cp's OEIS Frontend

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

%I A276835 #27 Aug 16 2025 23:53:27
%S A276835 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,
%T A276835 31,112,3,105,1,50,37,5,1,27,41,81,43,10,1,105,47,539,7,77,15,4,265,2,
%U A276835 3,520,3,351,59,945
%N A276835 Numerator of a modified exponentiated von Mangoldt function defined recursively.
%C A276835 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.
%C A276835 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.
%C A276835 Setting x = 1 gives ratios A276835(n)/A276836(n) equal to n when n is equal to the greater of the twin primes A006512.
%C A276835 Setting x = 2 gives ratios A276835(n)/A276836(n) equal to n when n is equal to A046132.
%C A276835 Setting x = 3 gives ratios A276835(n)/A276836(n) equal to n when n is equal to A046117.
%C A276835 Setting x = 4 gives ratios A276835(n)/A276836(n) equal to n when n is equal to A092402, and so on.
%H A276835 Robert G. Wilson v, <a href="/A276835/b276835.txt">Table of n, a(n) for n = 1..1000</a>
%F A276835 From _Mats Granvik_, Sep 20 2016, Sep 29 2016: (Start)
%F A276835 Recurrence for the ratio A276835(n)/A276836(n):
%F A276835 Let:
%F A276835 x = 1;
%F A276835 T(1, 1) = 1;
%F A276835 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.
%F A276835 Then A276835(n)/A276836(n) = T(n,1). (End)
%e A276835 The ratio A276835/A276836 starts: 1, 2, 3, 2, 5, 1/2, 7, 1/3, 3, 1/4, 11/3, 1/5, 13,...
%e A276835 The greater twin primes A006512 start: 5,7,13,... where the ratio is equal to n.
%t A276835 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 *)
%Y A276835 Cf. A276836, A006512, A014963.
%K A276835 nonn,frac
%O A276835 1,2
%A A276835 _Mats Granvik_, Sep 20 2016