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.

Previous Showing 21-30 of 30 results.

A325609 Unsorted q-signature of n!. Irregular triangle read by rows where T(n,k) is the multiplicity of q(k) in the factorization of n! into factors q(i) = prime(i)/i.

Original entry on oeis.org

1, 2, 1, 4, 1, 5, 2, 1, 7, 3, 1, 9, 3, 1, 1, 12, 3, 1, 1, 14, 5, 1, 1, 16, 6, 2, 1, 17, 7, 3, 1, 1, 20, 8, 3, 1, 1, 22, 9, 3, 1, 1, 1, 25, 9, 3, 2, 1, 1, 27, 11, 4, 2, 1, 1, 31, 11, 4, 2, 1, 1, 33, 11, 4, 3, 1, 1, 1, 36, 13, 4, 3, 1, 1, 1, 39, 13, 4, 3, 1, 1, 1, 1
Offset: 1

Views

Author

Gus Wiseman, May 12 2019

Keywords

Comments

Every positive integer has a unique q-factorization (encoded by A324924) into factors q(i) = prime(i)/i, i > 0. For example:
11 = q(1) q(2) q(3) q(5)
50 = q(1)^3 q(2)^2 q(3)^2
360 = q(1)^6 q(2)^3 q(3)
Row n is the sequence of nonzero exponents in the q-factorization of n!.
Also the number of terminal subtrees with Matula-Goebel number k of the rooted tree with Matula-Goebel number n!.

Examples

			We have 10! = q(1)^16 q(2)^6 q(3)^2 q(4), so row n = 10 is (16,6,2,1).
Triangle begins:
  {}
   1
   2  1
   4  1
   5  2  1
   7  3  1
   9  3  1  1
  12  3  1  1
  14  5  1  1
  16  6  2  1
  17  7  3  1  1
  20  8  3  1  1
  22  9  3  1  1  1
  25  9  3  2  1  1
  27 11  4  2  1  1
  31 11  4  2  1  1
  33 11  4  3  1  1  1
  36 13  4  3  1  1  1
  39 13  4  3  1  1  1  1
  42 14  5  3  1  1  1  1
		

Crossrefs

Row lengths are A000720.
Row sums are A325544(n) - 1.
Column k = 1 is A325543.
Matula-Goebel numbers: A007097, A061775, A109129, A196050, A317713, A324935.
Factorial numbers: A000142, A011371, A022559, A071626, A115627, A325276.
q-factorization: A324922, A324923, A324924, A325614, A325615, A325660.

Programs

  • Mathematica
    difac[n_]:=If[n==1,{},With[{i=PrimePi[FactorInteger[n][[1,1]]]},Sort[Prepend[difac[n*i/Prime[i]],i]]]];
    Table[Length/@Split[difac[n!]],{n,20}]

A226460 Let m! have prime factorization Product (p_j^e_j); a(n) = number of distinct prime factors p_j such that e_j = n has no solution for any m!.

Original entry on oeis.org

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

Views

Author

Naohiro Nomoto, Jun 08 2013

Keywords

Comments

If n belongs to A048247 then a(n) is equal to zero.
For a given prime p and n satisfying p^k + p^(k-1) + ... + 1 <= n < p^(k+1) + ... + 1 for some k, let r_k = n mod (p^k + p^(k-1) + ... + 1), r_(k-1) = r_k mod (p^(k-1) + ... + 1), and so on down to r_1 = r_2 mod (p + 1). Then, p^n appears in a factorial m! iff none of the r_i is congruent to -1. - Charlie Neder, Nov 03 2018

Examples

			For n = 11, there are three distinct prime factors (3, 5, 11) in factorization of m!.
3^10 divides 26! ( 26! is not divisible by 3^11).
3^13 divides 27!.
5^10 divides 49! ( 49! is not divisible by 5^11).
5^12 divides 50!.
11^10 divides 120! ( 120! is not divisible by 11^11).
11^12 divides 121!.
The exponent of three distinct prime factors never becomes equal to 11. (It searches for all the exponent of prime factorization of factorials [A000142].)
Therefore a(11)=3.
		

Crossrefs

Programs

  • PARI
    is(k, p) = my(c, s); while(sk;
    a(n) = sum(p=2, n, isprime(p)&&is(n, p)); \\ Jinyuan Wang, Aug 22 2021

A266620 a(n) = least non-divisor of n!.

