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.

A175304 A positive integer n is included if d(n+d(n)) = d(n), where d(n) is the number of divisors of n.

Original entry on oeis.org

3, 5, 6, 10, 11, 12, 17, 22, 29, 34, 35, 41, 44, 51, 58, 59, 60, 65, 70, 71, 72, 82, 84, 87, 91, 92, 96, 101, 102, 107, 111, 115, 118, 119, 125, 128, 129, 130, 137, 141, 142, 147, 149, 155, 174, 179, 182, 183, 191, 197, 201, 202, 205, 209, 213, 214, 215, 217, 222
Offset: 1

Views

Author

Leroy Quet, Mar 24 2010

Keywords

Comments

The sequence contains the smaller member of every pair of twin primes (A001359) and all squarefree semiprimes m such that m+4 is also a squarefree semiprime (A255746). Can one prove that this is an infinite sequence? - Vladimir Shevelev, Jul 11 2015
The sequence does not contain perfect squares. Indeed, let a(m)=k^2. Then d(k^2+d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (A046522) that d(k^2)<2*k. Hence, (k+1)^2 - k^2 > d(k^2). Thus k^2Vladimir Shevelev, Feb 10 2017
If p is prime and t+1 is odd prime, then p^t is not in the sequence. Indeed, if d(p^t+t+1)=t+1, then p^t+t+1=q^t, where q is prime > p (if p^t+t+1= say q^l*r^m, then (l+1)*(m+1)=t+1 which is impossible by the condition). But q>=p+2 and p^t+t+1>=p^t+2*t*p^(t-1) or t+1>=2*t*p^(t-1) which trivially has only solution t=1; however, by the condition t>=2. - Vladimir Shevelev, Feb 18 2017
If an odd integer k is in this sequence, so is 2k. - Charlie Neder, Jan 14 2019

Examples

			10 is in the sequence because d(10)=4 and d(10+d(10))=d(14)=4. - _Emeric Deutsch_, Apr 08 2010
		

Crossrefs

Positions of zeros in A286530.

Programs

  • Maple
    with(numtheory): a := proc (n) if tau(n+tau(n)) = tau(n) then n else end if end proc: seq(a(n), n = 1 .. 230); # Emeric Deutsch, Apr 08 2010
  • Mathematica
    Select[Range@ 224, Function[n, DivisorSigma[0, n + #] == # &@ DivisorSigma[0, n]]](* Michael De Vlieger, Sep 27 2015 *)
    Position[#, 0][[All, 1]] &@ Table[DivisorSigma[0, n + DivisorSigma[0, n]] - DivisorSigma[0, n], {n, 222}] (* Michael De Vlieger, May 21 2017 *)
  • PARI
    is(n)=numdiv(n+n=numdiv(n))==n \\ M. F. Hasler, Sep 27 2015

Extensions

More terms from Emeric Deutsch, Apr 08 2010

A286480 Compound filter (prime signature of n & prime signature of n+d(n)): a(n) = P(A046523(n), A286479(n)), where P(n,k) is sequence A000027 used as a pairing function and d(n) is number of divisors of n (A000005).

Original entry on oeis.org

2, 12, 5, 14, 5, 61, 12, 179, 109, 61, 5, 265, 23, 142, 27, 226, 5, 607, 23, 148, 42, 61, 12, 1509, 109, 601, 44, 148, 5, 625, 23, 698, 27, 61, 61, 1117, 23, 601, 27, 2509, 5, 850, 80, 265, 148, 142, 12, 1426, 109, 607, 61, 148, 23, 430, 27, 3765, 27, 61, 5, 8575, 80, 601, 148, 2144, 61, 625, 23, 148, 27, 1741, 5, 8587, 80, 601, 363, 148, 216, 625, 138, 5719
Offset: 1

Views

Author

Antti Karttunen, May 21 2017

Keywords

Crossrefs

Programs

  • Mathematica
    f[n_] := If[n == 1, 1, Times @@ MapIndexed[Prime[First[#2]]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]]]; Table[(2 + (#1 + #2)^2 - #1 - 3 #2)/2 & @@ {f@ n, f[n + DivisorSigma[0, n]]}, {n, 80}] (* Michael De Vlieger, May 21 2017 *)
  • PARI
    A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); };  \\ This function from Charles R Greathouse IV, Aug 17 2011
    A286479(n) = A046523(n+numdiv(n));
    A286480(n) = (1/2)*(2 + ((A046523(n)+A286479(n))^2) - A046523(n) - 3*A286479(n));
    
  • Python
    from sympy import factorint, divisor_count
    def T(n, m): return ((n + m)**2 - n - 3*m + 2)/2
    def P(n):
        f = factorint(n)
        return sorted([f[i] for i in f])
    def a046523(n):
        x=1
        while True:
            if P(n) == P(x): return x
            else: x+=1
    def a286479(n): return a046523(n + divisor_count(n))
    def a(n): return T(a046523(n), a286479(n)) # Indranil Ghosh, May 21 2017
  • Scheme
    (define (A286480 n) (* (/ 1 2) (+ (expt (+ (A046523 n) (A286479 n)) 2) (- (A046523 n)) (- (* 3 (A286479 n))) 2)))
    

Formula

a(n) = (1/2)*(2 + ((A046523(n)+A286479(n))^2) - A046523(n) - 3*A286479(n)).

A282354 Positive j such that d(j) = d(j + 2*d(j)), where d(j) is the number of divisors of j.

Original entry on oeis.org

3, 6, 7, 13, 14, 19, 20, 24, 26, 27, 32, 37, 38, 40, 43, 54, 57, 60, 63, 67, 69, 72, 74, 77, 79, 84, 85, 86, 87, 88, 97, 103, 108, 109, 111, 114, 115, 125, 126, 127, 132, 133, 134, 136, 138, 154, 158, 163, 170, 174, 177, 193, 194, 200, 201, 204, 205, 206, 209
Offset: 1

Views

Author

Vladimir Shevelev, Feb 13 2017

Keywords

Comments

The sequence contains the smaller member of every pair of cousin primes (A023200).
The sequence contains no perfect squares. Indeed, let a(m) = k^2 for some m. Then, by the definition, d(k^2 + 2*d(k^2)) = d(k^2). Note that d(k^2) is odd. On the other hand, it is known (cf. A046522) that d(k^2) < 2*k. Hence (k+2)^2 - k^2 = 4*k + 4 > 2*d(k^2). Thus k^2 < k^2 + 2*d(k^2) < (k+2)^2. Since, evidently, k^2 + 2*d(k^2) cannot be (k+1)^2, then k^2 + 2*d(k^2) cannot be a square. Therefore, d(k^2 + 2*d(k^2)) is even, which is a contradiction.

Crossrefs

Programs

  • Mathematica
    Select[Range@ 210, Function[d, DivisorSigma[0, # + 2 d] == d]@ DivisorSigma[0, #] &] (* Michael De Vlieger, Feb 13 2017 *)
  • PARI
    is(n)=my(d=numdiv(n)); d==numdiv(n+2*d) \\ Charles R Greathouse IV, Feb 14 2017

Extensions

More terms from Peter J. C. Moses, Feb 13 2017
Showing 1-3 of 3 results.