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 1-3 of 3 results.

A340159 a(n) is the smallest number m such that numbers m, m + 1, m + 2, ..., m + n - 1 have k, 2*k, 3*k, ..., n*k divisors respectively.

Original entry on oeis.org

1, 1, 61, 421, 211082, 11238341, 16788951482, 41126483642
Offset: 1

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Author

Jaroslav Krizek, Dec 29 2020

Keywords

Comments

a(n) is the smallest number m such that tau(m) = tau(m + 1)/2 = tau(m + 2)/3 = tau(m + 3)/4 = ... = tau(m + n - 1)/n, where tau(k) = the number of divisors of k (A000005).
Corresponding values of tau(a(n)): 1, 1, 2, 2, 4, 4, 4, ...
a(8) <= 41126483642. - David A. Corneth, Dec 31 2020
Any subsequent terms are > 10^11. - Lucas A. Brown, Mar 18 2024

Examples

			a(3) = 61 because 61, 62 and 63 have 2, 4, and 6 divisors respectively and there is no smaller number having this property.
		

Crossrefs

Cf. A294528 for similar sequence with primes.

Programs

  • PARI
    isok(m, n) = {my(k=numdiv(m)); for (i=1, n-1, if (numdiv(m+i) != (i+1)*k, return (0));); return(1);}
    a(n) = my(m=1); while(!isok(m, n), m++); m; \\ Michel Marcus, Dec 30 2020
    
  • Python
    # see LINKS

Extensions

a(7) from Jinyuan Wang, Dec 31 2020
a(8) from Lucas A. Brown, Mar 18 2024

A340158 Numbers m such that m, m + 1, m + 2, m + 3 and m + 4 have k, 2k, 3k, 4k and 5k divisors respectively.

Original entry on oeis.org

211082, 2364062, 2774165, 3379802, 3743573, 4390682, 5651042, 5845442, 6708578, 7326122, 7371482, 8566394, 8839202, 9056282, 10154642, 10301333, 10325621, 10446242, 10540202, 11238341, 11719562, 11978762, 12377282, 12871058, 13456202, 16840058, 16954562, 17155141
Offset: 1

Views

Author

Jaroslav Krizek, Dec 29 2020

Keywords

Comments

Numbers m such that tau(m) = tau(m + 1)/2 = tau(m + 2)/3 = tau(m + 3)/4 = tau(m + 4)/5, where tau(k) = the number of divisors of k (A000005).
Quintuples of [tau(a(n)), tau(a(n) + 1), tau(a(n) + 2), tau(a(n) + 3), tau(a(n) + 4)] = [tau(a(n)), 2*tau(a(n)), 3*tau(a(n)), 4*tau(a(n)), 5*tau(a(n))]: [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], [8, 16, 24, 32, 40], [4, 8, 12, 16, 20], [4, 8, 12, 16, 20], ...
Corresponding values of numbers k: 4, 4, 4, 8, 4, 4, 4, 4, 4, 8, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 4, 4, 4, 4, 4, 4, 4, ...
1524085621 is the smallest prime term (see A294528).
Subsequence of A063446, A339778 and A340157.

Examples

			tau(211082) = 4, tau(211083) = 8, tau(211084) = 12, tau(211085) = 16, tau(211086) = 20.
		

Crossrefs

Programs

  • Magma
    [m: m in [1..10^6] | #Divisors(m) eq #Divisors(m + 1)/2 and #Divisors(m) eq #Divisors(m + 2)/3 and #Divisors(m) eq #Divisors(m + 3)/4 and #Divisors(m) eq #Divisors(m + 4)/5]
    
  • Mathematica
    Select[Range[5*10^6], Equal @@ (DivisorSigma[0, # + {0, 1, 2, 3, 4}]/{1, 2, 3, 4, 5}) &] (* Amiram Eldar, Dec 30 2020 *)
  • PARI
    isok(m) = my(k = numdiv(m)); (numdiv(m+1) == 2*k) && (numdiv(m+2) == 3*k) && (numdiv(m+3) == 4*k) && (numdiv(m+4) == 5*k); \\ Michel Marcus, Jan 16 2021

A340871 Primes p such that p, p + 1, p + 2 and p + 3 have 2, 4, 6 and 8 divisors respectively.

Original entry on oeis.org

421, 30661, 50821, 54421, 130021, 195541, 423781, 635461, 1003381, 1577941, 1597381, 1883941, 2070421, 2100661, 2162581, 2534821, 2585941, 2666581, 2851621, 3296581, 3658021, 3800581, 4657381, 4969141, 5739541, 5962741, 6188821, 6537301, 6556741, 7090261
Offset: 1

Views

Author

Jaroslav Krizek, Jan 24 2021

Keywords

Comments

Primes from A340157.
Term 1524085621 is the smallest prime p such that p, p + 1, p + 2, p + 3 and p + 4 have 2, 4, 6, 8 and 10 divisors respectively. No such run exists for any 6 consecutive integers with prime p with this property (see A294528).

Examples

			tau(421) = 2, tau (422) = 4, tau (423) = 6, tau (424) = 8.
		

Crossrefs

Programs

  • Magma
    [m: m in [1..10^7] | IsPrime(m) and #Divisors(m + 1) eq 4 and #Divisors(m + 2) eq 6 and #Divisors(m + 3) eq 8];
  • Mathematica
    Select[Range[10^6], DivisorSigma[0, # + {0, 1, 2, 3}] == {2, 4, 6, 8} &] (* Amiram Eldar, Jan 25 2021 *)
    Select[Prime[Range[490000]],DivisorSigma[0,#+{1,2,3}]=={4,6,8}&] (* Harvey P. Dale, Oct 02 2021 *)
Showing 1-3 of 3 results.