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 11-19 of 19 results.

A154263 Number of triples such that p+F_s+(F_t)^3=n, where p is an odd prime, s and t are greater than one and F_s or F_t is odd.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 06 2009

Keywords

Comments

Zhi-Wei Sun conjectured that a(n)>0 for every n=5,6,...; in other words, any integer n>4 can be written as the sum of an odd prime, a positive Fibonacci number and a cube of a positive Fibonacci number, with one of two Fibonacci numbers odd. He has verified this up to 3*10^7.
Zhi-Wei Sun has offered a monetary reward for settling this conjecture.

Examples

			For n=14 the a(14)=4 solutions are 3+F_4+(F_3)^3, 5+F_2+(F_3)^3, 5+F_6+(F_2)^3, 11+F_3+(F_2)^3
		

References

  • R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.
  • Z. W. Sun and M. H. Le, Integers not of the form c(2^a+2^b)+p^{alpha}, Acta Arith. 99(2001), 183-190.

Crossrefs

Programs

  • Mathematica
    PQ[m_]:=m>2&&PrimeQ[m] RN[n_]:=Sum[If[(Mod[n,2]==0||Mod[x,3]>0)&&PQ[n-(Fibonacci[x])^3-Fibonacci[y]],1,0], {x,2,2*Log[2,n^(1/3)+1]},{y,2,2*Log[2,Max[2,n-(Fibonacci[x])^3]]}] Do[Print[n," ",RN[n]];Continue,{n,1,50000}]

A154364 Number of ways to express n as the sum of an odd prime, a positive Pell number and a companion Pell number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 07 2009

Keywords

Comments

This is inspired by the sequence A154290 and related conjectures of Sun. On Jan 08 2009, Zhi-Wei Sun and Qing-Hu Hou conjectured that a(n)>0 for n=6,7,...; in other words, any integer n>5 can be written as the sum of an odd prime, a positive Pell number and a companian Pell number. The Pell numbers are defined by P_0=0, P_1=1 and P_{n+1}=2P_n+P_{n-1} (n=1,2,3,...) and the companion Pell numbers are given by Q_0=Q_1=2 and Q_{n+1}=2Q_n+Q_{n-1} (n=1,2,3...). Note that for n>5 both P_n and Q_n are greater than 2^n.
D. S. McNeil disproved the conjecture by finding the 4 initial counterexamples: 169421772576, 189661491306, 257744272674, 534268276332. - Zhi-Wei Sun, Jan 17 2009
On Feb 01 2009, Zhi-Wei Sun observed that these 4 counterexamples are divisible by 42 and guessed that all counterexamples to the conjecture of Sun and Hou should be multiples of 42. - Zhi-Wei Sun, Feb 01 2009

Examples

			For n=10 the a(10)=3 solutions are 3+5+2, 3+1+6, 7+1+2.
		

Crossrefs

Programs

  • Maple
    Pell:=proc(n) if n=0 then return(0); elif n=1 then return(1); else return( 2*Pell(n-1) + Pell(n-2) ); fi; end proc: comp_Pell:=proc(n) if n=0 then return(2); elif n=1 then return(2); else return( 2*comp_Pell(n-1) + comp_Pell(n-2) ); fi; end proc: for n from 1 to 10^5 do rep_num:=0; for i from 1 while Pell(i)2) and isprime(p) then rep_num:=rep_num+1; fi; od; od; printf("%d %d\n", n, rep_num); od:
  • Mathematica
    nmax = 10^3;
    Pell[n_] := Pell[n] = If[n == 0, Return[0], If[n == 1, Return[1], Return[2* Pell[n - 1] + Pell[n - 2]]]];
    compPell[n_] := compPell[n] = If[n == 0, Return[2], If[n == 1, Return[2],  Return[2*compPell[n - 1] + compPell[n - 2]]]];
    Reap[For[n = 1, n <= nmax, n++, repnum = 0; For[i = 1, Pell[i] < n, i++, For[j = 1, Pell[i] + compPell[j] < n, j++, p = n - Pell[i] - compPell[j]; If[p > 2 && PrimeQ[p], repnum++]]]; Sow[repnum]]][[2, 1]] (* Jean-François Alcover, Dec 13 2017, translated from Maple *)

