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'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT]
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'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT]
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'√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT]
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02-19-2020, 06:34 PM
(This post was last modified: 02-19-2020 06:36 PM by Gerson W. Barbosa.)
Post: #8
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RE: '√√√√(2*√2*π*(10^3*(10^3+π/3^4)+1/(π*(π^5+π^2+1)+1/(√(2*π)))))' [NT]
Here’s a double pandigital expression:
'9/5-1/(30*672*√√4+8)+√(9/5-1/(30*672*√√4+8))' →NUM -> 3.14159265362 ( From Ramanujan’s '9/5 + √(9/5) = 3.1416407865' ) These are as easy to find as they are useless and futile. |
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