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 1-9 of 9 results.

A303265 Least y for which x^3 + y^4 = z^5 for some x > 1 and z = A300565(n).

Original entry on oeis.org

64, 625, 11664, 19208, 23328, 134456, 331776, 331776, 531441, 923521
Offset: 1

Views

Author

M. F. Hasler, Apr 23 2018

Keywords

Comments

The values listed here are the y-values corresponding to the z-values listed in A300565. The x-values are then readily computed as (z^6 - y^5)^(1/4).
See the main entry A300565 for all further information.

Examples

			A300565(1) = 32 is the smallest z such that z^5 = x^3 + y^4 for some x, y > 1, and the smallest such y is a(1) = 64. It then follows that x = (32^5 - 64^4)^(1/3) = (2^24)^(1/3) = 256.
A300565(2) = 250 is the second smallest z such that z^5 = x^3 + y^4 for some x, y > 1, and the smallest corresponding y is a(2) = 625. It then follows that x = (250^5 - 625^4)^(1/3) = 9375.
A300565(3) = 1944 is the next larger z such that z^5 = x^3 + y^4 for some x, y > 1, and the smallest corresponding y is a(2) = 11664. It then follows that x = (1944^5 - 11664^4)^(1/3) = 209952.
		

A100291 Numbers of the form a^4 + b^3 with a, b > 0.

Original entry on oeis.org

2, 9, 17, 24, 28, 43, 65, 80, 82, 89, 108, 126, 141, 145, 206, 217, 232, 257, 264, 283, 297, 320, 344, 359, 381, 424, 472, 513, 528, 593, 599, 626, 633, 652, 689, 730, 745, 750, 768, 810, 841, 968, 985, 1001, 1016, 1081, 1137, 1256, 1297, 1304, 1323, 1332
Offset: 1

Views

Author

T. D. Noe, Nov 18 2004

Keywords

Crossrefs

Cf. A100271 (primes of the form a^4 + b^3).
Cf. A055394 (a^2 + b^3: contains this as subsequence), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100292 (a^5 + b^2), A100293 (a^5 + b^3), A100294 (a^5 + b^4), A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303374 (a^4 + b^6), A303375 (a^5 + b^6).
Roots of 5th powers are listed in A300565 (z^5 = x^3 + y^4); see also A300564 (z^4 = x^2 + y^3) and A242183, A300566 (z^6 = x^4 + y^5), A300567 (z^7 = x^6 + y^5), A302174.

Programs

  • Mathematica
    lst={}; Do[p=a^4+b^3; If[p<2000, AppendTo[lst, p]], {a, 64}, {b, 256}]; Union[lst]
    With[{nn=20},Select[Union[#[[1]]^4+#[[2]]^3&/@Tuples[Range[20],2]],#<= nn^3+1&]] (* Harvey P. Dale, May 27 2020 *)
  • PARI
    is(n)=for(a=1, sqrtnint(n-1, 4), ispower(n-a^4, 3) && return(a)) \\ Returns a > 0 if n is in the sequence, or 0 otherwise. - M. F. Hasler, Apr 25 2018
    
  • PARI
    list(lim)=my(v=List());for(b=1,sqrtnint(lim\=1,3), my(b3=b^3); for(a=1,sqrtnint(lim-b3,4), listput(v,a^4+b3))); Set(v) \\ Charles R Greathouse IV, Jul 26 2021

Extensions

Edited by M. F. Hasler, Apr 25 2018

A100294 Numbers of the form a^5 + b^4 with a, b > 0.

