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Three hard to prove conjectures from Alexander R. Povolotsky

1) n! + prime(n) != m^k (so far proven only for the case when k=2)

2) n! + n^2 != m^2 (so far proven only for the case when n is prime number)

3) n! + Sum(j^2, j=1, j=n) != m^2 (so far no proof)

where != means "not equal" and k,m,n are integers


7901234568 / 9876543210 * 1234567890 = 0987654312


24/Pi = sum((30*k+7)binom(2k,k)^2(Hypergeometric2F1[1/2 - k/2, -k/2, 1, 64])/(-256)^k, k=0...infinity)

Another version of this identity is: Sum[(30*k+7)Binomial[2k,k]^2(Sum[Binomial[k-m,m]*Binomial[k,m]*16^m,{m,0,k/2}])/(-256)^k,{k,0,infinity}]


Sep
24
awarded  Autobiographer
Sep
23
awarded  Nice Answer
Jul
24
revised Is there an integral that proves $\pi > 333/106$?
added 1853 characters in body
Jul
22
comment Seeking proof for the formula relating Pi with its convergents
Obviously parameters in the formula somehow depend on "n". The most straight forward dependency on "n" is observed in what you call "m'" (and I call "p") ;{2,3,4,4,6,6} ... It appears that when "n" -> infinity - then the integral should come to 0 ...
Jul
22
comment Seeking proof for the formula relating Pi with its convergents
Pi - 103993/33102 -> a:=124360;b:=77159;p:=6;c:=2;int((x^(c+2*p)*(1-x)^(2*p)*(a+bx^2))/((a-b)*2^(p-2‌​‌​)*((-1)^(c)*(1+x^2))),x=0...1) 355/113 - Pi -> a:=25;b:=816;p:=4;c:=0;int((x^(c+2*p)*(1-x)^(2*p)*(a+bx^2))/((a-b)*2^(p-2)*((-1‌​‌​)^(c)*(1+x^2))),x=0...1) Pi - 333/106 -> a:=197;b:=462;p:=3;c:=-1;int((x^(c+2*p)*(1-x)^(2*p)(a+bx^2))/((a-b)*2^(p-2)*((-‌​‌​1)^(c)*(1+x^2))),x=0...1) 22/7 - Pi -> a:=1;b:=0;p:=2;c:=0;int((x^(c+2*p)*(1-x)^(2*p)*(a+bx^2))/((a-b)*2^(p-2)*((-1)‌​‌​^(c)*(1+x^2))),x=0...1)
Jul
20
revised Question on linear partial difference equation with three independent variables $n$, $m$, $k$.
deleted 4 characters in body
Jul
20
asked Question on linear partial difference equation with three independent variables $n$, $m$, $k$.
Jul
18
comment Seeking proof for the formula relating Pi with its convergents
@(Matt B.) In the next two comments I list parameters in your formula for all cases (I replaced alpha by a, bets by b, epsilon by c, m' by p) 104348/33215 - Pi -> a:=1349;b:=-1060;p:=6;c:=0;int((x^(c+2*p)*(1-x)^(2*p)*(a+b*x^2))/((a-b)*2^(p-2)*‌​((-1)^(c)*(1+x^2))),x=0...1)
Jul
17
awarded  Benefactor
Jul
17
accepted Seeking proof for the formula relating Pi with its convergents
Jul
17
awarded  Self-Learner
Jul
16
comment Permutation identities similar to $(7901234568 / 9876543210) \cdot 1234567890 = 0987654312$
@Gerry Myerson - systematic search ... That is what was done programmatically ...
Jul
16
comment Permutation identities similar to $(7901234568 / 9876543210) \cdot 1234567890 = 0987654312$
@Gerry Myerson Thanks - I think that you have proved that there are at least two pairs - yes ? If so, is there any way to evaluate how many (more than 2) such pairs exist for the given base with n>2 ?
Jul
16
awarded  Nice Question
Jul
15
comment Why this recursively defined sequence of real numbers converges to -Pi?
@nbubis - thanks for the info - it worked out that way anyway.
Jul
15
comment Why this recursively defined sequence of real numbers converges to -Pi?
@nbubis - I was hoping though that you will return to address additional questions (see my comments made on July 6 and July 9 ) towards which I really issued the bounty :-)
Jul
15
comment Why this recursively defined sequence of real numbers converges to -Pi?
@nbubis But it went to you automatically (from how I understand the rules) - correct ?
Jul
15
awarded  Revival
Jul
15
revised Permutation identities similar to $(7901234568 / 9876543210) \cdot 1234567890 = 0987654312$
added 3 characters in body
Jul
15
revised Permutation identities similar to $(7901234568 / 9876543210) \cdot 1234567890 = 0987654312$
added 51 characters in body