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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A340701 Upper of a pair of adjacent perfect powers, both with exponents > 2.

Original entry on oeis.org

32, 81, 128, 256, 1024, 1331, 2197, 50653, 59319, 195112, 279936, 456976, 614656, 3112136, 6436343, 22667121, 25412184, 38958219, 62748517, 96071912, 131096512, 418195493, 506261573, 741217625, 796597983, 1249243533, 2136750625, 2217373921, 5554637011, 5802888573
Offset: 1

Views

Author

Hugo Pfoertner, Jan 16 2021

Keywords

Examples

			See A340700.
		

Crossrefs

The corresponding lower members of the pairs are A340700.

Formula

a(n) = A340703(n)^A340705(n) = A340700(n) + A340706(n).

A340702 Root of the lower member A340700 of a pair of adjacent perfect powers, both with exponents > 2.

Original entry on oeis.org

3, 2, 5, 3, 10, 6, 3, 15, 3, 21, 23, 77, 85, 42, 186, 283, 71, 79, 89, 99, 107, 143, 150, 165, 168, 188, 1288, 1304, 273, 276, 1858, 2542, 2685, 396, 435, 4246, 612, 5619, 6109, 710, 2, 6549, 6573, 199, 201, 7082, 7563, 7888, 855, 7, 938, 11562, 1211, 1312, 1438
Offset: 1

Views

Author

Hugo Pfoertner, Jan 16 2021

Keywords

Examples

			See A340700.
		

Crossrefs

Formula

A340700(n) = a(n)^A340704(n).

A340703 Root of the upper member A340701 of a pair of adjacent perfect powers, both with exponents > 2.

Original entry on oeis.org

2, 3, 2, 2, 2, 11, 13, 37, 39, 58, 6, 26, 28, 146, 23, 69, 294, 339, 13, 458, 508, 53, 797, 905, 927, 1077, 215, 217, 1771, 1797, 283, 358, 373, 2908, 3296, 526, 5196, 649, 691, 6334, 6502, 728, 730, 6783, 6897, 772, 811, 837, 8115, 8786, 9182, 1115, 12908, 14363
Offset: 1

Views

Author

Hugo Pfoertner, Jan 16 2021

Keywords

Examples

			See A340700.
		

Crossrefs

Formula

A340701(n) = a(n)^A340705(n).

A340704 Exponent of the lower member A340700 of a pair of adjacent perfect powers, both with exponents > 2.

Original entry on oeis.org

3, 6, 3, 5, 3, 4, 7, 4, 10, 4, 4, 3, 3, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 4, 4, 3, 3, 3, 4, 4, 3, 4, 3, 3, 4, 38, 3, 3, 5, 5, 3, 3, 3, 4, 14, 4, 3, 4, 4, 4, 4, 3, 3, 4, 3, 3, 3, 3, 4, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 3, 3, 4, 4, 3, 3, 3, 3, 4, 4, 4, 4, 3
Offset: 1

Views

Author

Hugo Pfoertner, Jan 16 2021

Keywords

Examples

			See A340700.
		

Crossrefs

Formula

A340700(n) = A340702(n)^a(n).

A340705 Exponent of the upper member A340701 of a pair of adjacent perfect powers, both with exponents > 2.

Original entry on oeis.org

5, 4, 7, 8, 10, 3, 3, 3, 3, 3, 7, 4, 4, 3, 5, 4, 3, 3, 7, 3, 3, 5, 3, 3, 3, 3, 4, 4, 3, 3, 4, 4, 4, 3, 3, 4, 3, 4, 4, 3, 3, 4, 4, 3, 3, 4, 4, 4, 3, 3, 3, 4, 3, 3, 3, 3, 4, 4, 3, 4, 4, 4, 5, 3, 4, 4, 3, 4, 4, 4, 4, 4, 5, 4, 5, 4, 3, 4, 4, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 4
Offset: 1

Views

Author

Hugo Pfoertner, Jan 16 2021

Keywords

Examples

			See A340700.
		

Crossrefs

Formula

A340701(n) = A340703(n)^a(n).

A340706 Difference between upper and lower member of a pair of adjacent perfect powers A340700 and A340701, both with exponents > 2.

Original entry on oeis.org

5, 17, 3, 13, 24, 35, 10, 28, 270, 631, 95, 443, 531, 440, 1487, 1934, 503, 8138, 6276, 12311, 16911, 33892, 11573, 17000, 3807, 45197, 30753, 31457, 65170, 105597, 127209, 206808, 109516, 139456, 377711, 530040, 561600, 690742, 952332, 457704, 671064, 353107
Offset: 1

Views

Author

Hugo Pfoertner, Jan 16 2021

Keywords

Comments

The differences are expected to be bounded below by the Lang-Waldschmidt conjecture (see Waldschmidt 2013, p. 6, Conjecture 6).

