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 1-10 of 16 results. Next

A011778 Numbers k where A011776(k) grows.

Original entry on oeis.org

2, 6, 12, 24, 36, 40, 45, 48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 480, 630, 672, 720, 1080, 1120, 1260, 1344, 1440, 1890, 2016, 2160, 2240, 2520, 2880, 3024, 3360, 3780, 4032, 4320, 5040, 6048
Offset: 1

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Author

Keywords

Comments

Equivalently, 2 along with numbers A092427(j) where A092427(j) > A092427(j-1), for j > 0. - Derek Orr, Apr 16 2015
Is 45 the only odd number in this sequence? - Derek Orr, Apr 16 2015
a(184) = 212837625 and a(311) = 638512875 are the only other odd terms < 10^11. - Charlie Neder, Jan 27 2019

References

  • Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, Wiley NY 1991.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 251.

Crossrefs

Formula

a(n) = A061769(n) + 1.
a(n) = A092427(A061770(n)). - Derek Orr, Apr 16 2015

Extensions

Corrected by Vladeta Jovovic, Dec 01 2002

A373725 Numbers k such that A011776(k) = A011776(k+1).

Original entry on oeis.org

1, 2, 3, 4, 8, 9, 15, 27, 63, 195, 728, 1443, 3843, 5475, 6174, 11913, 13376, 24963, 37635, 77283, 98595, 113398, 158403, 178083, 209763, 293763, 294335, 319124, 376995, 406503, 438243, 454275, 538755, 574563, 770883, 996003, 1196835, 1331715, 1444803, 1473795
Offset: 1

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Author

Amiram Eldar, Jun 15 2024

Keywords

Comments

The corresponding values of A011776 are 1, 1, 1, 1, 2, 2, 3, 4, 10, 16, 60, ... .
All the terms above 3 are composite numbers since A011776(k) = 1 if and only if k = 4 or a prime.
Are there 3 consecutive integers above 8 that have an equal value of A011776? There are none below 10^10.
Conjecture: if p != 3 is a prime such that 2*p-1 is also a prime (p is in A005382 \ {3}), then 4*p^2 - 1 is a term of this sequence.

Crossrefs

Programs

  • Mathematica
    s[1] = 1; s[n_] := IntegerExponent[n!, n]; seq[kmax_] := Module[{v = {}, s1 = s[1]}, Do[s2 = s[k]; If[s1 == s2, AppendTo[v, k-1]]; s1 = s2, {k, 2, kmax}]; v]; seq[10^4]
  • PARI
    lista(kmax) = {my(s1 = 1, s2); for(k = 2, kmax, s2 = valuation(k!, k); if(s1 == s2, print1(k-1, ", ")); s1 = s2);}

A090622 Square array read by antidiagonals of highest power of k dividing n! (with n,k>1).

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 0, 1, 3, 0, 0, 1, 1, 4, 0, 1, 0, 1, 2, 4, 0, 0, 1, 1, 2, 2, 7, 0, 0, 0, 1, 1, 2, 2, 7, 0, 0, 1, 0, 2, 1, 3, 4, 8, 0, 0, 0, 1, 0, 2, 1, 3, 4, 8, 0, 0, 0, 0, 1, 1, 2, 1, 4, 4, 10, 0, 0, 0, 1, 1, 1, 1, 4, 2, 4, 5, 10, 0, 0, 1, 0, 1, 1, 2, 1, 4, 2, 5, 5, 11, 0, 0, 0, 1, 0, 1, 1, 2, 1, 4, 2, 5, 5, 11
Offset: 2

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Author

Henry Bottomley, Dec 06 2003

Keywords

Examples

			Square array starts:
1, 0, 0, 0, 0, 0, 0, ...
1, 1, 0, 0, 1, 0, 0, ...
3, 1, 1, 0, 1, 0, 1, ...
3, 1, 1, 1, 1, 0, 1, ...
4, 2, 2, 1, 2, 0, 1, ...
4, 2, 2, 1, 2, 1, 1, ...
7, 2, 3, 1, 2, 1, 2, ...
		

