|
|
|
|
|
|
|
|
|
#25byAlois P. Heinzat Tue Feb 14 08:21:11 EST 2023
|
|
|
|
|
|
|
|
#24byKevin Rydeat Tue Feb 14 03:36:48 EST 2023
|
|
|
|
|
|
Discussion
|
| Tue Feb 14
| 04:40
| Eric Simon Jacob:This (n%5 == 1) - (n%5 == 3) became [n == 1 (mod 5)] - [n == 3 (mod 5)] from you. If you think that is the best notation, we can do like that.
|
|
|
|
|
|
|
#23byKevin Rydeat Tue Feb 14 03:24:50 EST 2023
|
|
| FORMULA
|
(End)
FromEric Simon Jacob,Feb 14 2023: (Start)
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5) + (2/25)*sqrt(11/4-2/(sqrt(5)+3))*sin(2*Pi*(n+1)/5) + (sqrt(10)-5*sqrt(2))*sqrt(sqrt(5)+5)*sin(4*Pi*(n+1)/5)/(50*sqrt(5)+250) + (4/(5*sqrt(5)+25)-1/5)*(cos(2*Pi*(n+1)/5)-cos(4*Pi*(n+1)/5)).
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5 + (n%5 == 1) - (n%5 == 3)).
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5 + [n == 1 (mod 5)] - [n == 3 (mod 5)]). -Eric Simon Jacob,Feb 14 2023
|
|
| STATUS
|
proposed
editing
|
|
|
|
|
Discussion
|
| Tue Feb 14
| 03:26
| Kevin Ryde:Let's have the second one. Using A092202 would be fine too if desired.
|
|
|
|
|
|
|
#22byEric Simon Jacobat Tue Feb 14 02:40:15 EST 2023
|
|
|
|
|
|
|
|
#21byEric Simon Jacobat Tue Feb 14 02:39:29 EST 2023
|
|
| FORMULA
|
FromEric Simon Jacob,Feb 14 2023:(:(Start)
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5 + (n%5 == 1) - (n%5 == 3)))).
|
|
|
|
|
|
|
#20byEric Simon Jacobat Tue Feb 14 02:26:58 EST 2023
|
|
| FORMULA
|
(End)
FromEric Simon Jacob,Feb 14 2023:(Start)
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5) + (2/25)*sqrt(11/4-2/(sqrt(5)+3))*sin(2*Pi*(n+1)/5) + (sqrt(10)-5*sqrt(2))*sqrt(sqrt(5)+5)*sin(4*Pi*(n+1)/5)/(50*sqrt(5)+250) + (4/(5*sqrt(5)+25)-1/5)*(cos(2*Pi*(n+1)/5)-cos(4*Pi*(n+1)/5)).
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5 + (n%5 == 1) - (n%5 == 3))
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5) + (2/25)*sqrt(11/4-2/(sqrt(5)+3))*sin(2*Pi*(n+1)/5) + (sqrt(10)-5*sqrt(2))*sqrt(sqrt(5)+5)*sin(4*Pi*(n+1)/5)/(50*sqrt(5)+250) + (4/(5*sqrt(5)+25)-1/5)*(cos(2*Pi*(n+1)/5)-cos(4*Pi*(n+1)/5)). -Eric Simon Jacob,Feb 12 2023
|
|
|
|
|
Discussion
|
| Tue Feb 14
| 02:38
| Eric Simon Jacob:I am sorry, but i am physicist, not mathematician. The simplification is the final result to have a short formula for each particular case, but that is not the processus i use actually to build the mathematics theory giving any formula. I took a (Start) and (End) to not delete the original formula, but that will not be probably what you wish.
|
|
|
|
|
|
|
#19byKevin Rydeat Tue Feb 14 01:13:11 EST 2023
|
|
|
|
|
|
|
|
#18byEric Simon Jacobat Tue Feb 14 00:34:45 EST 2023
|
|
|
|
|
|
Discussion
|
| Tue Feb 14
| 00:42
| Jon E. Schoenfield:Thanks... but I don't understand. When you wrote "The simplification is making lose informations", did you mean that making the simplification I asked about would cause a loss of information? If so,... how??
|
|
| 01:13
| Kevin Ryde:This is not enough please. No reader can know this giant expression is simple cases according as n mod 5. Drop back to editing again.
|
|
|
|
|
|
|
#17byEric Simon Jacobat Tue Feb 14 00:33:44 EST 2023
|
|
| FORMULA
|
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5) +2*sin() + (2*Pi*(n+1)/5/25)*sqrt(11/25004-2/(625*sqrt(5)+1875)) + (3))*sin(2*Pi*(n+1)/5) + (sqrt(10)-5*sqrt(2))*sqrt(sqrt(5)+5)*sin(4*Pi*(n+1)/5)/(50*sqrt(5)+250) + (4/(5*sqrt(5)+25)-1/5)*(cos(2*Pi*(n+1)/5)-cos(4*Pi*(n+1)/5)). -Eric Simon Jacob,Feb 12 2023
|
|
|
|
|
|
|
#16byEric Simon Jacobat Mon Feb 13 21:00:38 EST 2023
|
|
| FORMULA
|
a(n) = (1/5)*(8*n^3 + 12*n^2 + 14*n + 5) + 2*sin(2*Pi*(n+1)/5)*sqrt(11/2500-2/(625*sqrt(5)+1875)) + (sqrt(10)-5*sqrt(2))*sqrt(sqrt(5)+5)*sin(4*Pi*(n+1)/5)/(50*sqrt(5)+250) + (404/(505*sqrt(5)+25025)-1/5)*(cos(2*Pi*(n+1)/5)-cos(4*Pi*(n+1)/5)). -Eric Simon Jacob,Feb 12 2023
|
|
|
|
|
|