Examples

			See A340700.
		

Crossrefs

Formula

a(n) = A340701(n) - A340700(n).

A076470 Perfect powers m^k where k >= 6.

Original entry on oeis.org

1, 64, 128, 256, 512, 729, 1024, 2048, 2187, 4096, 6561, 8192, 15625, 16384, 19683, 32768, 46656, 59049, 65536, 78125, 117649, 131072, 177147, 262144, 279936, 390625, 524288, 531441, 823543, 1000000, 1048576, 1594323, 1679616, 1771561
Offset: 1

Views

Author

Robert G. Wilson v, Oct 14 2002

Keywords

Comments

A necessary but not sufficient condition is that if p|n when at least p^6|n.

Crossrefs

Programs

  • Mathematica
    a = {1}; Do[ If[ Apply[ GCD, Last[ Transpose[ FactorInteger[n]]]] > 4, a = Append[a, n]; Print[n]], {n, 2, 1953124}]; a
  • Python
    from sympy import mobius, integer_nthroot
    def A076470(n):
        def f(x): return int(n+9+x-(sum(integer_nthroot(x,d)[0] for d in (6,10,15))<<1)-sum(integer_nthroot(x,d)[0] for d in (8,9,12,20,25))+sum(mobius(k)*(sum(integer_nthroot(x,k*i)[0] for i in range(1,6))-5) for k in range(6,x.bit_length())))
        kmin, kmax = 1,2
        while f(kmax) >= kmax:
            kmax <<= 1
        while True:
            kmid = kmax+kmin>>1
            if f(kmid) < kmid:
                kmax = kmid
            else:
                kmin = kmid
            if kmax-kmin <= 1:
                break
        return kmax # Chai Wah Wu, Aug 14 2024

Formula

Sum_{n>=1} 1/a(n) = 5 - zeta(2) - zeta(3) - zeta(4) - zeta(5) + Sum_{k>=2} mu(k)*(5 - zeta(k) - zeta(2*k) - zeta(3*k) - zeta(4*k) - zeta(5*k)) = 1.03342597171... . - Amiram Eldar, Dec 03 2022

A111231 Numbers which are perfect powers m^k equal to the sum of m distinct primes.

Original entry on oeis.org

0, 8, 16, 27, 32, 64, 81, 125, 128, 216, 243, 256, 343, 512, 625, 729, 1000, 1024, 1296, 1331, 1728, 2048, 2187, 2197, 2401, 2744, 3125, 3375, 4096, 4913, 5832, 6561, 6859, 7776, 8000, 8192, 9261, 10000, 10648, 12167, 13824, 14641, 15625, 16384, 16807
Offset: 1

Views

Author

Giovanni Teofilatto, Oct 28 2005

Keywords

Comments

Perfect powers m^k with k >= 3, m = 0 or m > 1.
Is a(n) = A076467(n) for all n > 1? - R. J. Mathar, May 22 2009
A sum of m distinct primes is >= A007504(m) ~ m^2(log m)/2 > m^2, also for small m, therefore the second condition excludes squares m^2. On the other hand, considering results related to Goldbach's conjecture (e.g., every even number >= 4 is the sum of at most 4 primes), it is increasingly improbable that some m^k with k >= 3 is not the sum of m primes. This explains the first comment - but can it be rigorously proved? - M. F. Hasler, May 25 2018

Examples

			a(1) = 0 because 0 = 0^2 = 0^3 is the sum of 0 primes;
a(2) = 8 because 8 = 2^3 = 3 + 5, sum of 2 primes;
a(3) = 16 because 16 = 2^4 = 3 + 13, sum of 2 primes.
a(4) = 27 because 27 = 3^3 = 3 + 11 + 13, sum of 3 primes.
		

Crossrefs

Programs

  • PARI
    is(n,d)={if(d=ispower(n), fordiv(d,e,e>1&&forvec(v=vector(d=sqrtnint(n,e)-1,i,[1,primepi((n-1)\2-d+3)]),prime(v[#v])<(d=n-vecsum(apply(i->prime(i),v)))&&isprime(d)&&return(1),2)), !n)} \\ M. F. Hasler, May 25 2018

Extensions

Offset corrected by R. J. Mathar, May 25 2009
Edited by M. F. Hasler, May 25 2018

A245713 Sorted imperfect powers b^p with b > 0, p > 2, with multiplicity.

