Alex
I made a conjecture that:
(1)^n*(Pi−A002485(n)/A002486(n))=(abs(i)*2^j)^(1)Int((x^l(1x)^(2*(j+2))*(k+(i+k)*x^2))/(1+x^2),x=0...1)
where A002485(n) and A002486(n) are integer sequences indexed by "n" and referenced in the OEIS.org and "i", "j", "k", "l"  are some integers, together representing the set, and to be experimentally found separately for the each value of "n"... Could my conjecture be proved analytically? PS There is a alleged (but not proved) possibility, that for each "n" an infinite number of {"i", "j", "k", "l"} sets could be found...  so the question arises whether there is a relationship between some two parameters in a {"i", "j", "k", "l"} sets, so one of those parameters could then be eliminated from the conjectured by me formula, thus reducing the number of parameters (members in the set) in it from 4 to 3?
I suggest the following two formulas for Heegner numbers (see OEIS A003173):
a) for the first four (smallest) Heegner numbers
a(n) = 1+((1 + sqrt(3))^(n1)  (1  sqrt(3))^(n1))/(2*sqrt(3)) for n = 1,2,3,4
b) for the last (largest) four Heegner numbers
a(n) = 19+24*((1 + sqrt(3))^(n6)  (1  sqrt(3))^(n6))/(2*sqrt(3)) for n = 6,7,8,9
In general
a(n) = a(k) + (a(k+1)a(k))*((1 + sqrt(3))^(nk)  (1  sqrt(3))^(nk))/(2*sqrt(3)) where for n =1,2,3,4 k=1 and for n =6,7,8,9 k=6
Below identity is quite trivial and may even be called superficial but to me it has some beauty in it ...
Pi^2 = (n*(n+1)*(2*n+1))*((sum(1/i^2,i=1...n))/(sum(i^2,i=1...n))), n>infinity
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[km,m]*Binomial[k,m]*16^m,{m,0,k/2}])/(256)^k,{k,0,infinity}]
BBP formula in a slight disguise sum((1/16)^k*(sum(((1)^(ceil(4/(2*n))))*(floor(4/n))/(8*k+n+floor(sqrt(n1))*(floor(sqrt(n1))+1)),n=1..4)),k=0..infinity)

Member for 5 years, 9 months

63 profile views

Last seen Mar 25 '17 at 20:58
Communities (12)
Top network posts
 25 Seeking proof for the formula relating Pi with its convergents
 20 Is formula valid for relating $\pi$ with ALL of its OEIS A002485(n)/A002486(n) convergents?
 18 Is there an integral that proves $\pi > 333/106$?
 14 Are the LIGO observations proof of a black hole merger, and what happened to the black holes?
 5 Discovery in Astronomy vs one in Physics  do they differ in required burden of evidence?
 5 Trends in the distribution of reordered digits of Pi (OEIS A096566)
 5 Relation between dark matter and supermassive black holes located in the center of galaxies
 View more network posts →
Top tags (3)
Badges (7)
Gold
Silver
Rarest

Nov 14
Bronze
Rarest

Jul 14 '14

Jul 12 '14

Aug 1 '14