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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A365966 Smallest prime factor of f(n) = 10^(2*n+1) + (10^n-1)/9.

Original entry on oeis.org

2, 7, 3, 11, 41, 3, 61, 7, 3, 11, 113, 3, 53, 7, 3, 11, 29, 3, 17, 7, 3, 11, 11111111111111111111111, 3, 41, 7, 3, 11, 53, 3, 661, 7, 3, 11, 17, 3, 2028119, 7, 3, 11, 83, 3, 173, 7, 3, 11, 40697, 3, 239, 7, 3, 11, 107, 3, 41, 7, 3, 11, 2836549, 3, 733, 7, 3, 11
Offset: 0

Views

Author

Jean-Marc Rebert, Sep 23 2023

Keywords

Comments

f(n) = 100..00011..11 is the least positive integer whose decimal digits are n+1 1's and n+1 0's.

Examples

			a(1) = 7, because the smallest prime factor of f(1) = 1001 = 7 * 11 * 13 is 7.
a(2) = 3, because the smallest prime factor of f(2) = 100011 = 3 * 17 * 37 * 43 is 3.

Links

Crossrefs

Cf. A020639, A365928, A365610, A067063.

Programs

  • Mathematica
    a[n_]:=Min[First/@FactorInteger[10^(2*n+1)+(10^n-1)/9]]; Array[a,64,0] (* Stefano Spezia, Sep 24 2023 *)
  • PARI
    a365966(n, limtd=10^9) = {my (x=10^(2*n+1)+(10^n-1)/9); forprime (p=2, limtd, if(x%p==0, return(p))); factor(x)[1,1]}; \\ Hugo Pfoertner, Nov 14 2023

Formula

a(n) = 3 iff n = 3k + 2, since f(n) is odd and has n+1 1 digits so that "casting out 9's" shows f(n) == n+1 (mod 3).
a(n) = 7 iff n = 6k + 1.
a(n) = 11 iff n = 6k + 3.

A365928 Smallest prime factor of f(n) = 10^(2*n) + (10^n - 1)/9.

Original entry on oeis.org

101, 3, 7, 7, 3, 317, 40637, 3, 7, 7, 3, 1487, 101, 3, 7, 7, 3, 39855301, 641, 3, 7, 7, 3, 162340676822011484150719, 101, 3, 7, 7, 3, 121068683, 47, 3, 7, 7, 3, 107, 71, 3, 7, 7, 3, 67, 695841737, 3, 7, 7, 3, 47, 101, 3, 7, 7, 3, 8933, 677, 3, 7, 7, 3, 10305833206337
Offset: 1

Views

Author

Jean-Marc Rebert, Sep 23 2023

Keywords

Examples

			a(1) = 101, because f(1) = 101 is prime.
a(2) = 3, because the smallest prime factor of f(2) = 10011 = 3 * 337 is 3.

Crossrefs

Cf. A020639, A067063, A365966, A102380.

Programs

  • Mathematica
    a[n_]:=Min[First/@FactorInteger[10^(2*n)+(10^n-1)/9]]; Array[a,59] (* Stefano Spezia, Sep 24 2023 *)
  • PARI
    a(n)=my(x=10^(2*n)+(10^n-1)/9);m=factor(x);return(m[1,1])
  • PARI
    a(n) = my(x=10^(2*n)+(10^n-1)/9, k=10); if (ispseudoprime(x), return(x)); while (1, m=factor(x, k); if (m[1,1]Michel Marcus, Sep 24 2023

Formula

a(3k + 2) = 3, a(6k + 3) = 7, a(6k + 4) = 7.

Extensions

a(60) from Jinyuan Wang, Sep 24 2023

A385579 Smallest prime factor that the repunit(n) = (10^n-1)/9 shares with at least one other binary vector of the same length in base 10, or 1 if they are coprime.

Original entry on oeis.org

1, 1, 1, 1, 11, 1, 3, 1, 11, 3, 11, 1, 3, 53, 11, 3, 11, 1, 3, 1, 11, 3, 11, 1, 3, 41, 11, 3, 11, 3191, 3, 2791, 11, 3, 11, 41, 3, 2028119, 11, 3, 11, 83, 3, 173, 11, 3, 11, 35121409, 3, 239, 11, 3, 11, 107, 3, 41, 11, 3, 11
Offset: 0

Views

Author

Dmytro Inosov, Jul 03 2025

Keywords

Comments

a(n) is the smallest prime factor that divides both the decimal repunit (10^n-1)/9 and at least one other smaller decimal number consisting of only 0's and 1's.
a(n)=1 iff n is a term in A385537 (indices of repunits coprime with all other binary vectors of the same length).

Examples

			a(3) = 1 because 111 = 3*37 is coprime with all other nonzero binary vectors of length 3, which are 001, 010, 011, 100, 101, 110. None of them is divisible by 3 or 37.
a(4) = 11 because 11 is the smallest prime factor of 1111 which it shares, for example, with the binary vector 0011.

Crossrefs

Cf. A067063, A003020, A378761, A385537, A385539.

Programs

  • Mathematica
    F[d_] := Min[Select[Table[Min[Transpose[FactorInteger[GCD[FromDigits[IntegerDigits[i,2]],(10^d-1)/9]]][[1]]], {i, 1, 2^d-2}], # > 1 &]];
Table[If[# < \[Infinity], #, 1] &[F[n]], {n, 0, 25}]
  • PARI
    a(n) = my(x=(10^n-1)/9, m=oo, b=0, z); for (i=1, 2^n-2, my(y=fromdigits(binary(i))); if ((z=gcd(y, x)) != 1, b=1; m = min(m, vecmin(factor(z)[,1])); ); ); if (b, m, 1); \\ Michel Marcus, Jul 03 2025

Formula

If a(n) > 1, A067063(n) <= a(n) <= A003020(n).
Previous Showing 11-13 of 13 results.

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