Below is conjectured by me formula relating π with ALL of its described via A002485(n)/A002486(n) OEIS integer sequences ratio
(-1)^n*(Pi−A002485(n)/A002486(n))=(Abs(i)2^j)^(-1)Int((x^l(1-x)^(2(j+2))*(k+(i+k)x^2))/(1+x^2),x=0..1)
where integer n>2 is the index for the terms in the OEIS A002485(n) & A002486(n) integer sequences;{i,j,k,l} are some signed integer parameters to be found experimentally or otherwise for each value of “n” when abs(l-j)=2m (“m" is some positive integer).
Also it appears that
(-1)^n*(Pi−A002485(n)/A002486(n))=((Abs(i))2^j)^(-1)Int((x^l(1-x)^(2(j+2))*(k+(i+k)x^2))/(1+x^2), x=0..1)
holds true for any n>2 when
i=(-1)^(n)3A002486(n);
k=(-1)^(n)(A363445(n-2+m)*A002486(n)-A363446(n-2+m)A002485(n))
j=2m;m >=0;l=0;
where A363445(n) & A363446(n) are OEIS integer sequences.
Below is conjectured by me formula relating LOG(2) (Ln(2)) with ALL of its convergents described via OEIS integer sequences A079942(n)/A079943(n) ratio.
(-1)^n*(LOG(2)−A079942(n)/A079943(n))=(Abs(i)2^j)^(-1)Int((x^l(1-x)^(2(j+2))*(k+(i+k)x^2))/(1+x^2),x=0...)
where integer n>2 is the index for the terms in the OEIS A079942(n) & A079943(n) integer sequences;{i,j,k,l} are some signed integer parameters to be found experimentally or otherwise for each value of “n”; when abs(l-j)=2m+1 (“m" is some positive integer).
Also it appears that
(-1)^n*(LOG(2)−A079942(n)/A079943(n))=((Abs(i))2^j)^(-1)Int((x^l(1-x)^(2(j+2))*(k+(i+k)x^2))/(1+x^2),x=0..1)
holds true for any n>2 when
i=(-1)^(n)3A079943(n);
k=(-1)^(n)(A363515(n-2+m)*A079943(n)-A363516(n-2+m)A079942(n))
j=2m+1 (m>=0);l=0
where A363515(n) & A363516(n) are OEIS integer sequences.
I found the identity
sqrt(exp(1))=16/31*(sum((1/2)^n*(1/2n^3+1/2n+1)/n!,n=1..inf)+1)
or
sqrt(e)=(16/31)(1+Sum_{n>=1}(1/2)^n(1/2n^3+1/2n+1)/n!)
http://www.strw.leidenuniv.nl/~mathar/public/mathar20071105.pdf
https://oeis.org/A019774
I suggested two formulas for Heegner numbers (OEIS A003173):
- for the first four smallest Heegner numbers
a(n)=1+((1+sqrt(3))^(n-1)-(1-sqrt(3))^(n-1))/(2*sqrt(3)) for n=1,2,3,4
- for the last largest four Heegner numbers
a(n)=19+24*((1+sqrt(3))^(n-6)-(1-sqrt(3))^(n-6))/(2sqrt(3)) for n=6,7,8,9
In general
a(n)=a(k)+(a(k+1)-a(k))((1+sqrt(3))^(n-k)-(1-sqrt(3))^(n-k))/(2*sqrt(3)) where for n=1,2,3,4 k=1 and for n=6,7,8,9 k=6
Three conjectures from me, Alexander R. Povolotsky
- n!+prime(n) != m^k -so far proven for the case when k=2
2)n!+n^2 != m^2 -so far proven for the case when n is prime
3)n!+Sum(j^2,j=1,j=n) != m^2 -so far no proof
!= means "not equal";k,m,n are integers
7901234568/9876543210*1234567890=0987654312
24/Pi=sum((30k+7)binom(2k,k)^2(Hypergeometric2F1[1/2-k/2,k/2,1,64])/(-256)^k,k=0..inf)
or
Sum[(30k+7)Binomial[2k,k]^2(Sum[Binomial[k-m,m]*Binomial[k,m]*16^m,{m,0,k/2}])/(-256)^k,{k,0,inf}]
Pi^2= lim(n*(n+1)(2n+1))*((sum(1/i^2,i=1..n))/(sum(i^2,i=1..n))), n->inf