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 1-5 of 5 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

A120503 Generalized meta-Fibonacci sequence a(n) with parameters s=0 and k=3.

Original entry on oeis.org

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

Views

Author

Frank Ruskey and Chris Deugau (deugaucj(AT)uvic.ca), Jun 20 2006

Keywords

References

  • Callaghan, Joseph, John J. Chew III, and Stephen M. Tanny. "On the behavior of a family of meta-Fibonacci sequences." SIAM Journal on Discrete Mathematics 18.4 (2005): 794-824. See T_{0,3} with initial values 0,0,1, and plotted in Fig. 1.5. This is essentially the same sequence. - N. J. A. Sloane, Apr 16 2014

Links

Crossrefs

Cf. A120514, A120525.

Programs

  • Maple
    a := proc(n)
option remember;
if n <= 1 then return 1 end if;
if n <= 3 then return n end if;
return add(a(n - i + 1 - a(n - i)), i = 1 .. 3)
end proc
  • Mathematica
    a[n_] := a[n] = If[1 <= n <= 3, n, Sum[a[n-i+1 - a[n-i]], {i, 1, 3}]];
Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Aug 02 2022 *)
  • PARI
    {a(n)=local(A); if(n<=3, max(0, n), A=vector(n, i, i); for(k=4, n, A[k]=A[k-A[k-1]]+A[k-1-A[k-2]]+A[k-2-A[k-3]]); A[n])} /* Michael Somos, Aug 31 2006 */
  • PARI
    apply( A120503(n)={my(s=sumdigits(n*=2, 3)\2); n\=3; while(s>0, s-=valuation(n++, 3)+1); n}, [1..99]) \\ M. F. Hasler, Dec 27 2019

Formula

If n = 1, a(n)=1. If 2 <= n <= 3, then a(n)=n. If n>3 then a(n)=a(n-a(n-1)) + a(n-1-a(n-2)) + a(n-2-a(n-3))
G.f.: A(z) = z / (1 - z) * prod( (1 - z^(3 * [i])) / (1 - z^[i]), i=1..infinity), where [i] = (3^i - 1) / 2.
a(n) = A007844(n)/3. - Michel Marcus, Aug 19 2013, conjectured. This is true: see the analogous sequence A007843 for a sketch of the proof. - M. F. Hasler, Dec 27 2019

A007845 Least positive integer k for which 5^n divides k!.

Original entry on oeis.org

1, 5, 10, 15, 20, 25, 25, 30, 35, 40, 45, 50, 50, 55, 60, 65, 70, 75, 75, 80, 85, 90, 95, 100, 100, 105, 110, 115, 120, 125, 125, 125, 130, 135, 140, 145, 150, 150, 155, 160, 165, 170, 175, 175, 180, 185, 190, 195, 200, 200, 205, 210, 215, 220, 225, 225, 230, 235, 240, 245
Offset: 0

Views

Author

Bruce Dearden and Jerry Metzger

Keywords

Comments

Also the smallest factorial having at least n trailing zeros. - Jud McCranie, Oct 05 2010
a(n) ~ 4n, a(n) > 4n. Every positive multiple of 5 occurs as much as the exponent of 5 in the prime factorization. - David A. Corneth, Jul 12 2016
Least k such that A027868(k) >= n. - Robert Israel, Jul 12 2016
See A007843 and A007844 for the analog for 2 and 3 instead of 5. - 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. A007843, A007844, A027868.

Programs

  • Maple
    1, seq(t $ padic:-ordp(t,5), t=5..1000, 5); # Robert Israel, Jul 12 2016
  • Mathematica
    lpi[n_]:=Module[{k=1,n5=5^n},While[!Divisible[k!,n5],k++];k]; Array[ lpi,60,0] (* Harvey P. Dale, Jun 19 2012 *)
  • PARI
    a(n) = {k = 1; while (valuation(k!, 5) < n, k++); k;} \\ Michel Marcus, Aug 19 2013
  • PARI
    a(n) = {my(ck = 4 * n, k = 5 * floor(ck/5), t = 0); if(ck > 0, t = sum(i = 1, logint(ck,5),ck\=5)); while(t < n, k+=5; t+=valuation(k,5));max(1,k)} \\ David A. Corneth, Jul 12 2016

Formula

a(n) = 5*A228297(n) for n > 0: see A007843. - M. F. Hasler, Dec 27 2019

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))).

A020646 Least positive integer k for which 7^n divides k!.

Original entry on oeis.org

1, 7, 14, 21, 28, 35, 42, 49, 49, 56, 63, 70, 77, 84, 91, 98, 98, 105, 112, 119, 126, 133, 140, 147, 147, 154, 161, 168, 175, 182, 189, 196, 196, 203, 210, 217, 224, 231, 238, 245, 245, 252, 259, 266, 273, 280, 287, 294, 294, 301, 308, 315, 322, 329, 336, 343, 343, 343, 350
Offset: 0

Views

Author

David W. Wilson

Keywords

References

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

Links

Crossrefs

Cf. A007843, A007844, A007845.

Programs

  • Mathematica
    lpi[n_]:=Module[{k = 1, n7 = 7^n}, While[! Divisible[k!, n7], k++]; k]; Array[lpi,60,0] (* Harvey P. Dale, Jun 29 2017 *)
  • PARI
    a(n) = {k = 1; while (valuation(k!, 7) < n, k++); k;} \\ Michel Marcus, Aug 19 2013
Showing 1-5 of 5 results.

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