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

1,2

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.

Links

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

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.

(End)

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

(End)

Example

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.

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 *)

Crossrefs

Cf. A276836, A006512, A014963.

Sequence in context: A214056 A368900 A014973 * A349630 A157753 A020500

Adjacent sequences: A276832 A276833 A276834 * A276836 A276837 A276838

Keyword

nonn,frac

Author

Mats Granvik, Sep 20 2016

Status

approved