A303374 Numbers of the form a^4 + b^6, with integers a, b > 0.

2, 17, 65, 80, 82, 145, 257, 320, 626, 689, 730, 745, 810, 985, 1297, 1354, 1360, 2025, 2402, 2465, 3130, 4097, 4112, 4160, 4177, 4352, 4721, 4825, 5392, 6497, 6562, 6625, 7290, 8192, 10001, 10064, 10657, 10729, 14096, 14642, 14705, 15370, 15626, 15641, 15706, 15881

Offset: 1

Author

M. F. Hasler, Apr 22 2018

Keywords

• nonn
• easy

Comments

A subsequence of A000404 (a^2 + b^2), A055394 (a^2 + b^3), A111925 (a^4 + b^2), A100291 (a^4 + b^3), A303372 (a^2 + b^6).

Although it is easy to produce many terms of this sequence, it is nontrivial to check whether a very large number is of this form. Maybe the most efficient way is to consider decompositions of n into sums of two positive squares (see sum2sqr in A133388), and check if one of the terms is a third power and the other a fourth power.

Crossrefs

Cf. A055394 (a^2 + b^3), 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).

Cf. A303372 (a^2 + b^6), A303373 (a^3 + b^6), A303375 (a^5 + b^6).

Programs

Mathematica

Take[Flatten[Table[a^4+b^6,{a,20},{b,20}]]//Union,50] (* Harvey P. Dale, Jul 17 2025 *)

PARI

is(n,k=4,m=6)=for(b=1,sqrtnint(n-1,m),ispower(n-b^m,k)&&return(b)) \\ Returns b > 0 if n is in the sequence, else 0.
is(n,L=sum2sqr(n))={for(i=1,#L,L[i][1]&&for(j=1,2,ispower(L[i][j],3)&&issquare(L[i][3-j])&&return(L[i][j])))} \\ See A133388 for sum2sqr(). Much faster than the above for n >> 10^30.
A303374(L=10^5,k=4,m=6,S=[])={for(a=1,sqrtnint(L-1,m),for(b=1,sqrtnint(L-a^m,k),S=setunion(S,[a^m+b^k])));S}