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

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

1,3

Comments

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.

References

Zhi-Wei Sun, On sums of primes and triangular numbers, J. Comb. Number Theory 1(2009), 65-76.

Links

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.

Example

 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.

Mathematica

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}]

Crossrefs

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

Sequence in context: A135551 A065531 A366574 * A239281 A024936 A144590

Adjacent sequences: A256555 A256556 A256557 * A256559 A256560 A256561

Keyword

nonn

Author

Zhi-Wei Sun, Apr 01 2015

Status

approved