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A303932
Number of ways to write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic residue modulo p, and k and m are nonnegative integers.
11
0, 1, 1, 1, 3, 4, 2, 3, 3, 1, 3, 5, 2, 1, 4, 2, 1, 4, 3, 4, 4, 2, 3, 7, 4, 2, 6, 3, 2, 4, 4, 3, 3, 2, 4, 6, 2, 1, 6, 2, 2, 6, 5, 6, 5, 5, 6, 8, 3, 5, 8, 5, 3, 7, 6, 5, 7, 6, 9, 7, 5, 7, 7, 3, 5, 9, 5, 7, 9, 6, 11, 10, 5, 11, 10, 4, 5, 13, 3, 5
OFFSET
1,5
COMMENTS
Conjecture 1. a(n) > 0 for all n > 1, i.e., any even number greater than two can be written as the sum of a power 2, a power of 3 and a prime p with 11 a quadratic residue modulo p.
Conjecture 2. For any integer n > 2, we can write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic nonresidue modulo p, and k and m are nonnegative integers.
We have verified Conjectures 1 and 2 for n up to 5*10^8.
LINKS
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)
EXAMPLE
a(2) = 1 since 2*2 = 2 + 2^0 + 3^0 with 11 a quadratic residue modulo the prime 2.
a(3) = 1 since 2*3 = 2 + 2^0 + 3^1 with 11 a quadratic residue modulo the prime 2.
a(10) = 1 since 2*10 = 7 + 2^2 + 3^2 with 11 a quadratic residue modulo the prime 7.
a(14) = 1 since 2*14 = 19 + 2^3 + 3^0 with 11 a quadratic residue modulo the prime 19.
a(17) = 1 since 2*17 = 5 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 5.
a(38) = 1 since 2*38 = 37 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 37.
MATHEMATICA
PQ[n_]:=PQ[n]=n==2||(n>2&&PrimeQ[n]&&JacobiSymbol[11, n]==1);
tab={}; Do[r=0; Do[If[PQ[2n-2^k-3^m], r=r+1], {k, 0, Log[2, 2n-1]}, {m, 0, Log[3, 2n-2^k]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, May 02 2018
STATUS
approved