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 1-5 of 5 results.

A351913 Least k such that A352483(k) = n, or -1 if no such k exists.

Original entry on oeis.org

3, 9, 5, 204, 7, 876, 20, 140, 11, 492, 13, 776, 32, 904, 17, 441, 19, 23364, 44, 2178, 23, 25, 27, 1544, 216, 3756, 29, 460, 31, 1928, 35, 2056, 280, 1644, 37, 5196, 117, 162, 41, 1089, 43, 2696, 92, 2824, 47, 49, 51, 6924, 153, 812, 53, 7524, 57, 3464, 116, 1521, 59, 940, 61
Offset: 1

Views

Author

Michel Marcus, Mar 18 2022

Keywords

Comments

What is the value of a(102)?
Conjecture: a(102) and all "Unknown" values in the a-file equal -1. - Paolo Xausa, Aug 16 2022

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{k = 3}, While[Denominator[k*(d = DivisorSigma[0, k])/(k - d)] != n, k++]; k]; Array[a, 60] (* Amiram Eldar, Mar 18 2022 *)
  • PARI
    f(n) = my(d=numdiv(n)); denominator(n*d/(n-d)); \\ A352483
    a(n) = {my(k=3); while (f(k) != n, k++); k;}
    
  • Python
    from math import gcd
    from sympy import divisor_count
    from itertools import count, islice
    def f(n): d = divisor_count(n); g = gcd(n-d, n*d); return (n-d)//g
    def agen():
        n, adict = 1, dict()
        for k in count(1):
            fk = f(k)
            if fk not in adict: adict[fk] = k
            while n in adict: yield adict[n]; n += 1
    print(list(islice(agen(), 60))) # Michael S. Branicky, Jul 23 2022

Formula

a(n) = n+2 iff n > 0 is a term of A040976. - Bernard Schott, Mar 24 2022

A352710 Least k such that A352483(k) = 2*n-1, or -1 if no such k exists.

Original entry on oeis.org

3, 5, 7, 20, 11, 13, 32, 17, 19, 44, 23, 27, 216, 29, 31, 35, 280, 37, 117, 41, 43, 92, 47, 51, 153, 53, 57, 116, 59, 61, 65, 520, 67, 207, 71, 73, 77, 616, 79, 164, 83, 87, 261, 89, 93, 95, 760, 97, 105, 101, 103, 212, 107, 109, 333, 113, 920, 119, 952, 123, 125, 1000, 127, 135, 131
Offset: 1

Views

Author

Michel Marcus, Mar 30 2022

Keywords

Crossrefs

Cf. A352483.
Bisection of A351913.

Programs

  • PARI
    f(n) = my(d=numdiv(n)); denominator(n*d/(n-d)); \\ A352483
    a(n) = {my(k=3, q=2*n-1); while (f(k) != q, k++); k; }

Formula

a(n) = A351913(2*n-1).

A146566 Numbers k such that k*sigma_0(k) is divisible by (k - sigma_0(k)).

Original entry on oeis.org

3, 4, 6, 8, 12, 18, 24, 36, 40, 60, 84, 156, 180, 600
Offset: 1

Views

Author

Ctibor O. Zizka, Nov 01 2008

Keywords

Comments

No other term < 10000000. - Michel Marcus, Jun 02 2013
No other term multiple of 12 below 10^9. - M. F. Hasler, Apr 16 2022

Crossrefs

Programs

  • Mathematica
    nsiQ[n_]:=Module[{s=DivisorSigma[0,n]},IntegerQ[(n*s)/(n-s)]]; Select[ Range[3,600],nsiQ] (* Harvey P. Dale, Dec 05 2019 *)
  • PARI
    isok(n) = {my(nd = numdiv(n)); type(n*nd/(n-nd)) == "t_INT"}  \\ Michel Marcus, Jun 02 2013
    
  • PARI
    is_A146566(n,d=numdiv(n))={n*d%(n-d)==0} \\ M. F. Hasler, Apr 16 2022

Formula

A352483(a(n)) = 1. - Bernard Schott, Mar 23 2022

Extensions

Corrected and extended by Michel Marcus, Jun 02 2013

A352482 Denominator of (n-d)/n*d where d = A000005(n) is the number of divisors of n.

Original entry on oeis.org

1, 1, 6, 12, 10, 12, 14, 8, 9, 20, 22, 12, 26, 28, 60, 80, 34, 9, 38, 60, 84, 44, 46, 12, 75, 52, 108, 84, 58, 120, 62, 96, 132, 68, 140, 12, 74, 76, 156, 10, 82, 168, 86, 132, 90, 92, 94, 240, 147, 75, 204, 156, 106, 216, 220, 28, 228, 116, 118, 15, 122, 124, 126, 448, 260, 264, 134
Offset: 1

Views

Author

Michel Marcus, Mar 18 2022

Keywords

Comments

The terms are of course the denominators of the fraction "in smallest terms", otherwise said: a(n) = n*d/gcd(n*d, n - d), which is unambiguous also for n = 1 and n = 2 where n - d = 0.

