A341681 Successive approximations up to 3^n for the 3-adic integer Sum_{k>=0} k!.

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

Offset

0,4

Comments

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)!

Links

Jianing Song, Table of n, a(n) for n = 0..1000

Formula

For n > 0, a(n) = (Sum_{k=0..m} k!) mod 3^n, where m = 2*(n + floor(log_3(2*n))).

Example

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.

Prog

(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))))

Crossrefs

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!).

Sequence in context: A101467 A162473 A138661 * A269538 A178256 A036426

Adjacent sequences: A341678 A341679 A341680 * A341682 A341683 A341684

Keyword

nonn

Author

Jianing Song, Feb 17 2021

Status

approved