A082648 Consider f(m) = Sum_{k=1..m} k! (A007489) when m is very large; a(n) = n-th digit from end.
3, 1, 3, 0, 4, 9, 0, 2, 4, 0, 2, 9, 8, 2, 5, 6, 3, 3, 2, 4, 4, 6, 5, 5, 2, 5, 0, 9, 3, 0, 5, 0, 1, 3, 9, 5, 3, 2, 3, 4, 0, 8, 4, 9, 9, 7, 0, 1, 1, 2, 6, 8, 3, 7, 4, 8, 6, 8, 7, 4, 9, 7, 4, 7, 4, 2, 2, 9, 0, 0, 4, 3, 3, 0, 5, 6, 5, 8, 6, 5
Offset: 1
Examples
Sum_{k=1..30} k! = 274410818470142134209703780940313.
The last 7 digits in reverse order give us the first 7 terms of this sequence: 3,1,3,0,4,9,0.
From _Jon E. Schoenfield_, Feb 16 2021: (Start)
The table below shows the 12 least-significant digits of Sum_{k=1..m} k! converging to the first 12 terms of this sequence (in reverse order) as m increases:
.
m Sum_{k=1..m} k! # corresponding digits
-- --------------- ----------------------
0 0 0
4 33 1
9 409113 2
14 93928268313 3
19 ...485935180313 4
24 ...567844940313 6
29 ...395300940313 7
34 ...323620940313 8
39 ...232420940313 9
44 ...080420940313 10
49 ...920420940313 12
...
oo ...920420940313
(End)
Programs
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Mathematica
Take[Reverse[IntegerDigits[Sum[n!, {n, 1, 500}]]], 100] (* generates first 100 terms *)
Comments