Original entry on oeis.org

2, 17, 33, 48, 82, 113, 244, 257, 259, 288, 324, 499, 626, 657, 868, 1025, 1040, 1105, 1280, 1297, 1328, 1539, 1649, 2320, 2402, 2433, 2644, 3126, 3141, 3206, 3381, 3425, 3750, 4097, 4128, 4339, 4421, 5120, 5526, 6562, 6593, 6804, 7221, 7585, 7777, 7792
Offset: 1

Views

Author

T. D. Noe, Nov 18 2004

Keywords

Comments

In view of computing A300566, it would be interesting to have an efficient way to check whether a given (large) n is in this sequence. - M. F. Hasler, Apr 25 2018

Crossrefs

Cf. A100274 (primes of the form a^5 + b^4).
Subsequence of A100292 (a^5 + b^2); see also A055394 (a^2 + b^3), A111925 (a^2 + b^4), A100291 (a^4 + b^3), A100293 (a^5 + b^3), A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303374 (a^4 + b^6), A303375 (a^5 + b^6).
Roots of 6th powers are listed in A300566 (z such that z^6 = x^5 + y^4 for some x, y >= 1); see also A300564 (z^4 = x^3 + y^2) and A242183, A300565 (z^5 = x^4 + y^3), A300567 (z^7 = x^6 + y^5), A302174.

Programs

  • Mathematica
    lst={}; Do[p=a^5+b^4; If[p<15000, AppendTo[lst, p]], {a, 16}, {b, 32}]; Union[lst]
  • PARI
    A100294_vec(L=10^6, k=4, m=5, S=List())={for(a=1, sqrtnint(L-1, m), for(b=1, sqrtnint(L-a^m, k), listput(S, a^m+b^k))); Set(S)} \\ all terms up to limit L. - M. F. Hasler, Apr 25 2018
    
  • PARI
    is(n, k=4, m=5)=for(a=1, sqrtnint(n-1, m), ispower(n-a^m,k) && return(a)) \\ Returns a > 0 if n is in the sequence, or 0 otherwise. - M. F. Hasler, Apr 25 2018

A300564 Numbers z such that there is a solution to x^2 + y^3 = z^4 with x, y, z >= 1.

Original entry on oeis.org

6, 9, 15, 35, 36, 42, 48, 57, 63, 71, 72, 75, 78, 90, 98, 100, 120, 135, 141, 147, 162, 195, 196, 204, 208, 215, 225, 243, 252, 260, 279, 280, 288, 289, 295, 300, 306, 336, 363, 364, 384, 405, 441, 450, 456, 462, 504, 510, 525, 537, 550, 568, 576, 600, 624, 630, 713, 720, 722, 735, 750, 784, 800, 819, 828, 841, 845, 847, 867, 875
Offset: 1

Views

Author

M. F. Hasler, Apr 16 2018

Keywords

Crossrefs

Cf. A242183, A242192, A300565 (z^5 = x^3 + x^4), A300566 (z^6 = x^4 + y^5).

Programs

  • PARI
    is(z)=for(y=1,sqrtnint(z^4,3),issquare(z^4-y^3,&x)&&x&&return(1))

Formula

Equals sequence A242183 with duplicates removed.

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

Original entry on oeis.org

8748, 10368, 342732
Offset: 1

Views

Author

M. F. Hasler, Apr 16 2018

Keywords

Comments

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

Examples

			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.
		

Crossrefs

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.

Programs

  • PARI
    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

A300567 Numbers z such that z^7 = x^5 + y^6 for some integers x, y >= 1.

Original entry on oeis.org

8192, 7593750, 8605184
Offset: 1

Views

Author

M. F. Hasler, Apr 16 2018

Keywords

Comments

Also in the sequence: 72900000 = 2^5*3^6*5^5, 51018336 = 2^5*3^13, 6083264512 = 2^14*13^5, 916132832 = 2^5*31^5, 6530347008 = 2^12*3^13, 16307453952 = 2^26*3^5, 233861123808 = 2^5*3^9*13^5, 850305600000 = 2^9*3^12*5^5.
Consider a solution (x,y,z) of x^5 + y^6 = z^7. For any m, (x*m^42, y*m^35, z*m^30) is also a solution. Reciprocally, if (x/m^42, y/m^35, z/m^30) 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.
When S = a^5 + b^12/4 is a square, then z = b^6/2 + sqrt(S) is a solution, with x = a*z and y = b*z. All known solutions are of this form. Sequence A303267 lists the y-values, thus equal to b*z where z = a(n), with corresponding x = a*z = (a(n)^7 - A303267(n)^6)^(1/5).