Original entry on oeis.org

2, 3, 4, 5, 7, 7, 11, 11, 11, 11, 13, 13, 17, 17, 17, 17, 19, 19, 23, 23, 23, 23, 29, 29, 29, 29, 29, 29, 31, 31, 37, 37, 37, 37, 37, 37, 41, 41, 41, 41, 43, 43, 47, 47, 47, 47, 53, 53, 53, 53, 53, 53, 59, 59, 59, 59, 59, 59, 61, 61, 67, 67, 67, 67, 67, 67, 71
Offset: 1

Views

Author

Jeffrey Shallit, Jan 01 2016

Keywords

Comments

It appears that a(n) = A151800(n) with the exception of n = 3. - Robert Israel, Jan 13 2016

Examples

			For n = 4 the least non-divisor of 4! = 24 = 2^3 * 3 is 5.
For n = 5 the least non-divisor of 5! = 120 = 2^3 * 3 * 5 is 7.
		

Crossrefs

Programs

  • Maple
    N:= 100: # to get a(1)..a(N)
    m:= 1 + numtheory:-pi(N):
    Primes:= [seq(ithprime(i),i=1..m)]:
    for i from 1 to m do pindex[Primes[i]]:= i od:
    V:= Vector(m):
    k:= 0:
    for n from 1 to N do
      for f in ifactors(n)[2] do
        q:= pindex[f[1]];
        V[q]:= V[q] + f[2];
        k:= max(k, q);
      od:
      a[n]:= min(seq(Primes[i]^(1+V[i]),i=1..k),Primes[k+1]);
    od:
    seq(a[n],n=1..N); # Robert Israel, Jan 13 2016
  • Mathematica
    Table[Complement[Range[2n], Divisors[n!]][[1]], {n, 30}] (* Alonso del Arte, Sep 23 2017 *)
    Table[Block[{m = n!, k = n + 1}, While[Divisible[m, k], k++]; k], {n, 67}] (* Michael De Vlieger, Sep 23 2017 *)
  • Python
    from sympy import nextprime
    def A266620(n): return 4 if n == 3 else nextprime(n) # Chai Wah Wu, Feb 22 2023

Formula

a(n) = min_{k >= 1} prime(k)^(1 + v(n!, prime(k))) where v(m, p) is the p-adic order of m. - Robert Israel, Jan 13 2016
a(n) = prime(pi(n) + 1) except for n = 3, in which case the least non-divisor of 3! is 4, not 5. - Alonso del Arte, Sep 23 2017

A325510 Number of non-isomorphic multiset partitions of the multiset of prime indices of n!.

Original entry on oeis.org

1, 1, 1, 2, 7, 16, 98, 269, 1397, 7582, 70520, 259906, 1677259, 5229112, 44726100, 666355170, 4917007185, 18459879921
Offset: 0

Views

Author

Gus Wiseman, May 08 2019

Keywords

Examples

			Non-isomorphic representatives of the a(2) = 1 through a(5) = 16 multiset partitions:
  {{1}}  {{12}}    {{1222}}        {{12333}}
         {{1}{2}}  {{1}{222}}      {{1}{2333}}
                   {{12}{22}}      {{12}{333}}
                   {{2}{122}}      {{13}{233}}
                   {{1}{2}{22}}    {{3}{1233}}
                   {{2}{2}{12}}    {{33}{123}}
                   {{1}{2}{2}{2}}  {{1}{2}{333}}
                                   {{1}{23}{33}}
                                   {{1}{3}{233}}
                                   {{3}{12}{33}}
                                   {{3}{13}{23}}
                                   {{3}{3}{123}}
                                   {{1}{1}{1}{23}}
                                   {{1}{2}{3}{33}}
                                   {{1}{3}{3}{23}}
                                   {{1}{2}{3}{3}{3}}
		

Crossrefs

Programs

  • PARI
    \\ Requires C(sig) from A318285.
    a(n)={if(n<2, 1, my(f=factor(n!)[,2], sig=vector(vecmax(f))); for(i=1, #f, sig[f[i]]++); C(sig))} \\ Andrew Howroyd, Jan 17 2023

Formula

a(n) = A317791(n!).
a(n) = A318285(A181819(n!)) = A318285(A325508(n)). - Andrew Howroyd, Jan 17 2023

Extensions

a(9)-a(17) from Andrew Howroyd, Jan 17 2023

A325511 Numbers whose prime signature is that of a factorial number.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 26, 29, 31, 33, 34, 35, 37, 38, 39, 40, 41, 43, 46, 47, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 65, 67, 69, 71, 73, 74, 77, 79, 82, 83, 85, 86, 87, 88, 89, 91, 93, 94, 95, 97, 101, 103, 104, 106
Offset: 1

Views

Author

Gus Wiseman, May 08 2019

Keywords

Comments

A181819(a(n)) belongs to A325508.

Examples

			The sequence of terms together with their prime indices begins:
   1: {}
   2: {1}
   3: {2}
   5: {3}
   6: {1,2}
   7: {4}
  10: {1,3}
  11: {5}
  13: {6}
  14: {1,4}
  15: {2,3}
  17: {7}
  19: {8}
  21: {2,4}
  22: {1,5}
  23: {9}
  24: {1,1,1,2}
  26: {1,6}
  29: {10}
  31: {11}
		

Crossrefs

Programs

  • Mathematica
    Select[Range[30],MemberQ[Table[Sort[Last/@FactorInteger[k!]],{k,#}],Sort[Last/@FactorInteger[#]]]&]

A115628 Factorization of n! into primorials.