Crossrefs

Programs

  • Maple
    f:= proc(n, p) local c, k; c, k:= 0, p;
           while n>=k do c:= c+iquo(n, k); k:= k*p od; c
        end:
    T:= (n, k)-> min(seq(iquo(f(n, i[1]), i[2]), i=ifactors(k)[2])):
    seq(seq(T(n, 2+d-n), n=2..d), d=2..20);  # Alois P. Heinz, Oct 04 2012
  • Mathematica
    f[n_, p_] := Module[{c = 0, k = p}, While[n >= k , c = c + Quotient[n, k]; k = k*p ]; c ]; t[n_, k_] := Min[ Table[ Quotient[f[n, i[[1]]], i[[2]]], {i, FactorInteger[k]}]]; Table[ Table[t[n, 2 + d - n], {n, 2, d}], {d, 2, 20}] // Flatten (* Jean-François Alcover, Oct 03 2013, translated from Alois P. Heinz's Maple program *)

Formula

For k=p prime: T(n,p) = [n/p] + [n/p^2] + [n/p^3] + .... For k = p^m a prime power: T(n,p^m) = [T(n,p)/m]. For k = b*c with b and c coprime: T(n,a*b) = min(T(n,a), T(n,b)). T(n,k) is close to, but below, n/A090624(k).

A060151 Number of base n digits required to write n!.

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 5, 6, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 19, 19, 20, 21, 22, 22, 23, 24, 25, 26, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 34, 35, 36, 37, 38, 38, 39, 40, 41, 42, 43, 43, 44, 45, 46, 47, 47, 48, 49, 50, 51, 51, 52, 53, 54, 55, 56, 56, 57
Offset: 1

Views

Author

Henry Bottomley, Mar 08 2001

Keywords

Examples

			a(6)=4 since 6!=720, which in base 6 is 3200.
		

Crossrefs

Cf. A011776 for number of final zeros of n! written in base n.

Programs

Formula

a(n) = 1 + floor(log(n!)/log(n)) = 1 + A039960(n) for n>1.
From Danny Rorabaugh, Apr 14 2015: (Start)
a(n) = 1 + log_n(A074182(n)) for n>1.
a(n) = A074184(n) = log_n(A074181(n)) for n>2.
(End)
a(n) = n - n/log n + O(1). - Charles R Greathouse IV, Oct 29 2016

A039960 For n >= 2, a(n) = largest value of k such that n^k is <= n! (a(0) = a(1) = 1 by convention).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 11, 12, 13, 14, 14, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 29, 30, 31, 32, 33, 33, 34, 35, 36, 37, 37, 38, 39, 40, 41, 42, 42, 43, 44, 45, 46, 46, 47, 48, 49, 50, 50, 51, 52, 53, 54, 55, 55, 56, 57
Offset: 0

Views

Author

Dan Bentley (bentini(AT)yahoo.com)

Keywords

Comments

Seems to be slightly more than (but asymptotic to) number of nonprimes less than or equal to n.

Examples

			a(7)=4 because 7! = 5040, 7^4 = 2401 but 7^5 = 16807.
a(6)=3 since 6^3.67195... = 720 = 6! and 6^3 <= 6! < 6^4, i.e., 216 <= 720 < 1296.
		

Crossrefs

Programs

  • Magma
    [1,1] cat [Floor(Log(Factorial(n))/Log(n)): n in [2..80]]; // Vincenzo Librandi, Apr 15 2015
    
  • Mathematica
    ds[x_, y_] :=y!-y^x; a[n_] :=Block[{m=1, s=ds[m, n]}, While[Sign[s]!=-1&&!Greater[m, 256], m++ ];m]; Table[a[n]-1, {n, 3, 200}]
    (* or *)
    Table[Count[Part[Sign[Table[Table[n!-n^j, {j, 1, 128}], {n, 1, 128}]], u], 1], {u, 1, 128}] (* Labos Elemer *)
    Join[{1,1},Table[Floor[Log[n,n!]],{n,2,80}]] (* Harvey P. Dale, Sep 24 2019 *)
  • PARI
    a(n)=if(n>3,lngamma(n+1)\log(n),1) \\ Charles R Greathouse IV, Sep 02 2015
  • Sage
    [1,1] + [floor(log(factorial(n))/log(n)) for n in range(2,75)] # Danny Rorabaugh, Apr 14 2015
    