A157218 Number of ways to write the n-th positive odd integer in the form p+2^x+7*2^y with p a prime congruent to 1 mod 6 and x,y positive integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 0, 2, 3, 1, 1, 3, 1, 1, 4, 2, 3, 2, 1, 3, 3, 2, 3, 5, 1, 2, 5, 2, 4, 5, 1, 4, 3, 1, 4, 7, 1, 5, 7, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 25 2009

Keywords

Comments

On Feb 24 2009, Zhi-Wei Sun conjectured that a(n)>0 for all n=18,19,...; in other words, any odd integer greater than 34 can be written as the sum of a prime congruent to 1 mod 6, a positive power of 2 and seven times a positive power of 2. Sun verified the conjecture for odd integers below 5*10^7, and Qing-Hu Hou continued the verification for odd integers below 1.5*10^8 (on Sun's request). Compare the conjecture with R. Crocker's result that there are infinitely many positive odd integers not of the form p + 2^x + 2^y with p an odd prime and x,y positive integers.

Examples

			For n=19 the a(19)=3 solutions are 2*19 - 1 = 7 + 2 + 7*2^2 = 7 + 2^4 + 7*2 = 19 + 2^2 + 7*2.
		

References

  • R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.

Crossrefs

Programs

  • Mathematica
    PQ[x_]:=x>1&&Mod[x,6]==1&&PrimeQ[x] RN[n_]:=Sum[If[PQ[2n-1-7*2^x-2^y],1,0], {x,1,Log[2,(2n-1)/7]},{y,1,Log[2,Max[2,2n-1-7*2^x]]}] Do[Print[n," ",RN[n]],{n,1,200000}]

Formula

a(n) = |{: p+2^x+7*2^y=2n-1 with p a prime congruent to 1 mod 6 and x,y positive integers}|.

A157237 Number of ways to write the n-th positive odd integer in the form p+2^x+11*2^y with p a prime congruent to 1 mod 6 and x,y positive integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 25 2009

Keywords

Comments

On Feb. 24, 2009, Zhi-Wei Sun conjectured that a(n)=0 if and only if n<16 or n=18, 21, 24, 51, 84, 1011, 59586; in other words, except for 35, 41, 47, 101, 167, 2021, 119171, any odd integer greater than 30 can be written as the sum of a prime congruent to 1 mod 6, a positive power of 2 and eleven times a positive power of 2. Sun verified the conjecture for odd integers below 5*10^7, and Qing-Hu Hou continued the verification for odd integers below 1.5*10^8 (on Sun's request). Compare the conjecture with Crocker's result that there are infinitely many positive odd integers not of the form p+2^x+2^y with p an odd prime and x,y positive integers.

Examples

			For n=19 the a(19)=2 solutions are 2*19-1=7+2^3+2*11=13+2+2*11.
		

References

  • R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.
  • Z. W. Sun and M. H. Le, Integers not of the form c(2^a+2^b)+p^{alpha}, Acta Arith. 99(2001), 183-190.

Crossrefs

Programs

  • Mathematica
    PQ[x_]:=x>1&&Mod[x,6]==1&&PrimeQ[x] RN[n_]:=Sum[If[PQ[2n-1-11*2^x-2^y],1,0], {x,1,Log[2,(2n-1)/11]},{y,1,Log[2,Max[2,2n-1-11*2^x]]}] Do[Print[n," ",RN[n]],{n,1,200000}]

Formula

a(n)=|{: p+2^x+11*2^y=2n-1 with p a prime congruent to 1 mod 6 and x,y positive integers}|

A154421 Number of ways to express n as the sum of an odd prime, a positive Fibonacci number and an even Lucas number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 09 2009

Keywords

Comments

On Jan 09 2009, Zhi-Wei Sun conjectured that a(n)>0 for all n=6,7,.... ; in other words, any integer n>5 can be written in the form p+F_s+L_{3t} with p an odd prime, s positive and t nonnegative. [Compare this with the conjecture related to the sequence A154290.] Sun verified the above conjecture up to 5*10^6 and Qing-Hu Hou continued the verification up to 2*10^8. If we set v_0=2, v_1=4 and v_{n+1}=4v_n+v_{n-1} for n=1,2,3,..., then L_{3t}=v_t is at least 4^t for every t=0,1,2,.... On Jan 17 2009, D. S. McNeil found that 36930553345551 cannot be written as the sum of a prime, a Fibonacci number and an even Lucas number.

Examples

			For n=8 the a(8)=3 solutions are 3 + F_4 + L_0, 3 + F_2 + L_3, 5 + F_2 + L_0.
		

References

  • R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.

