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A341681
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Successive approximations up to 3^n for the 3-adic integer Sum_{k>=0} k!.
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5
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0, 1, 1, 10, 64, 145, 145, 874, 3061, 3061, 42427, 42427, 396721, 928162, 4116808, 4116808, 4116808, 4116808, 262397134, 1037238112, 1037238112, 8010806914, 8010806914, 8010806914, 196297164568, 478726701049, 2173303919935, 2173303919935, 2173303919935, 25050096374896, 162310851104662
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OFFSET
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0,4
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COMMENTS
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a(n) == Sum_{k>=0} k! (mod 3^n). Since k! mod 3^n is eventually zero, a(n) is well-defined.
In general, for every prime p, the p-adic integer x = Sum_{k>=0} k! is well-defined. To find the approximation up to p^n (n > 0) for x, it is enough to add k! for 0 <= k <= m and then find the remainder of the sum modulo p^n, where m = (p - 1)*(n + floor(log_p((p-1)*n))). This is because p^n divides (m+1)!
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LINKS
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FORMULA
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For n > 0, a(n) = (Sum_{k=0..m} k!) mod 3^n, where m = 2*(n + floor(log_3(2*n))).
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EXAMPLE
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For n = 11, since 3^11 divides 27!, we have a(11) = (Sum_{k=0..26} k!) mod 3^11 = 42427.
For n = 24, since 3^24 divides 54!, we have a(24) = (Sum_{k=0..53} k!) mod 3^24 = 196297164568.
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PROG
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(PARI) a(n) = my(p=3); if(n==0, 0, lift(sum(k=0, (p-1)*(n+logint((p-1)*n, p)), Mod(k!, p^n))))
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CROSSREFS
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Cf. A341685 (digits of Sum_{k>=0} k!).
Successive approximations for the p-adic integer Sum_{k>=0} k!: A341680 (p=2), this sequence (p=3), A341682 (p=5), A341683 (p=7).
Cf. A007844 (least positive integer k for which 3^n divides k!).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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