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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A357300 a(n) is the smallest number m with exactly n divisors whose first digit equals the first digit of m.

Original entry on oeis.org

1, 10, 100, 108, 120, 180, 1040, 1020, 1170, 1008, 1260, 1680, 10010, 10530, 10200, 10260, 10560, 10800, 11340, 10920, 12600, 10080, 15840, 18480, 15120, 102060, 104400, 101640, 100320, 102600, 100980, 117600, 114660, 107100, 174240, 113400, 105840, 100800, 120120, 143640
Offset: 1

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Author

Bernard Schott, Sep 23 2022

Keywords

Comments

a(m) <= a(551) = 18681062400 for m < 555. All terms with values up to 2*10^10 start with 1. Do there exist a(n) starting with any other digit? - Charles R Greathouse IV, Sep 25 2022

Examples

			Of the twelve divisors of 108, four have their first digit equals to the first digit of 108: 1, 12, 18 and 108, and there is no such smaller number, hence a(4) = 108.
		

Crossrefs

Cf. A335491 (with last digit), A206287, A355592, A357299.
Similar, but with: A333456 (Niven numbers), A335038 (Zuckerman numbers).

Programs

  • Mathematica
    f[n_] := IntegerDigits[n][[1]]; s[n_] := Module[{fn = f[n]}, DivisorSum[n, 1 &, f[#] == fn &]]; seq[len_, nmax_] := Module[{v = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = s[n]; If[i <= len && v[[i]] == 0, c++; v[[i]] = n]; n++]; v]; seq[40, 10^6] (* Amiram Eldar, Sep 23 2022 *)
  • PARI
    f(n) = my(fd=digits(n)[1]); sumdiv(n, d, digits(d)[1] == fd); \\ A357299
    a(n) = my(k=1); while (f(k)!=n, k++); k; \\ Michel Marcus, Sep 23 2022
    
  • PARI
    v=vector(1000); v[1]=r=1; forfactored(n=2, 10^11, t=a(n[1],n[2],r); if(t>r && v[t]==0, v[t]=n[1]; print(t" "n[1]" = "n[2]); while(v[r],r++); r--)) \\ Charles R Greathouse IV, Sep 25 2022

Extensions

More terms from Michel Marcus, Sep 23 2022

A335491 a(n) is the smallest number m with exactly n divisors whose last digit equals the last digit of m.

Original entry on oeis.org

1, 11, 40, 60, 160, 120, 640, 240, 360, 480, 8064, 600, 18144, 1920, 1440, 1200, 72576, 1800, 52416, 2400, 5760, 30720, 183456, 3600, 12960, 122880, 9000, 9600, 602784, 7200, 445536, 8400, 92160, 798336, 51840, 12600, 2159136, 576576, 368640, 16800, 2935296, 28800
Offset: 1

Views

Author

Bernard Schott, Jun 11 2020

Keywords

Comments

a(n) exists for any n >= 1. Indeed, the number 5*2^n (see A020714), n >= 1, has exactly n divisors (5*2^1, 5*2^2, ..., 5*2^n), with the last digit 0. - Marius A. Burtea, Jun 12 2020
It's always the case when a(n) ends in 0 then a(n) = 10 * A005179(n). Proof: Let v be the list of divisors of a(n) that end in 0. We then have |v| = n and lcm(v) = a(n) as a(n) is in v and all other terms in v divide a(n). We then have lcm(v)/10 = a(n)/10 where a(n)/10 has exactly n divisors. The least positive integer that has exactly n divisors is A005179(n). - Bernard Schott and David A. Corneth, Jun 12 2020
For some p_i^e_i||a(n) and p_j^e_j||a(n) where p_i and p_j are primes such that p_i < p_j, 10 | (p_j - p_i) and t^k||u denotes t^k|u but t^(k + 1) doesn't divide u i.e. gcd(t, u/t^k) = 1 denotes then e_i >= e_j. For example k * 11^2 * 31^3 where gcd(k, 11*31) = 1 can't be a term as the multiplicity of 11 is less than the multiplicity of 31. - David A. Corneth, Jun 12 2020

Examples

			Of the twelve divisors of 60, four have their last digit equals to the last digit of 60: 10, 20, 30 and 60, and there is no smaller number k with four divisors whose last digit equals the last digit of k, hence a(4) = 60.
		

Crossrefs

Similar with: A333456 (Niven numbers), A335038 (Zuckerman numbers).

Programs

  • Magma
    a:=[]; for n in [1..30] do k:=1; while #[d:d in Divisors(k)|k mod 10 eq d mod 10] ne n do k:=k+1; end while; Append(~a,k); end for; a; // Marius A. Burtea, Jun 12 2020
  • Mathematica
    d[n_] := DivisorSum[n, 1 &, Mod[# - n, 10] == 0 &]; mx = 20; c = 0; n = 1; s = Table[0, {mx}]; While[c < mx, i = d[n]; If[i <= mx && s[[i]] == 0, c++; s[[i]] = n]; n++]; s (* Amiram Eldar, Jun 12 2020 *)
  • PARI
    f(n) = my(u=n%10); sumdiv(n, d, (d%10) == u);
    a(n) = my(k=1); while(f(k) != n, k++); k; \\ Michel Marcus, Jun 12 2020
    

Extensions

Corrected and extended by Marius A. Burtea, Jun 12 2020

A333457 a(n) is the smallest number with exactly n divisors that are Moran numbers, or -1 if no such number exists.

Original entry on oeis.org

18, 42, 84, 126, 252, 756, 1998, 1596, 2394, 4662, 4788, 9324, 18648, 23940, 46620, 93240, 139860, 177156, 559440, 354312, 708624, 1062936, 885780, 4606056, 1771560, 3543120, 5314680, 10629360, 38974320, 23030280, 46060560, 69090840, 138181680, 506666160
Offset: 1

Views

Author

Marius A. Burtea, May 03 2020

Keywords

Comments

m is a Moran number if (m / sum of digits of m) is prime (A001101).
Conjecture: For every n there is at least one number k with n divisors Moran numbers.
Conjecture: The terms are divisible by 6.
a(1) = 18, a(2) = 42 and a(3) = 84 are Moran numbers. Not all terms in the sequence are Moran numbers. For example: a(4) = 126 has digsum(126) = 9 and 126 / 9 = 14. Also, the terms a(5) - a(34) are not Moran numbers.

Examples

			Of the divisors of 18 (1, 2, 3, 6, 9, 18), only 18 is a Moran number: 18 / digsum (18) = 2.
Of the divisors of 84 (1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84), only 21, 42 and 84 are Moran numbers: 21 / digsum (21) = 7, 42 / digsum (42) = 7 and 84 / digsum (84) = 7.
		

Crossrefs

Cf. A001101, A007953 (sum of digits), A333456.

Programs

  • Magma
    a:=[]; for n in [1..20] do m:=1; while #[d:d in Divisors(m)|d mod &+Intseq(d) eq 0 and IsPrime(d div &+Intseq(d))] ne n do m:=m+1; end while; Append(~a,m); end for; a;
  • Mathematica
    numDiv[n_] := DivisorSum[n, 1 &, PrimeQ[#/Plus @@ IntegerDigits[#]] &]; a[n_] := Module[{k = 1}, While[numDiv[k] != n, k++]; k]; Array[a, 20] (* Amiram Eldar, May 11 2020 *)
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