A341683 Successive approximations up to 7^n for the 7-adic integer Sum_{k>=0} k!.

0, 6, 48, 97, 440, 14846, 31653, 31653, 1678739, 1678739, 122739560, 1535115805, 3512442548, 58877591352, 155766601759, 2190435820306, 30675804879964, 97141666019166, 97141666019166, 3353968861840064, 48949549603332636, 288326348496168639

Offset

0,2

Comments

a(n) == Sum_{k>=0} k! (mod 7^n). Since k! mod 7^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 7^n, where m = 6*(n + floor(log_7(6*n))).

Example

For n = 7, since 7^7 divides 49!, we have a(7) = (Sum_{k=0..48} k!) mod 7^7 = 31653.
For n = 55, since 7^55 divides 343!, we have a(55) = (Sum_{k=0..342} k!) mod 7^55 = 7563765912082524448071111141811678897409320968.

Prog

(PARI) a(n) = my(p=7); if(n==0, 0, lift(sum(k=0, (p-1)*(n+logint((p-1)*n, p)), Mod(k!, p^n))))

Crossrefs

Cf. A341687 (digits of Sum_{k>=0} k!).

Successive approximations for the p-adic integer Sum_{k>=0} k!: A341680 (p=2), A341681 (p=3), A341682 (p=5), this sequence (p=7).

Cf. A020646 (least positive integer k for which 7^n divides k!).

Sequence in context: A000252 A078237 A274131 * A259121 A052651 A153796

Adjacent sequences: A341680 A341681 A341682 * A341684 A341685 A341686

Keyword

nonn

Author

Jianing Song, Feb 17 2021

Status

approved