Original entry on oeis.org

1, 0, 1, 2, 1, 2, 0, 1, 2, 1, 1, 2, 1, 0, 1, 5, 1, 0, 1, 3, 3, 0, 1, 4, 2, 1, 1, 4, 2, 1, 0, 1, 5, 3, 1, 0, 1, 5, 3, 1, 0, 0, 1, 6, 3, 0, 1, 0, 1, 5, 3, 1, 1, 0, 1, 9, 3, 1, 1, 0, 1, 9, 3, 1, 1, 0, 0, 1, 8, 5, 1, 1, 0, 0, 1, 8, 5, 1, 1, 0, 0, 0, 1, 10, 4, 2, 1, 0, 0, 0, 1, 9, 5, 1, 2, 0, 0, 0, 1, 10, 5, 1, 1, 1
Offset: 2

Views

Author

Keywords

Comments

Factorial of n has a unique representation as a product of primorials: n! = 2#^T(n,1)*3#^T(n,2)*...*P_{Pi(n)}#^T(n,Pi(n)).

Examples

			Rows start: 1; 0,1; 2,1; 2,0,1; 2,1,1; 2,1,0,1; 5,1,0,1; 3,3,0,1.
		

Crossrefs

Cf. A115627, A000142, A002110. Row lengths are A000720.

Formula

T(n, k) = A115727(n, k)-A115727(n, k+1).

A176383 Triangle of the powers of the prime factorization of n! in row n.

Original entry on oeis.org

2, 2, 3, 8, 3, 8, 3, 5, 16, 9, 5, 16, 9, 5, 7, 128, 9, 5, 7, 128, 81, 5, 7, 256, 81, 25, 7, 256, 81, 25, 7, 11, 1024, 243, 25, 7, 11, 1024, 243, 25, 7, 11, 13, 2048, 243, 25, 49, 11, 13, 2048, 729, 125, 49, 11, 13, 32768, 729, 125, 49, 11, 13, 32768, 729, 125, 49, 11, 13, 17, 65536
Offset: 2

Views

Author

Vladimir Shevelev, Apr 16 2010

Keywords

Comments

Row n contains pi(n) = A000720(n) terms. The exponents are in A115627.
The first column contains the maximum power of 2 dividing n!, the second column the maximum power of 3 dividing n! etc.

Examples

			The irregular tables starts with n=2:
2; # =2! = 2
2*3; # =3! = 6
8*3; # =4! = 24
8*3*5; # =5! = 120
16*9*5; # =6!
16*9*5*7; # =7!
128*9*5*7; # =8!
128*81*5*7;
256*81*25*7;
		

Crossrefs

Programs

  • Maple
    with(numtheory):
    b:= proc(n) option remember; `if`(n=1, 1, b(n-1)+
          add(i[2]*x^pi(i[1]), i=ifactors(n)[2]))
        end:
    T:= n->(p->seq(ithprime(i)^coeff(p, x, i), i=1..pi(n)))(b(n)):
    seq(T(n), n=2..20);  # Alois P. Heinz, Jun 22 2014
  • Mathematica
    T[n_] := List @@ Power @@@ FactorInteger[n!];
    Array[T, 20, 2] // Flatten (* Jean-François Alcover, Mar 27 2017 *)
    Rest[Flatten[Table[#[[1]]^#[[2]]&/@FactorInteger[n!],{n,20}]]] (* Harvey P. Dale, Jan 04 2019 *)
  • PARI
    a(n)=my(i=2);while(n-primepi(i)>1,n-=primepi(i);i++);p=prime(n-1);p^sum(j=1,log(i)\log(p),i\=p) \\ David A. Corneth, Jun 21 2014

Extensions

Arbitrarily defined first 2 terms removed by R. J. Mathar, Apr 23 2010

A321362 a(n) is the least k such that in the prime power factorization of k! the exponents of primes p_1, ..., p_n are odd, while the exponent of p_(n+1) is even.

Original entry on oeis.org

2, 3, 5, 119, 57, 220, 1131, 2986, 1505, 3211, 21300, 26795, 11820, 14575, 67385, 221051, 33782, 132512, 819236, 1478432, 1630903, 26736550, 1095752, 41815849, 24813938, 31982450, 142574286, 860986855, 602660826, 2638930495, 2664421881, 1309662955, 33767540563
Offset: 1

Views

Author

Michel Marcus, Nov 07 2018

Keywords

Comments

Analog to A240537 where odd and even are swapped.

