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More on e^(pi*sqrt(163))
From
apovolot@gmail.com@21:1/5 to
All on Sun Dec 27 08:49:25 2015
Furthering my previous input, I suggest the following two formulas for
Heegner numbers (see 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))/(2*sqrt(3))
for n = 6,7,8,9
Then four almost integers (including famous Ramanujan's) could be
expressed as:
exp(Pi*sqrt(19+24*((1 + sqrt(3))^(n-6) - (1 -
sqrt(3))^(n-6))/(2*sqrt(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
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From
Alexander P-sky@21:1/5 to
All on Mon Jan 4 06:33:33 2016
Also the following below formulas (6), (7), (8), (9) allow to get
explicit expression for all nine terms of OEIS A003173
(6) a(n) = b(n) + (d(n))*((1 + sqrt(3))^(n-b(n)) - (1 - sqrt(3))^(n-b(n)))/(2*sqrt(3))
(7) b(n) = 3 + (-1)^c(n)*(4 + Exp(1)*ExpIntegralEi[-1]) +
Exp(1)*ExpIntegralE[3 + c(n),1]*Gamma[3 + c(n)]
(8) c(n) = (8100 - 3774*n + 875*n^2 - 185*n^3 + 25*n^4 - n^5)/(60*(129
- 50*n + 5*n^2))
(9) d(n) = a(b(n)+1)-a(b(n))
Also below recurrent formula (10) also holds true and it could be used
for verification of explicit identity derived from above formulas (6),
(7), (8), (9).
(10) d(n+2) = ((5-n)*d(n)+2*(n-4)*d(n+1))/(n-3), {d(1)=1,d(2)=1}
where formulas (6), (7), (8), (9) and (10) hold to be true for n=1,2,3,4,5,6,7,8,9
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From
Alexander P-sky@21:1/5 to
All on Mon Jan 4 11:26:38 2016
Also the following below formulas (6), (7), (8), (9) and (10) allow to
get explicit expression for all nine terms of OEIS A003173
(6) a(n) = b(n) + (d(n))*((1 + sqrt(3))^(n-f(n)) - (1 - sqrt(3))^(n-f(n)))/(2*sqrt(3))
(7) b(n) = 3 + (-1)^c(n)*(4 + Exp(1)*ExpIntegralEi[-1]) +
Exp(1)*ExpIntegralE[3 + c(n),1]*Gamma[3 + c(n)]
b(n) = 1 when c(n) = 1, that is for {n,1,4};
b(n) = 11 when c(n) = 2, that is for n=5;
b(n) = 19 when c(n) = 3, that is for {n,6,9}
(8) c(n) = (8100 - 3774*n + 875*n^2 - 185*n^3 + 25*n^4 - n^5)/(60*(129
- 50*n + 5*n^2))
Specifically
c(n) = {1,1,1,1,2,3,3,3,3} for {n,1,9}
(9) d(n) = -(480*(n^3-15*n^2+80*n-150))/(23*n^4-710*n^3+8005*n^2-39550*n+72552)
d(n) = 1 for {n,1,4};
d(n) = 0 for n = 5;
d(n). = 24 for {n,6,9}
(10) f(n) = -(24
(n^3-15*n^2+80*n-150))/(n^4-34*n^3+395*n^2-1970*n+3624)
f(n) = 1 for {n,1,4};
f(n) = 0 for n = 5;
f(n). = 6 for {n,6,9}
where formulas (6), (7), (8), (9) and (10) hold to be true for n=1,2,3,4,5,6,7,8,9
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