cp's OEIS Frontend

This sequence is noteworthy in light of the congruence relation shared by a(n) and prime(n). Namely, for n > 2, a(n) == prime(n) (mod 10). That is, the last digit of prime(n) is 'preserved' as the last digit of a(n). See A007652.

As well, extending the notion, one notes that for k == 1 (mod 4), Fibonacci(2^k * prime(n)) == prime(n) (mod 10).

For any prime number p, the Fibonacci number F_(2p) == -(2p/5) (mod p), where -(2p/5) is the Legendre or Jacobi symbol. - Yike Li, Aug 30 2022

Let na and nb represent the indices of the preceding and next A003226(n)'s beginning with a 9, and where (na - nb) >= 3 (note that the first such 'zone' begins with an exception for which the index A003226(na) = 1). Then for na < n < nb and such that n == (na + 1) mod 2, it appears that A003226(n) - a(n) = A003226(n+1) - a(n+1) = k.

In such cases, it also appears that a(n)*a(n+1) = k^2 - k.

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*_{p|n} 2^(p-1) * ((2^p+1) * 2^(n-p) - 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

...

It appears that these constitute all of the 'non-degenerate' cases for k less than 10^8. That is, k is not allowed to have leading zeros, but all 'legal' permutations of k, where k has length m, must be of length m as well. Therefore leading zeros are allowed in the construction of the total permutation product.

From David A. Corneth, Feb 15 2019: (Start)

Let S(m) be the sum of the distinct prime divisors of the product of all legal permutations of the digits of m.

Let Z(m) be a number where a zero is inserted after the first digit of m (m > 0). For example, Z(1) = 10, Z(19) = 109.

All terms with at most k digits can be found by iterating only over terms in A179239 with at most k digits.

For example, 345 is in A179239. S(345) = S(543), namely 543. As 543 is a permutation of 345, s = 543 is in the sequence.

Similarily, 445 is in A179239 and S(445) = 341, 445 doesn't produce a term. As S(445) = S(454) = S(544), all these number don't produce a term and do not have to be checked.

We have S(Z(m)) >= S(m). Proof: The permutations of Z(m) give the same distinct prime factors as m does, and maybe more. Therefore, S(Z(m)) >= S(m).

This can be applied to eliminate candidates. For example, S(10378) = 1447642. The largest possible value a number with digits of Z(10378) = 100378 can have is 873100. But 1447642 > 873100. So 100378 can't produce a term and doesn't have to be checked.

To perhaps quickly eliminate a candidate without checking all permutations one might let the last digit d of a permutation be such that gcd(d, 10) = 1 to hopefully get large prime factors (if there is such d). For example, when checking if 1378 gives a candidate, start with the 12 permutations ending in 1 or 3.

To find S(m) where m has digits zero, one might use a known value of S(m') where m' has a digit 0 removed from m and proceed finding S(m) with permutations having leading nonzero digits only. (End)

Beginning at prime(2)=3, group all primes into even/odd-indexed pairs, (prime(2n), prime(2n+1)). Then a(prime(2n)) and a(prime(2n+1)) are both equal to 2*A077133(n).

This sequence consists of runs of an even number of consecutive numbers. - David A. Corneth, Dec 18 2018