cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A341680 Successive approximations up to 2^n for the 2-adic integer Sum_{k>=0} k!.

Original entry on oeis.org

0, 0, 2, 2, 10, 26, 26, 26, 26, 26, 538, 538, 2586, 6682, 14874, 31258, 64026, 129562, 129562, 391706, 915994, 1964570, 4061722, 8256026, 8256026, 8256026, 8256026, 8256026, 142473754, 410909210, 947780122, 2021521946, 4169005594, 8463972890, 8463972890, 25643842074
Offset: 0

Views

Author

Jianing Song, Feb 17 2021

Keywords

Comments

a(n) == Sum_{k>=0} k! (mod 2^n). Since k! mod 2^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)!

Examples

			For n = 9, since 2^9 divides 12!, we have a(9) = (Sum_{k=0..11} k!) mod 2^9 = 26.
For n = 12, since 2^12 divides 16!, we have a(12) = (Sum_{k=0..15} k!) mod 2^12 = 2586.

Links

Crossrefs

Cf. A341684 (digits of Sum_{k>=0} k!).
Successive approximations for the p-adic integer Sum_{k>=0} k!: A341680 (this sequence), A341681 (p=3), A341682 (p=5), A341683 (p=7).
Cf. A007843 (least positive integer k for which 2^n divides k!).

Programs

  • PARI
    a(n) = my(p=2); if(n==0, 0, lift(sum(k=0, (p-1)*(n+logint((p-1)*n, p)), Mod(k!, p^n))))

Formula

For n > 0, a(n) = (Sum_{k=0..m} k!) mod 2^n, where m = n + floor(log_2(n)).

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

Original entry on oeis.org

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

Views

Author

Jianing Song, Feb 17 2021

Keywords

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)!

Examples

			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.

Links

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!).

Programs

  • 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))))

Formula

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

A341682 Successive approximations up to 5^n for the 5-adic integer Sum_{k>=0} k!.

Original entry on oeis.org

0, 4, 14, 64, 314, 2814, 2814, 2814, 237189, 1018439, 1018439, 10784064, 10784064, 743205939, 1963909064, 20274455939, 142344768439, 142344768439, 142344768439, 7771739299689, 64992198284064, 446461924846564, 1876973399455939, 4261159190471564, 16182088145549689, 16182088145549689
Offset: 0

Views

Author

Jianing Song, Feb 17 2021

Keywords

Comments

a(n) == Sum_{k>=0} k! (mod 5^n). Since k! mod 5^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)!

Examples

			For n = 7, since 5^7 divides 30!, we have a(7) = (Sum_{k=0..29} k!) mod 5^7 = 2814.
For n = 29, since 5^29 divides 125!, we have a(29) = (Sum_{k=0..124} k!) mod 5^29 = 173465698384532268439.

Links

Crossrefs

Cf. A341686 (digits of Sum_{k>=0} k!).
Successive approximations for the p-adic integer Sum_{k>=0} k!: A341680 (p=2), A341681 (p=3), this sequence (p=5), A341683 (p=7).
Cf. A007845 (least positive integer k for which 2^n divides k!).

Programs

  • PARI
    a(n) = my(p=5); if(n==0, 0, lift(sum(k=0, (p-1)*(n+logint((p-1)*n, p)), Mod(k!, p^n))))

Formula

For n > 0, a(n) = (Sum_{k=0..m} k!) mod 5^n, where m = 4*(n + floor(log_5(4*n))).

A341687 Expansion of the 7-adic integer Sum_{k>=0} k!.

Original entry on oeis.org

6, 6, 1, 1, 6, 1, 0, 2, 0, 3, 5, 1, 4, 1, 3, 6, 2, 0, 2, 4, 3, 5, 6, 3, 4, 3, 5, 0, 0, 4, 4, 0, 1, 0, 1, 6, 2, 0, 0, 3, 3, 5, 1, 4, 6, 1, 5, 1, 5, 4, 5, 5, 1, 5, 1, 6, 5, 6, 2, 2, 0, 2, 0, 5, 5, 0, 5, 5, 6, 5, 1, 4, 2, 2, 2, 1, 2, 2, 0, 5, 5, 5, 2, 6, 2, 0, 4, 0, 1, 3, 5
Offset: 0

Views

Author

Jianing Song, Feb 17 2021

Keywords

Comments

For every prime p, since valuation(k!,p) goes to infinity as k increases, Sum_{k>=0} k! is a well-defined p-adic constant.
Conjecture: this constant is transcendental, which means that it is not the root of any polynomial with integer coefficients.
Conjecture: this constant is normal, which means for every septenary (base-7) string s with length k, if we denote N(s,n) as the number of occurrences of s in the first n digits, then lim_{n->inf} N(s,n)/n = 1/7^k.

Examples

			Sum_{k>=0} k! = ...33002610104400534365342026314153020161166.

Links

Crossrefs

Cf. A341683 (successive approximations of Sum_{k>=0} k!).
Expansion of Sum_{k>=0} k! in p-adic integers: A341684 (p=2), A341685 (p=3), A341686 (p=5), this sequence (p=7).

Programs

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

Formula

a(n) = (A341683(n+1) - A341683(n))/7^n.
Showing 1-4 of 4 results.

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