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 1-3 of 3 results.

A007843 Least positive integer k for which 2^n divides k!.

Original entry on oeis.org

1, 2, 4, 4, 6, 8, 8, 8, 10, 12, 12, 14, 16, 16, 16, 16, 18, 20, 20, 22, 24, 24, 24, 26, 28, 28, 30, 32, 32, 32, 32, 32, 34, 36, 36, 38, 40, 40, 40, 42, 44, 44, 46, 48, 48, 48, 48, 50, 52, 52, 54, 56, 56, 56, 58, 60, 60, 62, 64, 64, 64, 64, 64, 64, 66, 68, 68, 70, 72, 72, 72, 74, 76, 76, 78
Offset: 0

Views

Author

Bruce Dearden and Jerry Metzger; R. Muller

Keywords

Comments

Obtained by writing every natural number n k times where 2^k divides n but 2^(k+1) does not divide n. - Amarnath Murthy, Aug 22 2002
An interval of the form (A007814(k!)-A007814(k), A007814(k!)] contains n >= 1 iff k = a(n). - Vladimir Shevelev, Mar 19 2012
It appears than for n > 0, a(n) is divisible by 2, and that the resulting sequence a(n)/2 is A046699 (ignoring first term, this is the Meta-Fibonacci sequence for s=0). - Michel Marcus, Aug 19 2013
The last part is proved in the Kullmann & Zhao preprint, Thm. 3.16. The first statement is obvious: to get a larger power of two in k!, k > 1 must be increased by 2, else the factor is odd and doesn't increase the 2-valuation of k!. The other part also follows from the comment in A046699: "n occurs A001511(n) times", where A001511 = A007814 + 1, A007814 = the number of powers of 2 in k. - M. F. Hasler, Dec 27 2019

References

  • H. Ibstedt, Smarandache Primitive Numbers, Smarandache Notions Journal, Vol. 8, No. 1-2-3, 1997, 216-229.

Links

Crossrefs

Cf. A007814, A007844, A007845, A020646, A048841-A048846.

Programs

  • Maple
    with(numtheory): ans := [ ]: p := ithprime(1): t0 := 1/p: for n from 0 to 50 do t0 := t0*p: t1 := 1: i := 1: while t1 mod t0 <> 0 do i := i+1: t1 := t1*i: od: ans := [ op(ans),i ]: od: ans;
# Alternative:
N:= 1000: # to get a(0) to a(N)
A:= Array(0..N):
A[0]:= 1:
A[1]:= 2:
B[2]:= 1:
for k from 4 by 2 do
  B[k]:= B[k-2] + padic:-ordp(k,2);
  A[B[k-2]+1..min(N,B[k])]:= k;
  if B[k] >= N then break fi;
od:
seq(A[i],i=0..N); # Robert Israel, Dec 07 2015
  • Mathematica
    a[n_] := (k=0; While[Mod[++k!, 2^n] > 0]; k); Table[a[n], {n, 0, 74}] (* Jean-François Alcover, Dec 08 2011 *)
Join[{1},Module[{nn=100,f},f=Table[{x!,x},{x,0,nn}];Table[ SelectFirst[ f,Divisible[#[[1]],2^n]&],{n,80}]][[All,2]]] (* Harvey P. Dale, Nov 20 2021 *)
  • PARI
    a(n)=if(n<0,0,s=1; while(s!%(2^n)>0,s++); s)
  • PARI
    a(n) = {k = 1; while (valuation(k!, 2) < n, k++); k;} \\ Michel Marcus, Aug 19 2013
  • PARI
    apply( A007843(n)={for(k=1,oo,(n-=valuation(k,2))>0||return(k))}, [0..99]) \\ This idea can also be used to compute most efficiently a vector a(0..N). - M. F. Hasler, Dec 27 2019
  • Python
    from itertools import count
def A007843(n):
    c = 0
    for k in count(1):
        c += (~k&k-1).bit_length()
        if c >= n:
            return k # Chai Wah Wu, Jul 08 2022

Formula

a(n) = A002034(2^n). For n>1, it appears that a(n+1) = a(n)+2 if n is in A005187. - Benoit Cloitre, Sep 01 2002
G.f.: 1 + 2*(x/(1-x))*Product_{k >= 1} (1+x^(2^k-1)). - Wadim Zudilin, Dec 07 2015
a(n) = 2*A046699(n) for n > 0. - Michel Marcus and M. F. Hasler, Dec 27 2019
a(2^i + r) = 2^i + a(r+1) for 0 <= r <= 2^i-2, and a(2^i + r) = 2^(i+1) for r = 2^i-1. - Kevin Ryde, Aug 06 2022

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

Original entry on oeis.org

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

Views

Author

Jianing Song, Feb 17 2021

Keywords

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)!

Examples

			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.

Links

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!).

Programs

  • 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))))

Formula

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

A228298 Generalized meta-Fibonacci sequence a(n) with parameters s=0 and k=7.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 12, 13, 14, 14, 15, 16, 17, 18, 19, 20, 21, 21, 22, 23, 24, 25, 26, 27, 28, 28, 29, 30, 31, 32, 33, 34, 35, 35, 36, 37, 38, 39, 40, 41, 42, 42, 43, 44, 45, 46, 47, 48, 49, 49, 49, 50, 51, 52, 53, 54, 55, 56, 56
Offset: 1

Views

Author

Michel Marcus, Aug 20 2013

Keywords

Comments

It appears than a(n) = A020646(n)/7 for n>0 (verified up to n=700).

Links

Crossrefs

Cf. A120503, A120507, A228297.

Programs

  • PARI
    a(n)= {local(A); if(n<=7, max(0, n), A=vector(n, i, i); for(k=8, n, A[k]= A[k-A[k-1]] + A[k-1-A[k-2]] + A[k-2-A[k-3]] + A[k-3-A[k-4]] + A[k-4-A[k-5]] + A[k-5-A[k-6]] + A[k-6-A[k-7]];); A[n];);}
Showing 1-3 of 3 results.

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