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 1-4 of 4 results.

A059958 Smallest number m such that m*(m+1) has at least n distinct prime factors.

Original entry on oeis.org

1, 2, 5, 14, 65, 209, 714, 7314, 38570, 254540, 728364, 11243154, 58524465, 812646120, 5163068910, 58720148850, 555409903685, 4339149420605, 69322940121435, 490005293940084, 5819629108725509, 76622240600506314
Offset: 1

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Author

Labos Elemer, Mar 02 2001

Keywords

Comments

The original definition left unclear whether "at least" or "exactly" n prime factors are required. Now the "at least" variant was chosen, for the other variant ("exactly"), see A069354: At least up to a(18), both criteria yield the same number, and therefore a(n) = A069354(n) - 1, since m and m+1 are always coprime. - M. F. Hasler, Jan 15 2014
10^13 < a(19) <= 69322940121435. - Giovanni Resta, Mar 24 2020
Terms a(1)-a(10) appear in Erdős and Nicolas (1978-1979). - Amiram Eldar, Jun 24 2023

Examples

			For n = 9, a(9)*(a(9) + 1) = 38570*38571 = (2*5*7*19*29)*(3*13*23*43) with 9 distinct prime factors.
		

Crossrefs

Programs

  • Mathematica
    With[{s = Map[PrimeNu[Times @@ #] &, Partition[Range[10^6], 2, 1]]}, Array[FirstPosition[s, n_/; n>=#][[1]] &, Max@ s]] (* Michael De Vlieger, Nov 02 2017 *)
  • PARI
    a(n) = my(m=1); while(omega(m*(m+1)) < n, m++); m; \\ Michel Marcus, Jul 09 2018

Formula

a(n) = Min_{ m | A001221(m*(m+1)) >= n }.
a(n) <= A002110(n) - 1 because A001221((q-1)*q) >= n+1 for q = A002110(n).
Conjecture: a(n) = A069354(n) - 1. - Robert G. Wilson v, Feb 18 2014

Extensions

More terms from William Rex Marshall, Mar 18 2001
Offset corrected and a(15)-a(16) from Donovan Johnson, Jan 31 2009
a(17) from Donovan Johnson, Sep 15 2010
a(18) from Don Reble, Jan 15 2014
Edited by M. F. Hasler, Jan 15 2014
a(19)-a(20) from Michael S. Branicky, Feb 08 2023
a(21) from Michael S. Branicky, Feb 10 2023
a(22) from Michael S. Branicky, Feb 23 2023

A359960 Smallest Niven (or Harshad) number (A005349) with exactly n distinct prime factors.

Original entry on oeis.org

1, 2, 6, 30, 210, 2310, 30030, 690690, 14804790, 223092870, 8254436190, 200560490130, 8222980095330, 304250263527210, 13082761331670030, 614889782588491410, 32589158477190044730, 1987938667108592728530, 117288381359406970983270, 7858321551080267055879090
Offset: 0

Views

Author

Bernard Schott, Jan 20 2023

Keywords

Comments

a(11) = 200560490130; a(13) = 304250263527210.
a(n) >= A002110(n) = prime(n)#.
Many terms are primorial numbers, see A360011.

Examples

			2310 = 2*3*5*7*11 is the smallest integer with 5 prime factors because it is a primorial number, as 2310 / (2+3+1+0) = 385, 2310 is a Niven number: a(5) = 2310.
		

Crossrefs

Similar: A060319 (Fibonacci), A083002 (oblong), A359961 (Zuckerman).

Programs

  • PARI
    a(n) = my(k=1); while ((k % sumdigits(k)) || (omega(k) != n), k++); k; \\ Michel Marcus, Jan 20 2023
    
  • PARI
    omega_niven(A, B, n) = A=max(A, vecprod(primes(n))); (f(m, p, j) = my(list=List()); forprime(q=p, sqrtnint(B\m, j), my(v=m*q, r=nextprime(q+1)); while(v <= B, if(j==1, if(v>=A && v%sumdigits(v) == 0, listput(list, v)), if(v*r <= B, list=concat(list, f(v, r, j-1)))); v *= q)); list); vecsort(Vec(f(1, 2, n)));
    a(n) = if(n==0, return(1)); my(x=vecprod(primes(n)), y=2*x); while(1, my(v=omega_niven(x, y, n)); if(#v >= 1, return(v[1])); x=y+1; y=2*x); \\ Daniel Suteu, Jan 22 2023

Extensions

a(8)-a(9) from Michel Marcus, Jan 20 2023
a(10)-a(19) from Daniel Suteu, Jan 22 2023

A359961 Smallest Zuckerman number (A007602) with exactly n distinct prime factors.

