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Alex
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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):

  1. 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
  2. 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

  1. 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

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Aug 5, 2014