Jerry Metzger - cp's OEIS frontend

User: Jerry Metzger

Jerry Metzger has authored 5 sequences.

A177356 a(n) is the index of the first 0 term in the rumor sequence with initial 0th term 1 and parameters b = 2 and n.

1, 2, 5, 10, 3, 18, 7, 24, 23, 22, 13, 4, 19, 18, 9, 6, 15, 374, 13, 12, 11, 370, 369, 32, 367, 366, 5, 28, 363, 8, 361, 360, 37, 358, 21, 356, 355, 354, 353, 16, 351, 100, 349, 98, 347, 346, 95, 344, 93, 92, 91, 340, 89, 10, 15, 336, 31, 6, 333, 82, 331, 80, 25, 328, 327, 326

Offset: 1

Jerry Metzger, Dec 10 2010

A rumor sequence (running modulus recurrence sequence) is defined as follows: fix integer parameters b > 1 and n > 0. Set z[0] = any integer, and, for k > 0, define z[k] to be the least nonnegative residue of b*z[k-1] modulo (k+n). The rumor sequence conjecture states that all such rumor sequences are eventually 0.

			For n = 15, the z-sequence terms are 1, 2, 4, 8, 16, 12, 3, 6, 12, 0, so a(15) = 9; that is, z[0] = 1, z[1] = 2, z[2] = 4, ..., z[8] = 12, and z[9] = 0. [Edited by _Petros Hadjicostas_, Dec 13 2019]

B. Dearden, J. Iiams, and J. Metzger, Rumor Arrays, J. Integer Seq. 16 (2013), #13.9.3.
B. Dearden and J. Metzger, Running Modulus Recursions, J. Integer Seq. 13(1) (2010), #10.1.6.

Cf. A208125, A330411.

For[n=1,n<50,n++,k=0;Clear[z];z[0]=1;z[k_]:=z[k]=Mod[2z[k-1],k+n];
While[z[k]>0,k++];Print[k];]

a(n) = inf{m > 0 | z[0] = 1, z[m] = 0, and z[k] = (2*z[k-1] mod (k + n)) for k = 1..m}. - Petros Hadjicostas, Dec 13 2019

More terms from Petros Hadjicostas, Dec 13 2019

A125849 a(n) = Sum_{m=1..n-1} floor(m(n-2)/2)^2.

Original entry on oeis.org

0, 0, 0, 1, 14, 62, 220, 547, 1260, 2444, 4560, 7685, 12650, 19466, 29484, 42567, 60760, 83672, 114240, 151689, 200070, 258070, 331100, 417131, 523204, 646372, 795600, 966797, 1171170, 1403234, 1676780, 1984655, 2343600, 2744496, 3207424

Offset: 0

Jerry Metzger, Jul 09 2008

nonn

for n from 0 to 15 do add( floor(m*(n-2)/2)^2,m=1..n-1) ; print(n,%) ; od: # R. J. Mathar

f[n_] := Sum[ Floor[m (n - 2)/2]^2, {m, n - 1}]; Table[ f@n, {n, 0, 35}] (* Robert G. Wilson v, Aug 03 2008 *)

G.f.: (x^3*(x^6+18*x^5+35*x^4+62*x^3+31*x^2+12*x+1))/((x+1)^4*(x-1)^6). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 28 2009

More terms from Robert G. Wilson v, Aug 03 2008

A007844 Least positive integer k for which 3^n divides k!.

Original entry on oeis.org

1, 3, 6, 9, 9, 12, 15, 18, 18, 21, 24, 27, 27, 27, 30, 33, 36, 36, 39, 42, 45, 45, 48, 51, 54, 54, 54, 57, 60, 63, 63, 66, 69, 72, 72, 75, 78, 81, 81, 81, 81, 84, 87, 90, 90, 93, 96, 99, 99, 102, 105, 108, 108, 108, 111, 114, 117, 117, 120, 123, 126, 126, 129, 132, 135, 135, 135

Offset: 0

Bruce Dearden and Jerry Metzger, R. Muller

nonn

It appears than for n>0, a(n) is divisible by 3, and that the resulting sequence a(n)/3 is A120503 (checked up to n=1000). - Michel Marcus, Aug 19 2013. [This is true: see A007843 for the idea of the proof. - M. F. Hasler, Dec 27 2019]

Also least positive integer k for which 6^n divides k!. - Michel Marcus, Aug 20 2013

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

T. D. Noe, Table of n, a(n) for n = 0..1000
F. Smarandache, Only Problems, Not Solutions!

Cf. A007843 (analog for 2), A007845 (analog for 5).

Cf. A120503 (Meta-Fibonacci, k = 3).

Array[Block[{k = 1}, While[Mod[k!, 3^#] != 0, k++]; k] &, 67, 0] (* Michael De Vlieger, Dec 29 2019 *)

a(n) = {k = 1; while (valuation(k!, 3) < n, k++); k;} \\ Michel Marcus, Aug 19 2013

apply( A007844(n)={my(s=sumdigits(n*=2,3)\2); n-=n%3; while(s>0, s-=valuation(n+=3,3)); n+!n}, [0..99]) \\ M. F. Hasler, Dec 27 2019

a(n) = 3*A120503(n) for n > 0, cf. A007843. - 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

Bruce Dearden and Jerry Metzger

nonn

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

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

T. D. Noe, Table of n, a(n) for n = 0..1000

Cf. A007843, A007844, A027868.

1, seq(t $ padic:-ordp(t,5), t=5..1000, 5); # Robert Israel, Jul 12 2016

lpi[n_]:=Module[{k=1,n5=5^n},While[!Divisible[k!,n5],k++];k]; Array[ lpi,60,0] (* Harvey P. Dale, Jun 19 2012 *)

a(n) = {k = 1; while (valuation(k!, 5) < n, k++); k;} \\ Michel Marcus, Aug 19 2013

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

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

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

Bruce Dearden and Jerry Metzger; R. Muller

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

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

T. D. Noe, Table of n, a(n) for n = 0..1000
Oliver Kullmann and Xishun Zhao, Parameters for minimal unsatisfiability: Smarandache primitive numbers and full clauses, arXiv preprint arXiv:1505.02318 [cs.DM], 2015.
Kevin Ryde, PARI/GP Code and Notes
F. Smarandache, Only Problems, Not Solutions!.
Wikipedia, Legendre's Formula.

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

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

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 *)

a(n)=if(n<0,0,s=1; while(s!%(2^n)>0,s++); s)

a(n) = {k = 1; while (valuation(k!, 2) < n, k++); k;} \\ Michel Marcus, Aug 19 2013

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

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

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

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User: Jerry Metzger

A177356 a(n) is the index of the first 0 term in the rumor sequence with initial 0th term 1 and parameters b = 2 and n.

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A125849 a(n) = Sum_{m=1..n-1} floor(m(n-2)/2)^2.

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A007844 Least positive integer k for which 3^n divides k!.

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

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

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