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Thomas Scheuerle, <a href="/A381464/b381464_1.txt">Table of n, a(n) for n = 1..5000</a>
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Thomas Scheuerle, <a href="/A381464/b381464_1.txt">Table of n, a(n) for n = 1..5000</a>
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Thomas Scheuerle, <a href="/A381464/b381464_1.txt">Table of n, a(n) for n = 1..5000</a>
approved
editing
proposed
approved
editing
proposed
The greatest common divisor of two consecutive Fibonacci numbers is 1, thus we know that (k-1)^F(m)*k^F(m+1) and (t-1)^F(n)*t^F(n+1) are all different for some m,n > 1 if k and t are chosen such that for m or n < 2 no solution for (k-1)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) exist, because this can not cannot be equal if (k-1)*k and (t-1)*t have different prime numbers as divisors and if the only difference is the exponent of the prime factors, then the distribution of these between (t-1) and t and thus their progression F(n) or F(n+1) is individually distinct. In this sequence we need also to consider the more general case (k-c)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) because sometimes we need to set a(k) = k+c. It is conjectured that in this case c is bounded to be < 3.
proposed
editing
editing
proposed
The greatest common divisor of two consecutive Fibonacci numbers is 1, thus we know that (k-1)^F(m)*k^F(m+1) and (t-1)^F(n)*t^F(n+1) are all different for some m,n > 1 if k and t are chosen such that for m or n < 2 no solution for (k-1)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) exist, because this can not be equal if (k-1)*k and (t-1)*t have different prime numbers as divisors and if the only difference is the exponent of the prime factors, then the distribution of these between (t-1) and t and thus their progression F(n) or F(n+1) is individually distinct. In this sequence we need also to consider the more general case (k-c)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) because sometimes we need to set a(k) = k+c. It is conjectured that in this case c is bounded to be < 3 by the known growth of this sequence.
The greatest common divisor of two consecutive Fibonacci numbers is 1, thus we know that (k-1)^F(m)*k^F(m+1) and (t-1)^F(n)*t^F(n+1) are all different for some m,n > 1 if k and t are chosen such that for m or n < 2 no solution for (k-1)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) exist. Obviously , because this will can not be equal if (k-1)*k and (t-1)*t have different prime numbers as divisors. If and if the only difference is the exponent of the prime factors, then the distribution of these between (t-1) and t and thus their progression F(n) or F(n+1) is individually distinct. In this sequence we need also to consider the more general case (k-c)^F(m)*k^F(m+1) = (t-1)^F(n)*t^F(n+1) because sometimes we need to set a(k) = k+c. It is conjectured that in this case c is bounded to be < 3 by the known growth of this sequence.