Examples

			a(1) = 8192 = 2^13 is in the sequence because (2^13)^7 = (2^18)^5 + (2^15)^6, using 18*5 = 15*6 = 90 = 13*7 - 1 and 1 + 1 = 2.
		

Crossrefs

Cf. A300564 (z^4 = x^2 + y^3) and A242183, A300565 (z^5 = x^3 + y^4), A300566 (z^6 = x^4 + y^5), A302174.
See A303267 for the y-values.
Cf. A303375 (numbers of the form a^5 + b^6).

Programs

  • PARI
    is(z)=for(y=1,sqrtnint(-1+z=z^7,6),ispower(z-y^6,5)&&return(y))
    /* Code below for illustration only, not guaranteed to give a complete list. */
    S=[]; N=1e5; forstep(b=1,99,1/6, forstep(a=1,N,1/6, issquare(b^12/4+a^5,&r)&& !frac(z=b^6/2+r)&& S=setunion(S,[z])); print1([b])); S
    
  • Python
    # See Hayden link. This code is built to identify valid z values based on specific conjectures outlined in the file.

A302174 Smallest solution x of x^n + y^(n+1) = z^(n+2), x, y, z >= 1.

Original entry on oeis.org

1, 2, 27, 256, 472392, 262144, 13759414272, 4294967296, 4057816381784064
Offset: 0

Views

Author

Jacques Tramu, Apr 07 2018

Keywords

Comments

From M. F. Hasler, Apr 13 2018: (Start)
Proofs for the upper limits given in the formula section:
For odd n = 2k-1, x = 2^(2*k^2) yields a solution with x^(2k-1) = y^(2k) = 1/2*z^(2k+1), y = 2^(k(2k-1)) and z = 2^(2k(k-1)+1), because 2*k^2*(2k-1) + 1 = (2k+1)*(2k(k-1)+1).
For even n = 2k, x = 2^(k*(2k+1))*3^(2k+2) yields a solution, with y = 2^(2*k^2)*3^(2k+1) and z = 2^(2*k^2-k+1)*3^(2k), because for the exponents of 3, (2k+1)^2 = (2k+2)2k + 1 and the factor 1+3 = 2^2 adds 2 to the (identical) exponent of 2 in x^(2k) and y^(2k+1), to factor as 2*k^2(2k+1) + 2 = (2k+2)(k(2k-1)+1). (End)

Examples

			1^0 + 3^1 = 2^2, therefore a(0) = 1.
2^1 + 5^2 = 3^3, so a(1) = 2. (No solution can have x = 1 because z^3 - 1 = (z - 1)(z^2 + z + 1) cannot be a square: if a = z - 1, then z^2 + z + 1 = a^2 + 3a + 3 is congruent to 3 modulo any factor of a, and a = 3b yields z^3 - 1 = 9*b*(3*b^2 + 3b + 1), the last factor being congruent to 1 modulo any factor of b, and cannot be a square.)
27^2 + 18^3 = 9^4, so a(2) = 27.
256^3 + 64^4 = 32^5, so a(3) = 256.
472392^4 + 52488^5 = 8748^6, so a(4) = 472392.
		

Crossrefs

Cf. A001105 (2*k^2), A060757 (4^k^2 = 2^(2k^2)), A000244 (3^k).
Conjectured to be a subsequence of A003586 (2^i*3^j).
Cf. A300564, A300565, A300566, A300567, A300568 (z^4 = x^2 + y^3, ..., z^8 = x^6 + y^7).

