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.

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A100716 Numbers k such that p^p divides k for some prime p.

Original entry on oeis.org

4, 8, 12, 16, 20, 24, 27, 28, 32, 36, 40, 44, 48, 52, 54, 56, 60, 64, 68, 72, 76, 80, 81, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 135, 136, 140, 144, 148, 152, 156, 160, 162, 164, 168, 172, 176, 180, 184, 188, 189, 192, 196, 200, 204, 208, 212
Offset: 1

Views

Author

Leroy Quet, Dec 10 2004

Keywords

Comments

Complement of A048103; A129251(a(n)) > 0; A051674 is a subsequence; A129254 = (terms a(k) such that a(k+1)=a(k)+1). - Reinhard Zumkeller, Apr 07 2007
A027748(a(n),k) <= A124010(a(n),k) for some k<=A001221(a(n)). - Reinhard Zumkeller, Apr 28 2012

Examples

			54 is included because 3^3 divides 54.
		

Crossrefs

Complement: A048103.
Positions of nonzeros in A129251.
Cf. A054744.
Cf. A051674 (a subsequence).
Subsequence of A276079 from which it differs for the first time at n=175, where a(175) = 628, while A276079(175) = 625, a value missing from here.

Programs

  • Haskell
    a100716 n = a100716_list !! (n-1)
    a100716_list = filter (\x -> or $
       zipWith (<=) (a027748_row x) (map toInteger $ a124010_row x)) [1..]
    -- Reinhard Zumkeller, Apr 28 2012
    (Scheme, with Antti Karttunen's IntSeq-library)
    (define A100716 (NONZERO-POS 1 1 A129251))
    ;; Antti Karttunen, Aug 18 2016
    
  • Mathematica
    fQ[n_] := Union[ Table[ #[[1]] <= #[[2]]] & /@ FactorInteger[n]][[ -1]] == True; Select[ Range[2, 215], fQ[ # ] &] (* Robert G. Wilson v, Dec 14 2004 *)
    f[n_] := Module[{aux=FactorInteger[n]}, Last@Union@Table[aux[[i,1]] <=  aux[[i,2]], {i,Length[aux]}] == True]; Select[Range[2,215], f] (* José María Grau Ribas, Jan 25 2012 *)
    Rest@ Select[Range@ 216, Times @@ Boole@ Map[First@ # > Last@ # &, FactorInteger@ #] == 0 &] (* Michael De Vlieger, Aug 19 2016 *)
  • PARI
    is(n)=forprime(p=2,default(primelimit),if(n%p^p==0,return(1));if(p^p>n,return(0))) \\ Charles R Greathouse IV, Jan 24 2012
    
  • Python
    from itertools import count, islice
    from sympy import factorint
    def A100716_gen(startvalue=1): # generator of terms >= startvalue
        return filter(lambda n:any(map(lambda d:d[1]>=d[0],factorint(n).items())),count(max(startvalue,1)))
    A100716_list = list(islice(A100716_gen(),30)) # Chai Wah Wu, Jan 05 2023

Formula

a(n) ~ k*n with k = 1/(1 - Product(1 - p^-p)) = 3.5969959469... where the product is over all primes p. - Charles R Greathouse IV, Jan 24 2012

Extensions

More terms from Robert G. Wilson v, Dec 14 2004

A376469 Starts of runs of 3 consecutive integers in which each member of the run has at least one divisor of the form p^e with p <= e, where p is a prime.

Original entry on oeis.org

71874, 109375, 156248, 181250, 228123, 265624, 409374, 446875, 493748, 518750, 565623, 603124, 746874, 784375, 831248, 856250, 903123, 940624, 1084374, 1121875, 1168748, 1193750, 1240623, 1278124, 1421874, 1459375, 1506248, 1531250, 1578123, 1615624, 1759374, 1796875
Offset: 1

Views

Author

Amiram Eldar, Sep 23 2024

Keywords

Comments

The start of the least run of 4 (and also 5) consecutive integers with this property is 3988418748.
The numbers of terms that do not exceed 10^k, for k = 5, 6, ..., are 1, 18, 178, 1783, 17845, 178458, ... . Apparently, the asymptotic density of this sequence exists and equals 0.00001784... .

Examples

			71874 = 2 * 3^3 * 11^3 is a term since it is divisible by 3^3, 71875 = 5^5 * 23 is divisible by 5^5, and 71876 = 2^2 * 7 * 17 * 151 is divisible by 2^2.
		

Crossrefs

Subsequence of A100716, A070258 and A129254.

Programs

  • Mathematica
    q[n_] := q[n] = AnyTrue[FactorInteger[n], First[#] <= Last[#] &]; Select[Range[2*10^6], q[#] && q[#+1] && q[#+2] &]
  • PARI
    is(n) = {my(f = factor(n)); for(i = 1, #f~, if(f[i,1] <= f[i,2], return(1))); 0;}
    lista(kmax) = {my(is1 = 0, is2 = 0, is3); for(k = 3, kmax, is3 = is(k); if(is1 && is2 && is3, print1(k-2, ", ")); is1 = is2; is2 = is3);}
Showing 1-2 of 2 results.