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On convolutions of Euler numbers
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On convolutions of Euler numbers
classification
math.NT
math.CO
keywords
eulerintegernumbersprimeconvolutionsdependingmoreovernumber
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We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive integer n and prime number p>2n+1 we have $$\sum_{k=0}^{p-1+2n}E_kE_{p-1+2n-k}=s(n) (mod p)$$ where s(n) is an integer only depending on n.
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