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.

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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]

A308403 Number of ways to write n as 6^i + 3^j + A008347(k), where i, j and k > 0 are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 25 2019

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 2. In other words, each n = 3,4,... can be written as 6^i + 3^j + prime(k) - prime(k-1) + ... + (-1)^(k-1)*prime(1), where i, j and k > 0 are nonnegative integers.
Conjecture 2: If {a,b} is among {2,m} (m = 3..14), {3,4}, {3,5}, then any integer n > 2 can be written as a^i + b^j + A008347(k) with i, j and k > 0 nonnegative integers.
Using Qing-Hu Hou's program, we have verified Conjectures 1 and 2 for n up to 10^9 and 10^7 respectively. - Zhi-Wei Sun, May 28 2019
Conjecture 1 verified up to 10^10. Conjecture 2 holds up to 10^10 for all cases except {2, 12} since 4551086841 cannot be written as 2^i + 12^j + A008347(k). - Giovanni Resta, May 28 2019

Examples

			a(3) = 1 with 3 - (6^0 + 3^0) = 1 = A008347(2).
a(4) = 1 with 4 - (6^0 + 3^0) = 2 = A008347(1).
a(24) = 1 with 24 - (6^0 + 3^0) = 22 = A008347(13).
a(234) = 1 with 234 - (6^1 + 3^3) = 201 = A008347(90).
a(1134) = 1 with 1134 - (6^2 + 3^0) = 1097 = A008347(322).
a(4330) = 1 with 4330 - (6^3 + 3^0) = 4113 = A008347(1016).
a(5619) = 1 with 5619 - (6^1 + 3^3) = 5586 = A008347(1379).
a(6128) = 1 with 6128 - (6^0 + 3^0) = 6126 = A008347(1499).
a(16161) = 1 with 16161 - (6^3 + 3^0) = 15944 = A008347(3445).
a(133544) = 1 with 133544 - (6^0 + 3^8) = 126982 = A008347(22579).
		

Crossrefs

Programs

  • Mathematica
    Pow[n_]:=Pow[n]=n>0&&IntegerQ[Log[3,n]];
    s[0]=0;s[n_]:=s[n]=Prime[n]-s[n-1];
    tab={};Do[r=0;Do[If[s[k]>=n,Goto[bb]];Do[If[Pow[n-s[k]-6^m],r=r+1],{m,0,Log[6,n-s[k]]}];Label[bb],{k,1,2n-1}];tab=Append[tab,r],{n,1,100}];Print[tab]

A305232 Number of ordered ways to write 2*n+1 as p + binomial(2k,k) + 2*binomial(2m,m), where p is an odd prime, and k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 27 2018

Keywords

Comments

The first value of n > 2 with a(n) = 0 is 15212443837. Neither 2*15212443837 + 1 = 30424887675 nor 2*15657981007 + 1 = 31315962015 can be written as the sum of a prime, a central binomial coefficient and twice a central binomial coefficient.

Examples

			a(3) = 1 since 2*3 + 1 = 7 = 3 + binomial(2*1,1) + 2*binomial(2*0,0) with 3 an odd prime.
a(368233372) = 1 since 2*368233372 + 1 = 736466745 = 735761311 + binomial(2*11,11) + 2*binomial(2*0,0) with 735761311 an odd prime.
a(5274658504) = 1 since 2*5274658504 + 1 = 10549317009 = 10549316083 + binomial(2*6,6) + 2*binomial(2*0,0) with 10549316083 an odd prime.
a(8722422187) = 1 since 2*8722422187 + 1 = 17444844375 = 17444844367 + binomial(2*2,2) + 2*binomial(2*0,0) with 17444844367 an odd prime.
a(10296844792) = 1 since 2*10296844792 + 1 = 20593689585 = 20593688659 + binomial(2*6,6) + 2*binomial(2*0,0) with 20593688659 an odd prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;k=0;Label[aa];k=k+1;If[Binomial[2k,k]>=2n+1`,Goto[cc]];m=0;Label[bb];If[2*Binomial[2m,m]>=2n+1-Binomial[2k,k],Goto[aa]]; If[PrimeQ[2n+1-Binomial[2k,k]-2*Binomial[2m,m]],r=r+1];m=m+1;Goto[bb];Label[cc];tab=Append[tab,r],{n,1,90}];Print[tab]
Previous Showing 11-13 of 13 results.