrational. The same conditions for normalcy apply.