Formula

a(n) = floor(log_n(n!)) for n > 1.
a(n) = A060151(n) - 1 for n > 1. - Henry Bottomley, Mar 08 2001
From Danny Rorabaugh, Apr 14 2015: (Start)
a(n) = log_n(A074182(n)) for n > 1.
a(n) = A074184 - 1 = log_n(A074181(n)) - 1 for n > 2. (End)
From Robert Israel, Apr 14 2015: (Start)
n*(1-1/log(n)) + 1 > log(n!)/log(n) > n*(1-1/log(n)) for n >= 7.
Thus a(n) is either floor(n*(1-1/log(n))) or ceiling(n*(1-1/log(n))) for n >= 7 (and in fact this is the case for n >= 3). (End)

Extensions

Corrected and extended by Henry Bottomley, Mar 08 2001
Edited by N. J. A. Sloane, Sep 26 2008 at the suggestion of R. J. Mathar

A090630 Greatest divisor d of n! such that d=m^k with k>1.

Original entry on oeis.org

1, 1, 1, 1, 8, 8, 144, 144, 576, 5184, 518400, 518400, 2073600, 2073600, 101606400, 914457600, 14631321600, 14631321600, 526727577600, 526727577600, 52672757760000, 221225582592000, 6373403688960000, 6373403688960000, 917770131210240000, 22944253280256000000, 3877578804363264000000
Offset: 0

Views

Author

Reinhard Zumkeller, Dec 13 2003

Keywords

Comments

a(n) is a square for all n except n = 4, 5 and 21 (Wilke, 1981). - Amiram Eldar, Jun 09 2022

Crossrefs

Programs

  • Maple
    f:= proc(n)
      local F,  k, d,r,s;
      F:= ifactors(n!)[2];
      r:= 1;
      for k from 2 to F[1][2] do
        r:= max(r, mul(f[1]^(k*floor(f[2]/k)),f=F))
      od:
    r
    end proc:
    1,1,seq(f(n), n=2..100); # Robert Israel, Dec 08 2014
  • Mathematica
    IsPower[n_] := If[n==1, True, GCD@@(Transpose[FactorInteger[n]][[2]])>1]; Table[Select[Divisors[n! ], IsPower][[ -1]], {n, 0, 25}]
  • PARI
    a(n)=my(f=factor(n!),m=1); for(i=2,if(#f~,f[1,2]), m=max(factorback(concat(Mat(f[,1]), f[,2]\i*i)),m)); m \\ Charles R Greathouse IV, Dec 09 2014

Formula

a(n)= n!/A251753(n). - Robert G. Wilson v, Dec 08 2014

Extensions

More terms from T. D. Noe, Oct 04 2004

A133481 a(1) = 1; for n > 1, a(n) is the least k such that k^n divides k! but k^(n+1) does not divide k!.

Original entry on oeis.org

1, 6, 15, 18, 12, 32, 24, 36, 40, 45, 48, 100, 84, 60, 154, 165, 72, 96, 80, 126, 90, 135, 286, 200, 312, 264, 168, 120, 297, 189, 160, 330, 544, 210, 144, 224, 300, 385, 396, 324, 252, 680, 350, 180, 280, 748, 572, 486, 400, 405, 315, 528, 320, 336, 450, 512, 288, 240, 715
Offset: 1

Views

Author

Masahiko Shin, Nov 29 2007

Keywords

Comments

Least k such that A011776(k) = n.
New record highs, by index: 1, 2, 3, 4, 6, 8, 9, 10, 11, 12, 15, 16, 23, 25, 32, 33, 42, 46, 63, 66, 79, 85, 100, 119, 128, 167, 188, 201, 213, 226, 240, 256, 335, 346, 348, 352, 360, 377, 385, 414, 426, 480, 481, 494, 504, 533, 555, 596, 656, 727, 883, 926, 938, 1026, 1094, ... - Robert G. Wilson v, Feb 28 2012
First 10000 terms are 163-smooth. - David A. Corneth, Mar 15 2019

Examples

			a(7)=24 because 24^7|24! and smaller numbers than 24 do not divide their factorials 7 times.
a(2) = 6 as 6^2|6! but 6! doesn't divide 6^(2 + 1) and 6 is the least positive integer with this property. - _David A. Corneth_, Mar 15 2019
		

References

  • Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 251.