Original entry on oeis.org

1, 2, 6, 132, 3276, 27132, 1117116, 111914712, 6111417312, 1113117121116, 1112712811322112, 11171121131111172
Offset: 0

Views

Author

Bernard Schott, Jan 21 2023

Keywords

Examples

			3276 = 2^2*3^2*7*13 is the smallest integer with 4 distinct prime factors that is also Zuckerman number as 3276 / (3*2*7*6) = 13, so a(4) = 3276.
		

Crossrefs

Similar: A060319 (Fibonacci), A083002 (oblong), A359960 (Niven).

Programs

  • PARI
    a(n) = my(k=1); while (!(p=vecprod(digits(k))) || (k % p) || (omega(k) != n), k++); k; \\ Michel Marcus, Jan 21 2023

Extensions

a(6)-a(7) from Michel Marcus, Jan 21 2023
a(8)-a(9) from Daniel Suteu, Jan 21 2023
a(10)-a(11) from Bert Dobbelaere, Jan 29 2023

A360301 Smallest exclusionary square (A029783) with exactly n distinct prime factors.

Original entry on oeis.org

2, 18, 84, 858, 31122, 3383898, 188841114, 68588585868, 440400004044, 7722272777722272
Offset: 1

Views

Author

Bernard Schott, Feb 02 2023

Keywords

Comments

There is no 5 in the prime factorization of these terms.
No other terms less than 10^14. - Michael S. Branicky, Feb 02 2023
1.69 * 10^15 < a(10) <= 7722272777722272. - Daniel Suteu, Feb 05 2023

Examples

			84 = 2^2 * 3 * 7 is the smallest integer with 3 distinct prime factors that is also an exclusionary square, because 84^2 = 7056, so a(3) = 84.
858 = 2 * 3 * 11 * 13 is the smallest integer with 4 distinct prime factors that is also an exclusionary square, because 858^2 = 736164, so a(4) = 858.
		

References

  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 60.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Revised Edition, 1997, page 144, entry 567.

Crossrefs

Cf. A029783.
Similar: A060319 (Fibonacci), A083002 (oblong), A359960 (Niven), A359961 (Zuckerman).

Programs

  • PARI
    omega_exclusionary_squares(A, B, n) = A=max(A, vecprod(primes(n))); (f(m, p, j) = my(list=List()); forprime(q=p, sqrtnint(B\m, j), if(q == 5, next); my(v=m*q); while(v <= B, if(j==1, if(v>=A && #setintersect(Set(digits(v)), Set(digits(v^2))) == 0, listput(list, v)), if(v*(q+1) <= B, list=concat(list, f(v, q+1, j-1)))); v *= q)); list); vecsort(Vec(f(1, 2, n)));
    a(n) = my(x=vecprod(primes(n)), y=2*x); while(1, my(v=omega_exclusionary_squares(x, y, n)); if(#v >= 1, return(v[1])); x=y+1; y=2*x); \\ Daniel Suteu, Feb 05 2023

Formula

Assuming a(n) exists, a(n) >= A002110(n+1)/5 >> exp((1 + o(1)) * n * log(n)). (The inequality is presumably strict for all n; for n > 34 it seems that all A002110(n) are pandigital.) - Charles R Greathouse IV, Feb 05 2023

Extensions

a(4)-a(7) from Amiram Eldar, Feb 02 2023
a(8)-a(9) from Michael S. Branicky, Feb 02 2023
a(10) from Michael S. Branicky, Feb 07 2023
Showing 1-4 of 4 results.