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.

Previous Showing 11-18 of 18 results.

A303949 Number of ways to write 2*n+1 as p + 2*(2^k+5^m) with p prime and 2^k+5^m a product of at most three distinct primes, where k and m are nonnegative integers.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 3, 5, 3, 3, 4, 3, 3, 4, 3, 4, 3, 4, 3, 4, 4, 5, 5, 4, 4, 5, 5, 6, 4, 3, 6, 7, 3, 6, 9, 7, 5, 8, 7, 6, 7, 9, 7, 8, 2, 8, 9, 5, 5, 6, 6, 7, 6, 6, 7, 10, 6, 7, 9, 5, 6, 8, 6, 3, 6, 7, 7, 8, 5, 10, 9, 8, 5, 9, 5, 7, 10, 5, 4, 10, 7, 6, 8, 6, 7, 8, 7, 6, 8, 6
Offset: 1

Views

Author

Zhi-Wei Sun, May 05 2018

Keywords

Comments

4787449 is the first value of n > 2 with a(n) = 0, and 2*4787449+1 = 9574899 has the unique representation as p + 2*(2^k+5^m): 9050609 + 2*(2^18+5^0) with 9050609 prime and 2^18+5^0 = 5*13*37*109.
See also A303934 and A304081 for related conjectures.

Examples

			a(3) = 1 since 2*3+1 = 3 + 2*(2^0+5^0) with 3 prime.
		

Crossrefs

Programs

  • Mathematica
    qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=3;
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+5^m]&&PrimeQ[2n+1-2(2^k+5^m)],r=r+1],{k,0,Log[2,n]},{m,0,Log[5,n+1/2-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304032 Number of ways to write 2*n as p + 2^k + 3^m with p prime and 2^k + 3^m a product of at most two distinct primes, where k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 4, 4, 4, 6, 6, 5, 8, 9, 4, 6, 7, 4, 9, 10, 6, 9, 10, 6, 11, 14, 7, 9, 11, 5, 10, 9, 6, 12, 10, 3, 11, 15, 7, 12, 16, 7, 9, 14, 9, 12, 14, 8, 12, 16, 5, 12, 18, 10, 12, 16, 9, 12, 19, 10, 13, 17, 6, 10, 15, 6, 10, 16, 10, 12, 15, 10, 17, 20, 8, 14, 15, 8, 11, 18, 9, 12
Offset: 1

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

The even number 58958 cannot be written as p + 2^k + 3^m with p and 2^k + 3^m both prime.
Clearly, a(n) <= A303702(n). We note that a(n) > 0 for all n = 2..5*10^8.
See also A304034 for a related conjecture.

Examples

			a(3) = 1 since 2*3 = 3 + 2^1 + 3^0 with 3 = 2^1 + 3^0 prime.
		

References

  • J. R. Chen, On the representation of a larger even integer as the sum of a prime and the product of at most two primes, Sci. Sinica 16(1973), 157-176.

Crossrefs

Programs

  • Mathematica
    qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=2;
    tab={};Do[r=0;Do[If[qq[2^k+3^m]&&PrimeQ[2n-2^k-3^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[3,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A218044 Numbers of the form 2^k + prime, with k > 0.

Original entry on oeis.org

4, 5, 6, 7, 9, 10, 11, 13, 15, 17, 18, 19, 21, 23, 25, 27, 29, 31, 33, 34, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 66, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119
Offset: 1

Views

Author

Michel Marcus, Oct 19 2012

Keywords

Comments

A039669 is included in this sequence.

Examples

			5 = 3 + 2 that is, a prime and a power of 2.
		

Crossrefs

Cf. A080340, A039669, A118955 (allows k=0).