Formula

For odd n, a(n) <= 2^((n+1)^2/2); for even n, a(n) <= 2^(n*(n+1)/2)*3^(n+2).
We may conjecture that, for n > 4, a(n) is given by these upper limits.

Extensions

Extended to a(0) = 1 by M. F. Hasler, Apr 13 2018

A300568 Numbers z such that z^8 = x^6 + y^7 for some integers x, y >= 1.

Original entry on oeis.org

47775744, 22143375000, 23328000000
Offset: 1

Views

Author

M. F. Hasler, May 04 2018

Keywords

Comments

Also in the sequence: 4001504141376, 4275897935643, 11284439629824, 100192997081088, ... (with b = 63, 51, 72, 96, ... cf. below).
Consider a solution (x,y,z) of x^6 + y^7 = z^8. For any m, (x*m^28, y*m^24, z*m^21) is also a solution. Reciprocally, if (x/m^28, y/m^24, z/m^21) 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.
When S = a^6 + b^14/4 is a square, then z = b^7/2 + sqrt(S) is a solution, with x = a*z and y = b*z. All known solutions are of this form.
Sequence A303268 lists the y-values, in the above case equal to b*z where z = a(n), with corresponding x = a*z = (a(n)^8 - A303268(n)^7)^(1/6).

Examples

			a(1) = 47775744 = 2^16*3^6 is in the sequence because for this (2^16*3^6)^8 =: z^8 = (12*z)^7 + (288*z)^6 = (2^18*3^7)^7 + (2^21*3^8)^6.
a(2) = 22143375000 and a(3) = 23328000000 are in the sequence because for these z, we have z^8 = (30*z)^7 + x^6 with x = 1350*z resp. x = 1800*z.
		

Crossrefs

Cf. A300564 (z^4 = x^2 + y^3) and A242183, A300565 (z^5 = x^3 + y^4), A300566 (z^6 = x^4 + y^5), A300567 (z^7 = x^5 + y^6), A302174.
See A303268 for the y-values.
Cf. A303376 (numbers of the form a^6 + b^7).

Programs

  • PARI
    is(z)=for(y=1,sqrtnint(-1+z=z^8,7),ispower(z-y^7,6)&&return(y)) \\ returns the y-value (cf. A303268) if z is in this sequence, 0 else.
    /* Code below for illustration only, not guaranteed to give a complete list. */
    S=[]; N=1e5; forstep(b=1,99,1/2, forstep(a=1,N,1/2, issquare(b^14/4+a^6,&r)&& !frac(z=b^7/2+r)&& !print("\n*** "z)&& S=setunion(S,[z])); print1([b])); S

A303268 Least y for which x^6 + y^7 = A300568(n)^8 for some x > 1.

Original entry on oeis.org

573308928, 664301250000, 699840000000
Offset: 1

Views

Author

M. F. Hasler, May 04 2018

Keywords

Comments

The values listed here are the y-values corresponding to the z-values listed in A300568. The x-values are then readily computed as (z^8 - y^7)^(1/6).
See the main entry A300567 for all further information.

Examples

			A300568(1) = 47775744 is the smallest z such that z^8 = x^6 + y^7 for some x, y > 1, and the smallest such y is a(1) = 12*z = 573308928. It then follows that x = (47775744^8 - 573308928^7)^(1/6) = 13759414272 = 288*z.
A300568(2) = 22143375000 is the second smallest z such that z^8 = x^6 + y^7 for some x, y > 1, and the smallest corresponding y is a(2) = 30*z = 664301250000. It then follows that x = (22143375000^8 - 664301250000^7)^(1/6) = 29893556250000 = 1350*z.
Similarly, a(3) = 30*A300568(2) = 699840000000 is the smallest y for which x = (A300568(3)^8 - y^7)^(1/6) is an integer, here x = 1800*A300568(3) = 60*a(3).
		

Crossrefs

Showing 1-9 of 9 results.