David Cobac - cp's OEIS frontend

User: David Cobac

David Cobac has authored 3 sequences.

A329181 a(n) = n if n is 1 or prime; otherwise (1) let m = (concatenation of the two divisors in the middle of rows of A027750(n)), (2) if m is prime then a(n) = m, otherwise return to (1) with n=m.

1, 2, 3, 211, 5, 23, 7, 223, 311, 773, 11, 21179, 13, 313, 1129, 3137, 17, 3449, 19, 59, 37, 211, 23, 223, 773, 3251, 313, 47, 29, 613, 31, 4373, 311, 21179, 1129, 3449, 37, 373, 313, 229, 41, 67, 43, 3137, 59, 223, 47, 4373, 131321, 33391, 317, 2333, 53

Offset: 1

David Cobac, Nov 07 2019

A term is a prime (or 1 for the first one) obtained by concatenating its two factors that are closest to its square root. Once they are concatenated (as strings), the process is iterated until concatenation gives a prime. Only composite numbers are processed.
At each iteration, we choose a couple (d, d') of divisors this way: n = d * d' and d = max({d >= 1 such that d|n and d<=sqrt(n)}), we replace n with the string concatenation of d and d' digits. The process ends with d = 1 (n is a prime).
This sequence is the balanced version of A316941.
Apparently it is only a conjecture that the process in the definition will alwats terminate. - N. J. A. Sloane, Feb 23 2020

			First three positive integers 1, 2, 3 do not change, so a(n)=n, for n <= 3.
4th term: sqrt(4)=2 and n=4=2*2 then n=22=2*11 and n=211 is a prime, so a(4)=211.
8th term: floor(sqrt(8))=2 and n=8=2*4 then n=24 but floor(sqrt(24))=4 so n=24=4*6 but floor(sqrt(46))=6 and 46's nearest factor to 6 is 2; thus 46=2*23 and 223 is a prime, so a(8)=223.
Thus a(8)=a(24)=a(46)=a(223)=223.

David Cobac, Table of n, a(n) for n = 1..590
David Cobac, C code

Same process as A316941 with different factors.

Cf. A002808 (the composite numbers), A027750 (the divisors of n).

C
```
/* See Cobac link. */
```

Array[If[! CompositeQ@ #, #, NestWhile[Block[{k = Floor@ Sqrt@ #}, While[Mod[#, k] != 0, k--]; FromDigits@ Flatten[IntegerDigits /@ {k, #/k}]] &, #, !PrimeQ@ # &]] &, 53] (* Michael De Vlieger, Nov 15 2019 *)

a(n) = {if (n==1, return (1)); if (isprime(n), return (n)); while (!isprime(n), my(d = divisors(n)); if (#d % 2 == 1, n = eval(concat(Str(d[#d\2+1]), Str(d[#d\2+1]))), n = eval(concat(Str(d[#d/2]), Str(d[#d/2+1]))));); n;} \\ Michel Marcus, Nov 15 2019

A306475 Smallest nonprime number <= 10^n (n>=1) with maximum distance from a prime.

Original entry on oeis.org

9, 93, 897, 9569, 31433, 492170, 4652430, 47326803, 436273150, 4302407536, 42652618575, 738832928197, 7177162612050, 90874329411895, 218209405436996, 1693182318746937, 80873624627235459, 804212830686678390

Offset: 1

David Cobac, Feb 18 2019

Each number is a mean of two consecutive primes.

Since, except 2, primes are odd numbers, this mean is an integer.

			For n=1: first prime numbers are 2, 3, 5, 7 and 11. Maximum difference between two consecutive primes is 4 between 7 and 11 thus a(1)=9.
For n=4: maximum difference between two primes less than 10^4 is 36, which occurs once: between 9551 and 9587. a(4)=(9551 + 9587)/2 = 9569.

Cf. A000040, A001223, A087378, A282690.

More terms (using the b-file at A002386) from Jon E. Schoenfield, Feb 19 2019

A323416 a(n) = (n-1)! * (10^n - 1) / 9.

Original entry on oeis.org

1, 11, 222, 6666, 266664, 13333320, 799999920, 55999999440, 4479999995520, 403199999959680, 40319999999596800, 4435199999995564800, 532223999999946777600, 69189119999999308108800, 9686476799999990313523200, 1452971519999999854702848000, 232475443199999997675245568000, 39520825343999999960479174656000

Offset: 1

David Cobac, Jan 13 2019

Take an n-digit number with distinct digits, add all permutations of the digits, divide by the sum of the digits: the result is a(n).
Proof from David A. Corneth, Jan 14 2019: (Start)
Let m be an n-digit number (without leading 0, where n > 0). Then n! permutations of digits can be formed.
So each digit occurs n!/n = (n-1)! times in each position. Therefore the total sum is (10^n - 1) * (n - 1)! * s where s is the sum of digits of n. Dividing this product by s gives a(n) = (10^n - 1) * (n - 1)!. QED (End)

			Example for n = 3:
Take the number 569.
Sum the permutations of its digits: 569 + 596 + 659 + 695 + 956 + 965 = 4440.
Add all its digits: 5 + 6 + 9 = 20.
Divide: 4440 / 20 = 222.
General proof for n = 3:
Number: abc where a,b,c are distinct.
The sum of the permutations is 200*(a+b+c) + 20*(a+b+c) + 2*(a+b+c) = 222*(a+b+c), so a(3) = 222.

David Cobac, Table of n, a(n) for n = 1..325
G. Villemin, Nombres permutés

Cf. A000142, A002275, A071267. Sum of digits A110728.

Table[(n-1)! (10^n-1)/9,{n,20}] (* Harvey P. Dale, Mar 15 2024 *)

a(n) = (10^n - 1) / 9 * (n-1)! \\ David A. Corneth, Jan 13 2019

f = lambda n:+(n==0) or n*f(n-1)
def seq(n):
   if n==0: return
   l = []
   for i in range(1, n + 1):
       # following line with a string repeat
       # s = int('1'*i)
       s = 0
       for j in range(i):
           s += 10 ** j
       l += [s*f(i-1)]
   return l

Recurrence relation: a(n+1) = n! * 10^n + n * a(n).
Proof: Assume R_n is a string of n 1's (repunit),
a(n) = (n-1)! * R_n so a(n+1) = n! * R_{n+1} = n! * (10^n + R_n);
Thus a(n+1) = n! * 10^n + n! * R_n = n! * 10^n + n * (n-1)! * R_n;
Hence a(n+1) = n! * 10^n + n * a(n).

Edited by N. J. A. Sloane, Jan 19 2019

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User: David Cobac

A329181 a(n) = n if n is 1 or prime; otherwise (1) let m = (concatenation of the two divisors in the middle of rows of A027750(n)), (2) if m is prime then a(n) = m, otherwise return to (1) with n=m.

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A306475 Smallest nonprime number <= 10^n (n>=1) with maximum distance from a prime.

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A323416 a(n) = (n-1)! * (10^n - 1) / 9.

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