A256544 -id:A256544 - cp's OEIS frontend

A256555 Number of ways to write n as the sum of two (unordered) distinct elements of the set {floor(p/3): p is prime}.

1, 1, 2, 2, 3, 3, 4, 3, 4, 4, 4, 4, 5, 5, 6, 6, 6, 5, 7, 6, 6, 7, 7, 8, 7, 8, 9, 7, 10, 7, 7, 9, 9, 9, 9, 12, 11, 10, 12, 8, 10, 10, 10, 9, 9, 13, 11, 10, 13, 11, 11, 12, 10, 10, 14, 14, 12, 12, 15, 13, 13, 13, 12, 14, 14, 15, 14, 13, 19, 13, 13, 15, 11, 13, 13, 15, 16, 17, 19, 16, 16, 15, 17, 15, 15, 17, 17, 16, 20, 16, 16, 20, 17, 19, 17, 18, 20, 17, 21, 18

Offset: 1

Zhi-Wei Sun, Apr 01 2015

nonn

Conjecture: For any integer m > 2, every positive integer can be written as the sum of two distinct elements of the set {floor(p/m): p is prime}.

Note that Goldbach's conjecture essentially asserts that any integer n > 1 can be written as floor(p/2) + floor(q/2) with p and q prime.

			 a(4) = 2 since 4 = 0 + 4 = 1 + 3 with 0,1,3,4 elements of the set {floor(p/3): p is prime}. Note that floor(2/3) = 0, floor(3/3) = 1, floor(11/3) = 3 and floor(13/3) = 4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000040, A002375, A256544.

S[n_]:=Union[Table[Floor[Prime[k]/3], {k,1,PrimePi[3n+2]}]]
L[n_]:=Length[S[n]]
Do[r=0;Do[If[Part[S[n],x]>=n/2,Goto[cc]];
If[MemberQ[S[n], n-Part[S[n],x]]==True,r=r+1]; Continue,{x,1,L[n]}];Label[cc];Print[n," ",r];Continue, {n,1,100}]

A256558 Number of ways to write n = p + floor(k*(k+1)/4), where p is a prime and k is a positive integer.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 01 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 1.
(ii) For any integer m > 4 not equal to 12, each integer n > 1 can be written as p + floor((k^2-1)/m), where p is a prime and k is a positive integer.
We also have some other conjectures on representations n = p + floor(k*(k+1)/m) with m > 4.

			 a(15) = 1 since 15 = 5 + floor(6*7/4) with 5 prime.
a(420) = 1 since 420 = 419 + floor(2*3/4) with 419 prime.
a(945) = 1 since 945 = 877 + floor(16*17/4) with 877 prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, On sums of primes and triangular numbers, arXiv:0803.3737 [math.NT], 2008-2009.
Zhi-Wei Sun, On sums of primes and triangular numbers, J. Comb. Number Theory 1(2009), 65-76.

Cf. A000040, A000217, A132399, A256544, A256555.

Do[r=0;Do[If[PrimeQ[n-Floor[k(k+1)/4]],r=r+1],{k,1,(Sqrt[16n+1]-1)/2}];Print[n," ",r];Continue,{n,1,100}]

A256587 Number of ways to write n = r + s + t, where r,s,t are elements of the set {floor(k*(k+1)/4): k = 1,2,3,...} with s odd and r <= s <= t.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 02 2015

nonn

Conjecture: (i) a(n) > 0 for all n > 1.
(ii) If the ordered pair (b,c) is among (1,2), (1,3), (1,4), (1,5), (1,6), (1,8), (1,9), (2,2) and (2,3), then each nonnegative integer can be written as r + b*s + c*t, where r,s,t belong to the set S = {floor(k*(k+1)/4): k = 1, 2, 3, ...}.
We have shown that if b and c are positive integers with b <= c such that every n = 0,1,2,... can be written as r + b*s + c*t with r,s,t in the above set S, then the ordered pair (b,c) must be among (1,1), (1,2,), (1,3), (1,4), (1,5), (1,6), (1,8), (1,9), (2,2) and (2,3).

			 a(27) = 1 since 27 = 0 + 5 + 22 = floor(1*2/4) + floor(4*5/4) + floor(9*10/4).
a(56) = 1 since 56 = 1 + 3 + 52 = floor(2*3/4) + floor(3*4/4) + floor(14*15/4).

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000217, A256544, A256558.

S[n_]:=Union[Table[Floor[k*(k+1)/4],{k,1,(Sqrt[16n+13]-1)/2}]]
L[n_]:=Length[S[n]]
Do[r=0;Do[If[Part[S[n],x]>n/3,Goto[cc]];Do[If[Part[S[n],x]+2*Part[S[n],y]>n,Goto[bb]];
If[Mod[Part[S[n],y],2]==1&&MemberQ[S[n], n-Part[S[n],x]-Part[S[n],y]]==True,r=r+1];
Continue,{y,x,L[n]}];Label[bb];Continue,{x,1,L[n]}];Label[cc];Print[n," ",r];Continue, {n,1,100}]

cp's OEIS Frontend

A256555 Number of ways to write n as the sum of two (unordered) distinct elements of the set {floor(p/3): p is prime}.

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A256558 Number of ways to write n = p + floor(k*(k+1)/4), where p is a prime and k is a positive integer.

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A256587 Number of ways to write n = r + s + t, where r,s,t are elements of the set {floor(k*(k+1)/4): k = 1,2,3,...} with s odd and r <= s <= t.

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