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On the number of representations of certain integers as sums of 11 or 13 squares

机译:关于以11或13平方和表示的某些整数的表示数量

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Let r_k(n) denote the number of representations of an integer n as a sum of k squares. We prove that for odd primes p, r_(11)(p~2) = (330)/(31)(p~9+1) - 22(-1)~((p-1)/2)p~4 + (352)/(31) H(p), where H( p) is the coefficient of qp in the expansion of q ∏_(j=1)~∞(1-(1-q)~j)~(16)(1-q~(2j))~4 + 32q~2 ∏_(j=1)~∞((1-q~(2j))~(28))/(1-(-q)~j)~8. This result, together with the theory of modular forms of half integer weight is used to prove that r_(11)(n) = r_(11)(n')(2~9[λ/2]+9)/(p~9-1)-p~4((-n')/p)(p~9[λ_p/2]-1)/(p~9-1)], wheren = 2~λ ∏_P p~(λp) is the prime factorisation of n, n' is the square-free part of n, and n' is of the form 8k+7. The products here are taken over the odd primes p, and (n/p) is the Legendre symbol. We also prove that for odd primes p, r_(13)(p~2) = (4030)/(691)(p~(11)/+1)-26p~5+(13936)/(691)τ(p), where (n) is Ramanujan's τ function, defined by q ∏_(j=1)~∞ (1-q~j)~(24) = ∑_(n=1)~∞ τ(n)q~n. A conjectured formula for r_(2k+1)(p~2) is given, for general k and general odd primes p.
机译:令r_k(n)表示整数n的表示数目作为k个平方的总和。我们证明对于奇质数p,r_(11)(p〜2)=(330)/(31)(p〜9 + 1)-22(-1)〜((p-1)/ 2)p〜 4 +(352)/(31)H(p),其中H(p)是q ∏_(j = 1)〜∞(1-(1-q)〜j)〜展开时qp的系数(16)(1-q〜(2j))〜4 + 32q〜2 ∏_(j = 1)〜∞((1-q〜(2j))〜(28))/(1-(-q) 〜j)〜8。该结果与半整数权重的模块化形式理论一起用于证明r_(11)(n)= r_(11)(n')(2〜9 [λ/ 2] +9)/(p 〜9-1)-p〜4((-n')/ p)(p〜9 [λ_p/ 2] -1)/(p〜9-1)],其中n = 2〜λ∏_P p〜( λp)是n的素因数分解,n'是n的无平方部分,n'的形式为8k + 7。此处的乘积取于奇质数p,并且(n / p)是Legendre符号。我们还证明对于奇质数p,r_(13)(p〜2)=(4030)/(691)(p〜(11)/ + 1)-26p〜5 +(13936)/(691)τ( p),其中(n)是拉马努詹的τ函数,由q ∏_(j = 1)〜∞(1-q〜j)〜(24)= ∑_(n = 1)〜∞τ(n)q定义〜n。对于一般k和一般奇质数p,给出了r_(2k + 1)(p〜2)的猜想公式。

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