By Florentin Smarandache

Articles, notes, generalizations, paradoxes, miscellaneous in arithmetic, linguistics, and schooling.

**Read Online or Download Collected Papers, Vol. I, second edition PDF**

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**Additional resources for Collected Papers, Vol. I, second edition**

**Example text**

K j takes values from 0 to at most m j − 1 ; 4 4 m k3 takes values from 0 to m3 − 1 = −1= − 1 = − 1 = 1; (2, 0, 0) 2 d3 ⎛ x ≡ 3k1 + 3k2 + 3(mod 4) ⎞ ⎜ ⎟ + 1(mod 4) ⎟ ; k3 = 0 ⇒ ⎜ y ≡ 2k1 ⎜z ≡ k2 (mod 4) ⎟⎠ ⎝ ⎛ 3k1 + 3k2 + 1⎞ ⎜ ⎟ +1⎟ k3 = 1 ⇒ ⎜ 2k1 ⎜ k2 ⎟⎠ ⎝ k1 takes values from 0 to at most 3. ⎛ 3k2 + 3 ⎞ ⎛ 3k2 + 1 ⎞ ⎛ 3k2 + 2 ⎞ ⎛ 3k2 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ k1 = 0 ⇒ ⎜ 1⎟ , ⎜ 1⎟ ; k1 = 1 ⇒ ⎜ 3⎟ , ⎜ 3 ⎟ ; ⎜ k ⎟ ⎜ ⎜ k ⎟ ⎜k ⎟ ⎟ ⎝ 2 ⎠ ⎝ k2 ⎝ 2 ⎠ ⎝ 2⎠ ⎠ for k1 = 2 and 3 we obtain the same expressions as for k1 = 1 and 0.

Pp. 41-47. [Published in “GAMMA”, Braşov, Anul X, No.. 1-2, October 1987, p. 8] 37 ON A THEOREM OF WILSON §1. In 1770 Wilson found the following result in the Number’s Theory: “If p is prime, then ( p − 1)! ≡ (−1mod p )”. Did you ever question yourself what happens if the module m is not anymore prime? It’s simple, one answers, “if m is not prime and m ≠ 4 then (m − 1)! ≡ 0 (mod m )”; for the proof see [4]. This is fine, I would continue, but if in the product from the left side of this congruence we consider only numbers that are prime with m ?

I =1 n −1 For example, if we note Sn = ∑ g j , we define two ( s + 1) × ( s + 1) matrixes such j =1 that: ⎡1 ⎢S n Bn = ⎢ ⎢ : ⎢ ⎣ Sn− s +1 0 0 ... 0 gn : gn−1 ... gn − s + 2 : ... : gn − s +1 gn − s ... gn− 2 s + 3 ⎤ gn − s +1 ⎥ ⎥, ⎥ : ⎥ gn − 2 s + 2 ⎦ 0 n ≥ 1 , and ⎡1 ⎢1 ⎢ M = ⎢: ⎢ ⎢1 ⎢⎣1 0 0 ... 1 1 ... : : ... 1 0 ... 1 0 ... 0⎤ 0 ⎥⎥ : ⎥, ⎥ 1⎥ 0 ⎥⎦ thus, we have analogously: Bn +1 = M n +1 , M r + c = M r ⋅ M c , whence Sr + c = Sr + gr Sc + gr −1Sc−1 + ... + gr − s +1Sc− s +1 , gr + c = gr gc + gr −1gc−1 + ...