Crossrefs

Programs

  • Mathematica
    PQ[m_]:=m>2&&PrimeQ[m] RN[n_]:=Sum[If[PQ[n-2*Fibonacci[3x+1]+Fibonacci[3x]-Fibonacci[y]],1,0], {x,0,Log[2,n]},{y,2,2*Log[2,Max[2,n-2*Fibonacci[3x+1]+Fibonacci[3x]]]}] Do[Print[n," ",RN[n]];Continue,{n,1,50000}]

Formula

a(n) = |{: p+F_s+L_{3t}=n with p an odd prime, s>1 and t nonnegative}|.

Extensions

McNeil's counterexample added by Zhi-Wei Sun, Jan 20 2009

A157225 Number of ways to write the n-th positive odd integer in the form p+2^x+7*2^y with p a prime congruent to 5 mod 6 and x,y positive integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 0, 2, 3, 1, 2, 4, 1, 2, 4, 2, 2, 3, 2, 2, 4, 2, 4, 4, 1, 5, 5, 2, 5, 7, 1, 3, 7, 2, 4, 8, 2, 4, 3, 2, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 25 2009

Keywords

Comments

On Feb. 24, 2009, Zhi-Wei Sun conjectured that a(n)=0 if and only if n<11 or n=13,16,992; in other words, except for 25, 31, 1983, any odd integer greater than 20 can be written as the sum of a prime congruent to 5 mod 6, a positive power of 2 and seven times a positive power of 2. Sun verified the conjecture for odd integers below 5*10^7, and Qing-Hu Hou continued the verification for odd integers below 1.5*10^8 (on Sun's request). Compare the conjecture with Crocker's result that there are infinitely many positive odd integers not of the form p+2^x+2^y with p an odd prime and x,y positive integers.

Examples

			For n=18 the a(18)=3 solutions are 2*18-1=5+2+7*2^2=5+2^4+7*2=17+2^2+7*2.
		

References

  • R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.
  • Z. W. Sun and M. H. Le, Integers not of the form c(2^a+2^b)+p^{alpha}, Acta Arith. 99(2001), 183-190.

Crossrefs

Programs

  • Mathematica
    PQ[x_]:=x>1&&Mod[x,6]==5&&PrimeQ[x] RN[n_]:=Sum[If[PQ[2n-1-7*2^x-2^y],1,0], {x,1,Log[2,(2n-1)/7]},{y,1,Log[2,Max[2,2n-1-7*2^x]]}] Do[Print[n," ",RN[n]],{n,1,200000}]

Formula

a(n)=|{: p+2^x+7*2^y=2n-1 with p a prime congruent to 5 mod 6 and x,y positive integers}|

A157242 Number of ways to write the n-th positive odd integer in the form p+2^x+11*2^y with p a prime congruent to 5 mod 6 and x,y positive integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 2, 1, 0, 2, 2, 0, 2, 2, 1, 3, 2, 2, 3, 1, 2, 4, 2, 2, 5, 1, 2, 5, 2, 2, 4, 2, 2, 3, 2, 3, 4, 4, 3, 6, 2, 3, 6, 5, 1, 7, 4, 2, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 25 2009

Keywords

Comments

On Feb. 24, 2009, Zhi-Wei Sun conjectured that a(n)=0 if and only if n<15 or n=17, 20, 23, 86, 124; in other words, except for 33, 39, 45, 171 and 247, any odd integer greater than 28 can be written as the sum of a prime p=5 (mod 6), a positive power of 2 and eleven times a positive power of 2. Sun verified the conjecture for odd integers below 5*10^7. Knowing the conjecture from Sun, Qing-Hu Hou and D. S. McNeil have continued the verification for odd integers below 1.5*10^8 and 10^12 respectively, and they have found no counterexample. Compare the conjecture with Crocker's result that there are infinitely many positive odd integers not of the form p+2^x+2^y with p an odd prime and x,y positive integers.

Examples

			For n=18 the a(18)=2 solutions are 2*18-1=5+2^3+2*11=11+2+2*11.
		

References

  • R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.
  • Z. W. Sun and M. H. Le, Integers not of the form c(2^a+2^b)+p^{alpha}, Acta Arith. 99(2001), 183-190.

