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.

A277075 Numbers n for which A276711(n) = 1.

Original entry on oeis.org

2, 4, 5, 8, 9, 10, 16, 22, 24, 25, 26, 36, 58, 64, 76, 82, 85, 120, 170, 196, 202, 214, 324, 328, 370, 412, 505, 562, 676, 706, 730, 799, 841, 1024, 1204, 1243, 1549, 1681, 1849, 2146, 2986, 10404, 46656, 52900, 112896, 122500
Offset: 1

Views

Author

Robert Israel, Sep 27 2016

Keywords

Comments

Is the sequence finite?
a(47) > 10^7 if it exists.

Examples

			122500 is in the sequence because 122500 = 121169 + 11^3 is the only way to write 122500 as the sum of a prime and a nonnegative perfect power.

Crossrefs

Cf. A276711.

Programs

  • Maple
    N:= 10^6: # to get all terms <= N
Primes:= select(isprime, [2,seq(i,i=3..N,2)]):
Pows:= {0,1,seq(seq(b^k,k=2..floor(log[b](N))),b=2..floor(sqrt(N)))}:
G:= expand(add(x^p,p=Primes)*add(x^r,r=Pows)):
B:= [seq(coeff(G,x,i),i=1..N)]:
select(t -> B[t]=1, [$1..N]);

A276717 Least prime p < n^2 such that n^2 - p = x^k for some integers x > 1 and k > 1, or 1 if such a prime p does not exist.

Original entry on oeis.org

1, 1, 5, 7, 17, 11, 13, 37, 17, 19, 89, 19, 41, 71, 29, 13, 73, 199, 37, 157, 41, 43, 17, 47, 113, 433, 53, 541, 809, 59, 61, 997, 89, 67, 1009, 71, 73, 113, 521, 79, 1553, 83, 1721, 1693, 89, 1873, 1697, 107, 97, 313, 101, 103, 761, 107, 109, 11, 113, 239, 1433, 2269
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 16 2016

Keywords

Comments

The conjecture in A276711 implies that a(n) > 1 for all n > 2 except for n = 11^3 = 1331.
Note that for any integer n > 2 neither n^2 nor n^2 - 1 could be a prime.

Examples

			a(2) = 1 since neither 2^2 - 2 nor 2^2 -3 has the form x^k with x and k integers greater than one.
a(3) = 5 since 5 is a prime with 3^2 - 5 = 2^2 but neither 3^2 - 2 nor 3^2 - 3 is a perfect power.
a(4913) = 23613281 since 23613281 is a prime with 4913^2 - 23613281 = 2^19, and 4913^2 - p is not a perfect power for any prime p < 23613281.

Links

Crossrefs

Cf. A000040, A001597, A276711.

Programs

  • Mathematica
    Do[Do[If[IntegerQ[(n^2-Prime[j])^(1/k)],Print[n," ",Prime[j]];Goto[aa]],{j,1,PrimePi[n^2-2]},{k,2,Log[2,n^2-Prime[j]]}];Print[n," ",1];Label[aa];Continue,{n,1,60}]

A276830 Number of ways to write n as ((p-1)/2)^2 + P_2, where p is an odd prime and P_2 is a product of at most two primes.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 2, 2, 1, 3, 3, 2, 2, 3, 3, 2, 1, 3, 2, 2, 1, 2, 3, 2, 1, 4, 3, 2, 2, 4, 2, 3, 1, 3, 4, 2, 2, 5, 4, 4, 2, 5, 3, 3, 2, 3, 5, 3, 1, 5, 3, 2, 2, 2, 3, 3, 2, 4, 4, 3, 2, 5, 3, 2, 3, 5, 3, 4, 3, 4, 5, 2, 3, 5, 4, 2, 3, 5, 2, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Sep 20 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 9, 17, 21, 25, 33, 49, 109, 169, 189, 361, 841, 961, 12769, 19321.
See also A276825 for a similar conjecture involving cubes, and some comments on x^2 + P_2.

Examples

			a(2) = 1 since 2 = ((3-1)/2)^2 + 1 with 3 prime.
a(3) = 1 since 3 = ((3-1)/2)^2 + 2 with 3 and 2 both prime.
a(4) = 1 since 4 = ((3-1)/2)^2 + 3 with 3 prime.
a(9) = 1 since 9 = ((5-1)/2)^2 + 5 with 5 prime.
a(17) = 1 since 17 = ((5-1)/2)^2 + 13 with 5 and 13 both prime.
a(21) = 1 since 21 = ((5-1)/2)^2 + 17 with 5 and 17 both prime.
a(25) = 1 since 25 = ((5-1)/2)^2 + 3*7 with 5, 3 and 7 all prime.
a(33) = 1 since 33 = ((5-1)/2)^2 + 29 with 5 and 29 both prime.
a(49) = 1 since 49 = ((13-1)/2)^2 + 13 with 13 prime.
a(109) = 1 since 109 = ((13-1)/2)^2 + 73 with 13 and 73 both prime.
a(169) = 1 since 169 = ((13-1)/2)^2 + 7*19 with 13, 7 and 19 all prime.
a(189) = 1 since 189 = ((5-1)/2)^2 + 5*37 with 5 and 37 both prime.
a(361) = 1 since 361 = ((37-1)/2)^2 + + 37 with 37 prime.
a(841) = 1 since 841 = ((37-1)/2)^2 + 11*47 with 37, 11 and 47 all prime.
a(961) = 1 since 961 = ((61-1)/2)^2 + 61 with 61 prime.
a(12769) = 1 since 12769 = ((109-1)/2)^2 + 59*167 with 109, 59 and 167 all prime.
a(19321) = 1 since 19321 = ((277-1)/2)^2 + 277 with 277 prime.

Links

Crossrefs

Cf. A000040, A000290, A037143, A235645, A276711, A276825.

Programs

  • Mathematica
    PP[n_]:=PP[n]=PrimeQ[Sqrt[n]]||(SquareFreeQ[n]&&Length[FactorInteger[n]]<=2)
Do[r=0;Do[If[PP[n-((Prime[k]-1)/2)^2],r=r+1;If[r>1,Goto[aa]]],{k,2,PrimePi[2*Sqrt[n]+1]}];Print[n," ",r];
Label[aa];If[Mod[n,50000]==0,Print[n]];Continue,{n,10^5,1000000}]
Showing 1-3 of 3 results.

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