Examples

			The number n = 1 has d = 1 divisors, so (n-d)/(n*d) = 0/1 has denominator a(1) = 1.
The number n = 2 has d = 2 divisors, so (n-d)/(n*d) = 0/4 = 0/1 has denominator a(2) =  1 when written in smallest terms.
The number n = 3 has d = 2 divisors, so (n-d)/(n*d) = 1/6 has denominator a(3) =  6.
The number n = 4 has d = 3 divisors, so (n-d)/(n*d) = 1/12 has denominator a(4) = 12.
The number n = 6 has d = 4 divisors, so (n-d)/(n*d) = 2/24 = 1/12 has denominator a(6) = 12.
		

Crossrefs

Cf. A000005, A049820, A146566, A352483 (numerator).

Programs

  • Mathematica
    a[n_] := Numerator[n*(d = DivisorSigma[0, n])/(n - d)]; Array[a, 100, 3] (* Amiram Eldar, Mar 18 2022 *)
  • PARI
    A352482(n,d=numdiv(n))=denominator((n-d)/(n*d))

Extensions

Edited and extended to offset 1 by M. F. Hasler, Apr 17 2022

A353012 Numbers N such that gcd(N - d, N*d) >= d^2, where d = A000005(N) is the number of divisors of N.

Original entry on oeis.org

1, 2, 136, 156, 328, 444, 584, 600, 712, 732, 776, 876, 904, 1096, 1164, 1176, 1308, 1544, 1864, 1884, 1928, 2056, 2172, 2248, 2316, 2504, 2601, 2696, 2748, 2824, 2892, 2904, 3208, 3240, 3249, 3272, 3324, 3464, 3592, 3656, 3756, 4044, 4056, 4168, 4188, 4476, 4552, 4616
Offset: 1

Views

Author

M. F. Hasler, Apr 15 2022

Keywords

Comments

As d^2 | N-d we have N = k*d^2 + d for some k >= 0 and d > 1. So gcd(k*d^2 + d - d, (N*d^2 + d)*d) = gcd(k*d^2, k*d^3 + d^2) = gcd(k*d^2, d^2) = d^2. So for any N such that d^2 | gcd(N - d, N*d) we have gcd(N - d, N*d) = d^2. - David A. Corneth, Apr 20 2022
Since gcd(N - d, N*d) is never larger than d^2 (if N = n*g, d = f*g with gcd(n,f) = 1, then gcd(N - d, N*d) = g*gcd(n-f,n*f*g) = g*gcd(n-f, f*f*g) <= g*g, since by assumption, no factor of f divides n), so one can also replace "=" by ">=" in the definition.

Examples

			N = 1 is in the sequence because d(N) = 1, gcd(1 - 1, 1*1) = 1 = d^2.
N = 2 is in the sequence because d(N) = 2, gcd(2 - 2, 2*2) = 4 = d^2.
N = 136 = 8*17 is in the sequence because d(N) = 4*2 = 8, gcd(8*17 - 8, 8*17*8) = gcd(8*16, 8*8*17) = 8*8 = d^2. Similarly for N = 8*p with any prime p = 8*k + 1.
N = 156 = 2^2*3*13 is in the sequence because d(n) = 3*2*2 = 12, gcd(12*13 - 12, 12*13*12) = gcd(12*12, 12*12*13) = 12*12 = d^2. Similarly for any N = 12*p with prime p = 12*k + 1.
More generally, when N = m*p^k with p^k == 1 (mod m) and m = (k+1)*d(m), then d(N) = d(m)*(k+1) = m and gcd(n - d, n*d) = gcd(m*p^k - m, m*p^k*m) = m*gcd(p^k - 1, p^k*m) = m^2. This holds for m = 8 and 12 with k = 1, for m = 9, 18 and 24 with k = 2, etc: see sequence A033950 for the m-values.
		

Crossrefs

Cf. A000005 (number of divisors), A352483 (numerator of (n-d)/(n*d)), A352482 (denominator), A049820 (n - d), A146566 (n*d is divisible by n-d), A033950 (refactorable or tau numbers: d(n) | n, supersequence of this).

Programs

  • Mathematica
    Select[Range[4650], GCD[#1 - #2, #1 #2] == #2^2 & @@ {#, DivisorSigma[0, #]} &] (* Michael De Vlieger, Apr 21 2022 *)
  • PARI
    select( {is(n, d=numdiv(n))=gcd(n-d,d^2)==d^2}, [1..10^4])

Formula

For all m in A033950, the sequence contains all numbers m*p^k with k = m/d(m) - 1, and p^k == 1 (mod m), in particular 8*A007519 and 12*A068228 (k = 1, m = 8 and 12), 9*A129805^2, 18*A129805^2 and 24*A215848^2 (k = 2, m = 9, 18 and 24, A^2 = {x^2, x in A}), etc.
Showing 1-5 of 5 results.