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 1-4 of 4 results.

A132399 Number of ordered ways of writing n = i + j, where i is 0 or a prime and j is a triangular number (A000217) >= 0.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Mar 23 2008

Keywords

Comments

Based on a posting by Zhi-Wei Sun to the Number Theory Mailing List, Mar 23 2008, where he conjectures that a(n) > 0 except for n = 216.
Zhi-Wei Sun has offered a monetary reward for settling this conjecture.
No counterexample below 10^10. - D. S. McNeil
Note that A076768 contains 216 and the numbers n whose only representation has 0 instead of a prime; all other integers appear to be the sum of a prime and a triangular number. Except for n=216, there is no other n < 2*10^9 for which a(n)=0.
It is clear that a(t) > 0 for any triangular number t because we always have the representation t = t+0. Triangular numbers tend to have only a few representations. Hence by not plotting a(n) for triangular n, the plot (see link) more clearly shows how a(n) slowly increases as n increases. This is more evidence that 216 is the only exception.
216 is the only exception less than 10^12. Let p(n) be the least prime (or 0 if n is triangular) such that n = p(n) + t(n), where t(n) is a triangular number. For n < 10^12, the largest value of p(n) is only 2297990273, which occurs at n=882560134401. - T. D. Noe, Jan 23 2009

Examples

			0 = 0+0, so a(0) = 1,
3 = 3+0 = 2+1 = 0+3, so a(3) = 3.
8 = 7+1 = 5+3 = 2+6, so a(8) = 3.
		

Crossrefs

Cf. A065397 (primes p whose only representation as the sum of a prime and a triangular number is p+0), A090302 (largest prime p for each n).
Cf. A154752 (smallest prime p for each n). - T. D. Noe, Jan 19 2009

Extensions

Corrected, edited and extended by T. D. Noe, Mar 26 2008
Edited by N. J. A. Sloane, Jan 15 2009

A117054 Number of ordered ways of writing n = i + j, where i is a prime and j is of the form k*(k+1), k > 0.

Original entry on oeis.org

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

Views

Author

N. J. A. Sloane, Jan 15 2009

Keywords

Comments

Based on a posting by Zhi-Wei Sun to the Number Theory Mailing List, Mar 23 2008, where he conjectures that a(2n+1) > 0 for n >= 2.
Zhi-Wei Sun has offered a monetary reward for settling this conjecture.
No counterexample below 10^10. - D. S. McNeil

Crossrefs

Programs

  • Maple
    t0:=array(0..300); for n from 0 to 300 do t0[n]:=0; od:
    t1:=[seq(ithprime(i),i=1..70)]; t2:=[seq(n*(n+1),n=1..30)];
    for i from 1 to 70 do for j from 1 to 30 do k:=t1[i]+t2[j]; if k <= 300 then t0[k]:=t0[k]+1; fi; od: od:
    t3:=[seq(t0[n],n=1..300)];

A238733 Number of primes p < n such that floor((n-p)/3) = (q-1)*(q-3)/8 for some prime q.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 03 2014

Keywords

Comments

Conjecture: (i) For any integers m > 2 and n > 2, there is a prime p < n such that floor((n-p)/m) has the form (q-1)*(q-3)/8 with q an odd prime.
(ii) If m > 2 and n > m + 1, then there is a prime p < n such that floor((n-p)/m) has the form (q^2 - 1)/8 with q an odd prime, except for the case m = 3 and n = 19.
Note that (q-1)*(q-3)/8 = r*(r+1)/2 with r = (q-3)/2. It seems that a(n) = 1 only for n = 3, 19, 25, 29, 31, 67, 79, 95, 96, 331, 373, 409.

Examples

			a(25) = 1 since floor((25-23)/3) = 0 = (3-1)*(3-3)/8 with 23 and 3 both prime.
a(96) = 1 since floor((96-11)/3) = 28 = (17-1)*(17-3)/8 with 11 and 17 both prime.
a(409) = 1 since floor((409-379)/3) = 10 = (11-1)*(11-3)/8 with 379 and 11 both prime.
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=PrimeQ[Sqrt[8n+1]+2]
    t[n_,k_]:=TQ[Floor[(n-Prime[k])/3]]
    a[n_]:=Sum[If[t[n,k],1,0],{k,1,PrimePi[n-1]}]
    Table[a[n],{n,1,80}]
  • PARI
    has(x)=issquare(8*x+1,&x) && isprime(x+2)
    a(n)=my(s); forprime(p=2,n-1,s+=has((n-p)\3)); s \\ Charles R Greathouse IV, Mar 03 2014

A335641 Number of ordered ways to write 2n+1 as p + x*(9x+7) with p prime and x an integer.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Oct 03 2020

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0. Also, a(n) = 1 only for n = 1, 4, 5, 7, 8, 18, 24, 25, 42, 68, 70, 85, 117, 118, 196, 238, 287, 497, 628, 677, 732.
We have verified a(n) > 0 for all n = 1..2*10^8.
Conjecture 2: Let f(x) be any of the polynomials x*(3x+1), x*(5x+1), 2x*(3x+1), 2x*(3x+2). Then, each odd integer greater than one can be written as p + f(x) with p prime and x an integer.

Examples

			a(68) = 1, and 2*68+1 = 137 + 0*(9*0+7) with 137 prime.
a(117) = 1, and 2*117+1 = 233 + (-1)*(9*(-1)+7) with 233 prime.
a(238) = 1, and 2*238+1 = 461 + 1*(9*1+7) with 461 prime.
a(287) = 1, and 2*287+1 = 293 + (-6)*(9*(-6)+7) with 293 prime.
a(732) = 1, and 2*732+1 = 673 + 9*(9*9+7) with 673 prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[PrimeQ[2n+1-x*(9*x+7)],r=r+1],{x,-Floor[(Sqrt[36(2n+1)+49]+7)/18],(Sqrt[36(2n+1)+49]-7)/18}];
    tab=Append[tab,r],{n,1,100}];Print[tab]
Showing 1-4 of 4 results.