Examples

			For k=2, 2!=2 with factorization 2 where the first exponent is odd.
For k=3, 3!=6 with factorization 2*3 where the first 2 exponents are odd.
For k=5, 5!=120 with factorization 2*3*5 where the first 3 exponents are odd.
In each case, there are no lesser numbers with the same property.
		

Crossrefs

Programs

  • C
    See Links section.
  • Mathematica
    aQ[k_,n_] := Module[{e=FactorInteger[k!][[;;,2]]}, Length[e]>=n && AllTrue[ e[[1;;n]], OddQ ] && If[Length[e]>n, EvenQ[e[[n+1]]], True]]; a[n_] := Module[ {k=2}, While[!aQ[k,n], k++]; k ]; Array[a,10] (* Amiram Eldar, Nov 07 2018 *)
  • PARI
    isok(v, n) = {if (#v < n, return (0)); for (i=1, n, if (!(v[i] % 2), return(0));); (#v == n) || !(v[n+1] % 2);}
    newv(v, i) = {if (isprime(i), return(concat(v, 1))); f = factor(i); for (k=1, #f~, v[primepi(f[k,1])] += f[k,2];); return (v);}
    a(n) = {my(v =[1], i = 2); while (!isok(v, n), i++; v = newv(v, i)); i;}
    

Extensions

a(16)-a(32) from Rémy Sigrist, Nov 08 2018
a(33) from Giovanni Resta, Nov 09 2018

A330706 Numbers m such that the prime factorization of m! contains no composite exponents.

Original entry on oeis.org

1, 2, 3, 4, 5, 8, 14
Offset: 1

Views

Author

Devansh Singh, Mar 29 2020

Keywords

Comments

This sequence is finite and a(7) = 14 is the last term. Nagura (see references) proves that for n >= 25, there is always a prime between n and 1.2*n. Hence, for any prime p > 25, there is always a number m between 4p and 4.8*p, and so floor(m/p) = 4. Since by assumption p > 4, floor(floor(m/p)/p) = 0 and so m! is divisible by p^4 but not p^5. It remains to check the primes up to 25 individually. - Charles R Greathouse IV, Apr 14 2020

Examples

			4 is a term since 4! = (2^3)*(3^1) and the multiplicity of 2 is 3 which is prime and the multiplicity of 3 is 1.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[100], !AnyTrue[FactorInteger[#!][[;;,2]], CompositeQ] &] (* Amiram Eldar, Mar 29 2020 *)
  • PARI
    ok(n)={my(f=factor(n!)[,2]); for(i=1, #f, if(f[i]<>1 && !isprime(f[i]), return(0))); 1}
    {select(ok, [1..100])} \\ Andrew Howroyd, Mar 29 2020
    
  • PARI
    f(m,p)=my(s); while(m\=p, s+=m); s;
    is(n)=forprime(p=2,n\4+1, if(!isprime(f(n,p)), return(0))); 1;
    select(is,[1..25]) \\ Charles R Greathouse IV, Apr 14 2020

A353068 Irregular triangle read by rows: T(n,k) = n - multiplicity of prime(k) as a divisor of n!.

Original entry on oeis.org

1, 2, 2, 1, 3, 2, 4, 4, 2, 4, 5, 3, 5, 6, 6, 1, 6, 7, 7, 2, 5, 8, 8, 2, 6, 8, 9, 3, 7, 9, 10, 10, 2, 7, 10, 11, 11, 3, 8, 11, 12, 12, 12, 3, 9, 12, 12, 13, 13, 4, 9, 12, 13, 14, 14, 1, 10, 13, 14, 15, 15, 2, 11, 14, 15, 16, 16, 16, 2, 10, 15, 16, 17, 17, 17, 3, 11, 16, 17, 18, 18, 18, 18
Offset: 2

Views

Author

Michel Marcus, Apr 21 2022

Keywords

Comments

This is the "s" in the American Mathematical Monthly problem.

Examples

			First few rows are:
  1;
  2, 2;
  1, 3;
  2, 4, 4;
  2, 4, 5;
  3, 5, 6, 6;
  1, 6, 7, 7;
  2, 5, 8, 8;
  2, 6, 8, 9;
  3, 7, 9, 10, 10;
  2, 7, 10, 11, 11;
  ...
		

Crossrefs

Cf. A115627, A000120 (column 1), A089792 (column 2).

Programs

  • Mathematica
    T[n_, k_] := n - IntegerExponent[n!, Prime[k]]; Table[T[n, k], {n, 2, 19}, {k, 1, PrimePi[n]}] // Flatten (* Amiram Eldar, Apr 21 2022 *)
  • PARI
    row(n) = vector(primepi(n), k, n-valuation(n!, prime(k)));

Formula

T(n,k) = n - A115627(n, k).
Previous Showing 21-30 of 30 results.