Programs

  • Maple
    q:= n-> ormap(isprime, [seq(n-2^k, k=1..ilog2(n))]):
    select(q, [$0..200])[];  # Alois P. Heinz, Feb 14 2020
  • Mathematica
    nn = 119; ps = Prime[Range[PrimePi[nn]]]; p2 = 2^Range[Log[2, nn]]; u = {}; Do[u = Union[u, ps + p2[[i]]], {i, Length[p2]}]; Select[u, # <= nn &] (* T. D. Noe, Oct 19 2012 *)
  • PARI
    isok(n) = {forprime(p=2, n, my(d = n - p); if ((d==2) || (ispower(d,,&k) && (k==2)), return(1));); 0;} \\ Michel Marcus, Apr 18 2016

A304122 Squarefree numbers of the form 2^k + 5^m, where k is a positive integer and m is a nonnegative integer.

Original entry on oeis.org

3, 5, 7, 13, 17, 21, 29, 33, 37, 41, 57, 65, 69, 89, 127, 129, 133, 141, 157, 253, 257, 281, 381, 517, 537, 627, 629, 633, 641, 689, 753, 881, 1049, 1137, 1149, 1649, 2049, 2053, 2073, 2173, 3127, 3129, 3133, 3157, 3189, 3253, 3637, 4097, 4101, 4121
Offset: 1

Views

Author

Zhi-Wei Sun, May 07 2018

Keywords

Comments

The conjecture in A304081 has the following equivalent version: Any even number greater than 4 can be written as the sum of a prime and a term of the current sequence, and also any odd number greater than 8 can be written as the sum of a prime and twice a term of the current sequence.

Examples

			a(1) = 3 since 3 = 2^1 + 5^0 is squarefree.
a(6) = 21 since 21 = 2^4 + 5^1 = 3*7 is squarefree.
		

Crossrefs

Programs

  • Mathematica
    V={}; Do[If[SquareFreeQ[2^k+5^m],V=Append[V,2^k+5^m]],{k,1,12},{m,0,5}];
    LL:=LL=Sort[DeleteDuplicates[V]];
    a[n_]:=a[n]=LL[[n]];
    Table[a[n],{n,1,50}]

A212292 Odd numbers not of the form p^2 + q^2 + r with p, q, and r prime.

Original entry on oeis.org

1, 3, 5, 7, 9, 17, 33, 43, 83, 179, 623, 713, 1019
Offset: 1

Views

Author

Keywords

Comments

The corresponding sequence with the restriction to primes removed is empty.
Wang shows that all but x^{9/20+e} members of this sequence up to x are congruent to 2 mod 3, for any e > 0.
There are no more terms < 10^7. - Donovan Johnson, Jun 27 2012
There are no more terms < 4*10^9. - Jud McCranie, Jun 09 2013
There are no more terms < 10^11. - Giovanni Resta, Jun 09 2013

References

  • Wang Mingqiang, On sums of a prime, and a square of prime, and a k-power of prime, Northeastern Mathematical Journal 18:4 (2002), pp. 283-286.

Crossrefs

Programs

  • PARI
    list(lim)=my(p1=vector(primepi(sqrt(lim-5.5)),i,prime(i)^2), p2=List(), v=List(), u=List([1,3,5,7,9]), t); for(i=1,#p1, for(j=i,#p1,t=p1[i]+p1[j]; if(t>lim, break, listput(p2,t)))); p2=vecsort(Vec(p2),,8); for(i=1,#p2,forprime(p=2,lim-p2[i],listput(v,p2[i]+p))); v=select(n->n%2, vecsort(Vec(v),,8)); for(i=2,#v,forstep(j=v[i-1]+2,v[i]-2,2,listput(u,j))); Vec(u)

A303997 Number of ways to write 2*n as p + 3^k + binomial(2*m,m), where p is a prime, and k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 4, 4, 4, 5, 4, 6, 6, 4, 6, 8, 6, 5, 8, 5, 5, 8, 5, 6, 10, 4, 4, 7, 5, 5, 7, 6, 4, 8, 4, 6, 11, 6, 5, 10, 8, 7, 9, 11, 7, 10, 7, 4, 11, 9, 9, 9, 10, 8, 12, 9, 9, 11, 9, 5, 8, 8, 4, 11, 8, 7, 8, 8, 7, 10, 8, 7, 6, 7, 5, 10, 9, 7, 12, 8, 5, 7, 9, 8, 9, 8, 6, 8, 11
Offset: 1

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

502743678 is the first value of n > 1 with a(n) = 0.