Crossrefs

Programs

  • Mathematica
    PQ[x_]:=x>1&&Mod[x,6]==5&&PrimeQ[x] RN[n_]:=Sum[If[PQ[2n-1-11*2^x-2^y],1,0], {x,1,Log[2,(2n-1)/11]},{y,1,Log[2,Max[2,2n-1-11*2^x]]}] Do[Print[n," ",RN[n]],{n,1,200000}]

Formula

a(n)=|{: p+2^x+11*2^y=2n-1 with p a prime congruent to 5 mod 6 and x,y positive integers}|

A144559 a(n) = number of triples [i,j,k] with i+j+k = n, i an odd prime, j an odd Fibonacci number and k a positive Fibonacci number.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 3, 2, 6, 3, 7, 3, 7, 4, 6, 5, 8, 5, 10, 5, 12, 5, 10, 5, 12, 7, 13, 6, 15, 4, 12, 6, 13, 7, 13, 5, 16, 5, 13, 8, 11, 8, 11, 7, 17, 8, 15, 6, 12, 8, 11, 10, 13, 7, 13, 6, 12, 9, 12, 8, 14, 7, 19, 8, 18, 10, 16, 9, 15, 9, 16, 6, 16, 9, 19, 11, 18, 7, 19, 8, 16, 10, 14, 7, 18, 8, 21
Offset: 1

Views

Author

N. J. A. Sloane, Jan 03 2009

Keywords

Comments

Zhi-Wei Sun conjectured on the Number Theory Mailing List that a(n) > 0 for all n > 4.
The conjecture has been verified by D. S. McNeil for all n < 10^13.

Examples

			5 = 3+1+1, 6 = 3+1+2, 7 = 5+1+1 = 3+3+1 = 3+1+3.
		

Crossrefs

See A154257 for a better version.

Programs

  • Maple
    with(combinat); F:=fibonacci; ans:=array(1..100); oF:=[]; pF:=[];
    for n from 1 to 100 do ans[n] := 0; od:
    for n from 2 to 12 do if F(n) mod 2 = 1 then oF:=[op(oF),F(n)]; fi; od;
    for n from 2 to 12 do pF:=[op(pF),F(n)]; od:
    for i from 2 to 30 do t1:=ithprime(i);
    for j from 1 to nops(oF) do t2:=t1+oF[j]:
    for k from 1 to nops(pF) do t3:=t2+pF[k];
    if t3 <= 100 then ans[t3]:=ans[t3]+1; fi;
    od: od: od: [seq(ans[n],n=1..100)];

A157372 Number of ways to write the (n+50)-th positive odd integer in the form p+2^x+51*2^y with p an odd prime and x,y positive integers.

Original entry on oeis.org

0, 0, 0, 1, 2, 2, 2, 2, 3, 1, 3, 4, 2, 2, 3, 1, 2, 3, 3, 2, 4, 1, 2, 5, 2, 3, 3, 1, 3, 2, 1, 3, 4, 1, 2, 5, 2, 2, 6, 3, 2, 3, 3, 2, 4, 1, 3, 3, 2, 1, 3, 2, 2, 6, 3, 4, 7, 4, 5, 7
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 28 2009

Keywords

Comments

Zhi-Wei Sun guessed that a(n)=0 if and only if n=1,2,3,127; in other words, except for 353, any odd integer greater than 105 can be written as the sum of an odd prime, a positive power of two and 51 times a positive power of two. D. S. McNeil has verified this for odd integers below 10^12. This is a part of the project for the form p+2^x+k*2^y with k=3,5,...,61 initiated by Zhi-Wei Sun in Jan. 2009.

Examples

			For n = 9 the a(9) = 3 solutions are: 2*59-1 = 7+2^3+51*2 = 11+2^2+51*2 = 13+2+51*2.
		

References

  • R. Crocker, On a sum of a prime and two powers of two, Pacific J. Math. 36(1971), 103-107.
  • Z. W. Sun and M. H. Le, Integers not of the form c(2^a+2^b)+p^{alpha}, Acta Arith. 99(2001), 183-190.

Crossrefs

Programs

  • Mathematica
    PQ[x_]:=x>1&&PrimeQ[x] RN[n_]:=Sum[If[PQ[2(n+50)-1-51*2^x-2^y],1,0], {x,1,Log[2,(2(n+50)-1)/51]},{y,1,Log[2,Max[2,2(n+50)-1-51*2^x]]}] Do[Print[n," ",RN[n]],{n,1,200000}]

Formula

a(n)=|{: p+2^x+51*2^y=2(n+50)-1 with p an odd prime and x,y positive integers}|
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