A276835 Numerator of a modified exponentiated von Mangoldt function defined recursively.

1, 2, 3, 2, 5, 1, 7, 1, 3, 1, 11, 1, 13, 1, 1, 6, 17, 3, 19, 4, 1, 3, 23, 90, 5, 5, 3, 3, 29, 12, 31, 112, 3, 105, 1, 50, 37, 5, 1, 27, 41, 81, 43, 10, 1, 105, 47, 539, 7, 77, 15, 4, 265, 2, 3, 520, 3, 351, 59, 945

Offset: 1

Author

Mats Granvik, Sep 20 2016

Keywords

• nonn
• frac

Comments

Conjecture: For n>3: If and only if the ratio A276835(n)/A276836(n) is equal to n then n is equal to the greater of the twin primes A006512.

Justification: Whenever n is equal to the greater of the twin primes then in the recurrence that defines the table t(n,k) at k=1 the Product_{i=1..n-1} t(n,k+i)=1, and Product_{i=1..n-1} t(n-2,k+i) = 1 because by definition of a prime the only divisors are 1 (at n=k in table t(n,k)) and the prime itself (at k=1 in the table t(n,k)) and thereby n/Product_{i=1..n-1}t(n,k+i)/Product_{i=1..n-1}t(n-2,k+i) = n. Since the exponentiated von Mangoldt function is the unique arithmetic function such that when multiplied over the divisors, is equal to n, and since the exponentiated von Mangoldt function is equal to n at prime numbers only, and since at n not equal to the greater of the twin primes the modified recurrence for the exponentiated von Mangoldt function by recursion messes with the output so much that the output cannot possibly be equal to n at any other numbers than at n equal to the greater of the twin primes.

Setting x = 1 gives ratios A276835(n)/A276836(n) equal to n when n is equal to the greater of the twin primes A006512.

Setting x = 2 gives ratios A276835(n)/A276836(n) equal to n when n is equal to A046132.

Setting x = 3 gives ratios A276835(n)/A276836(n) equal to n when n is equal to A046117.

Setting x = 4 gives ratios A276835(n)/A276836(n) equal to n when n is equal to A092402, and so on.

Examples

			The ratio A276835/A276836 starts: 1, 2, 3, 2, 5, 1/2, 7, 1/3, 3, 1/4, 11/3, 1/5, 13,...
The greater twin primes A006512 start: 5,7,13,... where the ratio is equal to n.
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A276836, A006512, A014963.

Programs

Mathematica

Clear[t, x]; (*setting x=1 gives ratio equal to n when n is the greater of the twin primes, x=2 gives ratio equal to n when n is the greater of the cousin primes and so on.*) x = 1; nn = 60; t[1, 1] = 1; t[n_, k_] := t[n, k] = If[k == 1, n/Product[t[n - 2*x, k + i], {i, 1, n - 2*x}]/Product[t[n, k + i], {i, 1, n - 1}], If[Mod[n, k] == 0, t[n/k, 1], 1], 1]; Monitor[a = Table[t[n, 1], {n, 1, nn}];, n]; Numerator[a] (* Mats Granvik, Sep 20 2016, Sep 29 2016 *)

Formula

From Mats Granvik, Sep 20 2016, Sep 29 2016: (Start)

Recurrence for the ratio A276835(n)/A276836(n):

Let:

x = 1;

T(1, 1) = 1;

T(n, k) = If k = 1 then n/Product_{i=1..n-2*x}(T(n-2*x, k + i))/Product_{i=1..n-1}(T(n, k + i)) else if Mod(n, k) = 0 then T(n/k, 1) else 1 else 1.

Then A276835(n)/A276836(n) = T(n,1). (End)