Examples

			a(2) = 1 since 2*2 = 2 + 3^0 + binomial(2*0,0) with 2 prime.
a(3) = 2 since 2*3 = 3 + 3^0 + binomial(2*1,1) = 2 + 3^1 + binomial(2*0,0) with 3 and 2 both prime.
		

Crossrefs

Programs

  • Mathematica
    c[n_]:=c[n]=Binomial[2n,n];
    tab={};Do[r=0;k=0;Label[bb];If[c[k]>=2n,Goto[aa]];Do[If[PrimeQ[2n-c[k]-3^m],r=r+1],{m,0,Log[3,2n-c[k]]}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,90}];Print[tab]

A303998 Number of ways to write 2*n+1 as p + 2^k + binomial(2*m,m), where p is a prime, and k and m are positive integers.

Original entry on oeis.org

0, 0, 1, 2, 3, 4, 4, 5, 3, 6, 5, 6, 8, 7, 5, 7, 7, 6, 8, 11, 5, 8, 9, 5, 10, 8, 7, 8, 7, 5, 7, 10, 6, 9, 9, 5, 11, 12, 8, 13, 12, 9, 8, 15, 9, 11, 12, 11, 7, 10, 9, 10, 14, 9, 12, 12, 11, 11, 12, 9, 9, 12, 8, 5, 13, 9, 10, 14, 10, 13, 9, 15, 10, 12, 9, 12, 11, 9, 11, 13
Offset: 1

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 2.
This has been verified for n up to 10^9.

Examples

			a(3) = 1 since 2*3+1 = 3 + 2^1 + binomial(2*1,1) with 3 prime.
a(4) = 2 since 2*4+1 = 3 + 2^2 + binomial(2*1,1) = 5 + 2^1 + binomial(2*1,1) with 3 and 5 both prime.
		

Crossrefs

Programs

  • Mathematica
    c[n_]:=c[n]=Binomial[2n,n];
    tab={};Do[r=0;k=1;Label[bb];If[c[k]>2n,Goto[aa]];Do[If[PrimeQ[2n+1-c[k]-2^m],r=r+1],{m,1,Log[2,2n+1-c[k]]}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A367186 Numbers that can be written as 2^k + prime in more than one way.

Original entry on oeis.org

4, 6, 7, 9, 11, 13, 15, 18, 19, 21, 23, 25, 27, 31, 33, 35, 37, 39, 43, 45, 47, 49, 51, 55, 57, 61, 63, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 91, 93, 95, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 121, 123, 125, 129, 131, 133, 135, 139, 141, 143, 145, 147, 151, 153, 155
Offset: 1

Views

Author

Yuda Chen, Nov 08 2023

Keywords

Comments

Numbers m such that A109925(m) > 1.

Examples

			4 is a term since 4 = 2^0 + 3 = 2^1 + 2 which is 2 ways.
6 is a term since 6 = 2^0 + 5 = 2^2 + 2.
		

Crossrefs

Subsequence of A118955.

Programs

  • PARI
    isok(m) = sum(k=0, logint(m,2), isprime(m-2^k)) > 1; \\ Michel Marcus, Nov 10 2023
    
  • Python
    from itertools import count, islice
    from sympy import isprime
    def A367186_gen(startvalue=1): # generator of terms >= startvalue
        for n in count(max(startvalue,1)):
            c = 0
            for i in range(n.bit_length()-1,-1,-1):
                if isprime(n-(1<1:
                    yield n
                    break
    A367186_list = list(islice(A367186_gen(),30)) # Chai Wah Wu, Nov 29 2023
Previous Showing 11-18 of 18 results.