Crossrefs

Programs

  • Mathematica
    kdn[n_]:=Module[{k=2},While[!Divisible[k!,k^n]||Divisible[k!, k^(n+1)], k++];k]; Join[{1},Array[kdn,60,2]] (* Harvey P. Dale, Feb 27 2012 *)
  • PARI
    a(n)=if(n<2,1,my(k=2);while(valuation(k!,k)!=n,k++);k) \\ Charles R Greathouse IV, Feb 27 2012
    
  • PARI
    See Corneth link \\ David A. Corneth, Mar 15 2019

Extensions

Edited by N. J. A. Sloane using material from A011777, Nov 29 2007

A011777 a(n) = least k>1 such that k^n divides k!.

Original entry on oeis.org

2, 6, 15, 18, 12, 32, 24, 36, 40, 45, 48, 100, 84, 60, 154, 165, 72, 96, 80, 126, 90, 135, 286, 200, 312, 264, 168, 120, 297, 189, 160, 330, 544, 210, 144, 224, 300, 385, 396, 324, 252, 680, 350, 180, 280, 748, 572, 486, 400, 405, 315, 528, 320, 336, 450, 512, 288, 240, 715
Offset: 1

Views

Author

Keywords

Comments

For n >= 2, least k such that A011776(k)=n.

References

  • Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 251.

Crossrefs

See A133481 for a better version. Cf. A011776, A011778.

A074845 Numbers k such that S(k) = largest difference between consecutive divisors of k (ordered by size), where S(k) is the Kempner function (A002034).

Original entry on oeis.org

6, 8, 9, 10, 14, 22, 26, 34, 38, 46, 58, 62, 74, 82, 86, 94, 106, 118, 122, 134, 142, 146, 158, 166, 178, 194, 202, 206, 214, 218, 226, 254, 262, 274, 278, 298, 302, 314, 326, 334, 346, 358, 362, 382, 386, 394, 398, 422, 446, 454, 458, 466, 478, 482, 502, 514
Offset: 1

Views

Author

Jason Earls, Sep 10 2002

Keywords

Comments

It appears that terms > 6 are simply given by: composite k such that k^2 doesn't divide A000254(k). - Benoit Cloitre, Mar 09 2004
It appears that A011776(a(k)) = 2. - Gionata Neri, Jul 31 2017
It appears that this sequence consists of the numbers k such that A045763(k) > 0 and k does not divide A070251(k). - Isaac Saffold, Jun 01 2018

Crossrefs

Programs

  • Mathematica
    Select[Range@ 514, Function[n, Module[{m = 1}, While[! Divisible[m!, n], m++]; m] == Max@ Differences@ Divisors@ n]] (* Michael De Vlieger, Jul 31 2017 *)
  • PARI
    K(n) = my(s=1); while(s!%n>0, s++); s;
    dd(n) = my(vd=divisors(n)); vecmax(vector(#vd-1, k, vd[k+1] - vd[k]));
    isok(n) = K(n) == dd(n); \\ Michel Marcus, Aug 03 2017

A346418 a(n) is the exponent of the largest power of n that divides the least common multiple of {1,2,...,n} (A003418). a(1) = 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jul 16 2021

Keywords

Examples

			a(2) = 1 since A003418(2) = 2, and 2^1|A003418(2).
a(30) = 2 since A003418(30) = 2329089562800 = 30^2 * 2587877292, and 30^2|A003418(30).
		

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := IntegerExponent[LCM @@ Range[n], n]; Array[a, 100]
  • PARI
    a(n) = if (n==1, 1, valuation(lcm([1..n]), n)); \\ Michel Marcus, Jul 17 2021

Formula

a(n) <= omega(n), and a(n) < omega(n) whenever omega(n) > 1.
Max_{k=2..n} a(k) ~ log(n)/(log(log(n)) + o(1)) (Erdős, 1992).
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