A257999 -id:A257999 - cp's OEIS frontend

A300566 Numbers z such that there is a solution to x^4 + y^5 = z^6 with x, y, z >= 1.

8748, 10368, 342732

Offset: 1

M. F. Hasler, Apr 16 2018

Also in the sequence: 810000 = 2^4*3^4*5^4, 1361367 = 3^4*7^5, 3240000 = 2^6*3^4*5^4, 9335088 = 2^4*3^5*7^4, 25312500 = 2^2*3^4*5^7, 31505922 = 2*3^8*7^4, 43740000 = 2^5*3^7*5^4, 512578125 = 3^8*5^7, 1215000000 = 2^6*3^5*5^7, 1701708750 = 2*3^4*5^4*7^5, 2196150000 = 2^4*3*5^5*11^4, 2431012500 = 2^2*3^4*5^5*7^4, 4269246912 = 2^6*3^4*7^7, 4447203750 = 2*3^5*5^4*11^4, 36015000000 = 2^6*3*5^7*7^4, 48717927500 = 2^2*5^4*11^7, 75969140625 = 3^4*5^8*7^4, 91116682272 = 2^5*3^4*7^4*11^4. - Jacques Tramu, Apr 17 2018
Consider a solution (x,y,z) of x^4 + y^5 = z^6. For any m, (x*m^15, y*m^12, z*m^10) is also a solution. Reciprocally, if (x/m^15, y/m^12, z/m^10) is a triple of integers for some m, then this is also a solution. We call primitive a solution for which there is no such m > 1. - M. F. Hasler, Apr 17 2018
Observation: a(n) = A054744(n+38) = A257999(n+32), at least for 1 <= n <= 2 in both cases. - Omar E. Pol, Apr 17 2018
These relations hold only for n = 1 and 2. The next larger known term 342732 = 2^2*3*13^4 shows that in general the terms don't belong to A054744 nor A257999, although the earlier comment implies that each term gives rise to infinitely many non-primitive terms in A054744. - M. F. Hasler, Apr 19 2018
When S = a^4 + b^10/4 is a square, then z = b^5/2 + sqrt(S) is a solution, with x = a*z and y = b*z. All known solutions and further solutions 8957952, 10616832, 52200625, 216486432, ... are of this form (with rational a, b). - M. F. Hasler, Apr 19 2018

			a(1) = 8748 = 2^2*3^7 is in the sequence because 8748^6 = (2^3*3^8)^5 + (2^3*3^10)^4, using 2^3 + 1 = 3^2. Similarly, all z = 4*3^(10k-3) are in the sequence for k >= 1, with x = 8*3^(15k-5) and y = 8*3^(12k-4).
a(2) = 10368 = 2^7*3^4 is in the sequence because 10368^6 = (2^8*3^5)^5 + (2^10*3^6)^4, using 3 + 1 = 2^2. Similarly, any z = 2^7*3^(10k+4) is in the sequence for k >= 0, with x = 2^10*3^(15k+6) and y = 2^8*3^(12k+5).
z = 342732 = 2^2*3*13^4 is in the sequence because (2^2*3*13^4)^6 = (2^3*13^5)^5 + (2^3*5*13^6)^4, using 2^3*13 + 5^4 = 3^6.
z = 810000 = 2^4*3^4*5^4 is in the sequence because z^6 = x^4 + y^5 with x = 2^5*3^6*5^6 and y = 2^4*3^5*5^5 (using 1 + 3*5 = 2^4).
z = 1361367 = 3^4*7^5 is in the sequence because z^6 = x^4 + y^5 with x = 3^5*7^8 and y = 2*3^4*7^6.

Cf. A300564 (z^4 = x^2 + y^3) and A242183, A300565 (z^5 = x^3 + y^4), A302174.
Cf. A100294: numbers of the form a^5 + b^4.
See A303266 for the y-values.

is(z)=for(y=1,sqrtnint(-1+z=z^6,5),ispower(z-y^5,4)&&return(y))
/* Code below for illustration only, not guaranteed to give a complete list. Half-integral values give the additional term 31505922 for b = 63/2. Third-integral values give the additional solution z = 342732 for b = 26/3. */
S=[]; N=1e5; forstep(b=1,9,1/3, forstep(a=1,N,1/3, issquare(b^10+a^4<<2,&r)&& !frac(z=b^5/2+r/2)&& !print1(z",")&&S=setunion(S,[z])); print1([b])); S

A036667 Numbers of the form 2^i*3^j, i+j even.

Original entry on oeis.org

1, 4, 6, 9, 16, 24, 36, 54, 64, 81, 96, 144, 216, 256, 324, 384, 486, 576, 729, 864, 1024, 1296, 1536, 1944, 2304, 2916, 3456, 4096, 4374, 5184, 6144, 6561, 7776, 9216, 11664, 13824, 16384, 17496, 20736, 24576, 26244, 31104, 36864, 39366

Offset: 1

N. J. A. Sloane

nonn

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Complement of A257999 with respect to A003586.
Intersection of A028260 and A003586.
Cf. A025620 (subsequence), A069352, A022328, A022329.

a036667 n = a036667_list !! (n-1)
a036667_list = filter (even . flip mod 2 . a001222) a003586_list
-- Reinhard Zumkeller, May 16 2015

max = 40000;
Reap[Do[k = 2^i 3^j; If[k <= max && EvenQ[i+j], Sow[k]], {i, 0, Log[2, max] // Ceiling}, {j, 0, Log[3, max] // Ceiling}]][[2, 1]] // Union (* Jean-François Alcover, Aug 04 2018 *)

from sympy import integer_log
def A036667(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum((x//3**i).bit_length()+(i&1^1)>>1 for i in range(integer_log(x, 3)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Jan 30 2025

A069352(a(n)) mod 2 = 0. - Reinhard Zumkeller, May 16 2015

Sum_{n>=1} 1/a(n) = 7/4. - Amiram Eldar, Feb 18 2021

Offset corrected by Reinhard Zumkeller, May 16 2015

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A300566 Numbers z such that there is a solution to x^4 + y^5 = z^6 with x, y, z >= 1.

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A036667 Numbers of the form 2^i*3^j, i+j even.

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