Original entry on oeis.org

1, 8, 16, 27, 32, 64, 64, 81, 125, 128, 216, 243, 256, 256, 343, 512, 512, 625, 729, 729, 1000, 1024, 1024, 1296, 1331, 1728, 2048, 2187, 2197, 2401, 2744, 3125, 3375, 4096, 4096, 4096, 4096, 4913, 5832, 6561, 6561, 6859, 7776, 8000, 8192, 9261
Offset: 1

Views

Author

Anatoly E. Voevudko, Jul 30 2014

Keywords

Comments

No multiple terms for b=1.
This sequence strictly follows requirements of the Beal conjecture.
Less than 550 of these powers satisfy 196 Beal's conjecture equations.

Crossrefs

Programs

  • Maple
    N:= 10^5: # to get all terms <= N
    L:= [1, seq(seq(b^p, p=3..floor(log[b](N))),b=2..floor(N^(1/3)))]:
    sort(L); # Robert Israel, Nov 09 2015
  • Mathematica
    mx = 10000; Join[{1}, Sort@ Flatten@ Table[b^p, {b, 2, Sqrt@ mx}, {p, 3, Log[b, mx]}]] (* Robert G. Wilson v, Nov 09 2015 *)
  • PARI
    A245713(lim)={my(L=List(1),lim2=logint(lim,2));for(p=3,lim2, for(b=2,sqrtnint(lim,p),listput(L, b^p);));listsort(L); print(L);} \\ Anatoly E. Voevudko, Sep 21 2015

A304433 Numbers n such that n^3 is the sum of two distinct perfect powers > 1 (x^k + y^m; x, y, k, m >= 2).

Original entry on oeis.org

5, 7, 8, 10, 12, 13, 14, 17, 20, 25, 26, 28, 29, 32, 33, 34, 37, 40, 41, 45, 48, 50, 52, 53, 56, 57, 58, 61, 63, 65, 68, 71, 72, 73, 74, 78, 80, 82, 85, 89, 90, 97, 98, 100, 101, 104, 105, 106, 109, 112, 113, 114, 116, 117, 122, 125, 126, 128
Offset: 1

Views

Author

M. F. Hasler, May 12 2018

Keywords

Comments

Motivated by the search of solutions to a^n + b^(2n+2)/4 = (perfect square), which arises when searching solutions to x^n + y^(n+1) = z^(n+2) of the form x = a*z, y = b*z. It turns out that many solutions are of the form a^n = d (b^(n+1) + d), where d is a perfect power.

Examples

			5^3 = 125 = 4^2 + 11^2; 7^3 = 10^2 + 3^5; 8^3 = 13^2 + 7^3, ...
		

Crossrefs

Cf. A001597 (perfect powers), A076467 (cubes and higher powers), A304434, A304435, A304436 (analog for n^4, n^5, n^6).

Programs

  • Maple
    N:= 200: # to get terms <= N
    N3:= N^3:
    P:= {seq(seq(x^k, k=3..floor(log[x](N3))), x=2..N)}:
    filter:= proc(n) local n3, Pp, x, y;
      n3:= n^3;
      if remove(t -> subs(t, x)<=1 or subs(t, y)<=1 or subs(t, x-y)=0, [isolve(x^2+y^2=n3)]) <> [] then return true fi;
      Pp:= map(t ->n3-t, P minus {n3, n3/2});
       (Pp intersect P <> {}) or (select(issqr, Pp) <> {})
    end proc:
    select(filter, [$2..N]); # Robert Israel, Jun 01 2018
  • Mathematica
    M = 200;
    M3 = M^3;
    P = Union@ Flatten@ Table[Table[x^k, {k, 3, Floor[Log[x, M3]]}], {x, 2, M}];
    filterQ[n_] := Module[{n3, Pp, x, y}, n3 = n^3; If[Solve[x > 1 && y > 1 && x != y && x^2 + y^2 == n3, {x, y}, Integers] != {}, Return[True]]; Pp = n3 - (P ~Complement~ {n3, n3/2}); (Pp ~Intersection~ P) != {} || Select[ Pp, IntegerQ[Sqrt[#]]&] != {}];
    Select[Range[2, M], filterQ] (* Jean-François Alcover, Jun 21 2020, after Robert Israel *)
  • PARI
    L=200^3; P=List(); for(x=2, sqrtnint(L,3), for(k=3, logint(L, x), listput(P, x^k))); #P=Set(P) \\ This P = A076467 \ {1} = A111231 \ {0} up to limit L.
    is(n,e=3)={for(i=1, #s=sum2sqr(n=n^e), vecmin(s[i])>1 && s[i][1]!=s[i][2] && return(1)); for(i=1, #P, n>P[i]||return; ispower(n-P[i])&& P[i]*2 != n && return(1))} \\ The above P must be computed up to L >= n^3. For sum2sqr() see A133388.
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