A354501 - OEIS

%I #14 Aug 15 2022 23:32:47

%S 82,9,106,213,48,54,165,56,191,64,163,158,129,243,215,251,124,227,57,

%T 130,155,47,255,135,52,142,67,68,196,222,233,203,84,123,148,50,166,

%U 194,35,61,238,76,149,11,66,250,195,78,8,46,161,102,40,217,36,178,118,91,162,73,109

%N The inverse Rijndael S-box used in the Advanced Encryption Standard (AES); inverse permutation of A354500.

%H Jianing Song, <a href="/A354501/b354501.txt">Table of n, a(n) for n = 0..255</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Rijndael_S-box">Rijndael S-box</a>

%F a(n) = ivgenpoly((((x^6+x^3+x)*genpoly(n) + x^2 + 1) mod (x^8+1))^254 mod (x^8+x^4+x^3+x+1)), where ivgenpoly and genpoly are the notations introduced in A355891. Beware that all the operations are done in GF(2)[x].

%F To be more concretely, to obtain a(n):

%F - Write the binary expansion of n and view it as a polynomial p(x) in GF(2)[x]; (E.g., 103 = 1100111_2 => x^6 + x^5 + x^2 + x + 1)

%F - Compute q(x) = ((x^6+x^3+x)*p(x) + x^2 + 1) mod (x^8+1) in GF(2)[x]; (E.g., x^6 + x^5 + x^2 + x + 1 => x^5 + x^3 + 1)

%F - Compute r(x) = q(x)^254 mod (x^8+x^4+x^3+x+1) in GF(2)[x]; (E.g., x^5 + x^3 + 1 => x^3 + x)

%F - To get a(n), view r(x) as a number. (E.g., x^3 + x => 2^3 + 2 = 10)

%F This is the inverse to the process described in A354500.

%e The inverse Rijndael S-box written in hexadecimal:

%e      +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +A +B +C +D +E +F

%e   00 52 09 6A D5 30 36 A5 38 BF 40 A3 9E 81 F3 D7 FB

%e   10 7C E3 39 82 9B 2F FF 87 34 8E 43 44 C4 DE E9 CB

%e   20 54 7B 94 32 A6 C2 23 3D EE 4C 95 0B 42 FA C3 4E

%e   30 08 2E A1 66 28 D9 24 B2 76 5B A2 49 6D 8B D1 25

%e   40 72 F8 F6 64 86 68 98 16 D4 A4 5C CC 5D 65 B6 92

%e   50 6C 70 48 50 FD ED B9 DA 5E 15 46 57 A7 8D 9D 84

%e   60 90 D8 AB 00 8C BC D3 0A F7 E4 58 05 B8 B3 45 06

%e   70 D0 2C 1E 8F CA 3F 0F 02 C1 AF BD 03 01 13 8A 6B

%e   80 3A 91 11 41 4F 67 DC EA 97 F2 CF CE F0 B4 E6 73

%e   90 96 AC 74 22 E7 AD 35 85 E2 F9 37 E8 1C 75 DF 6E

%e   A0 47 F1 1A 71 1D 29 C5 89 6F B7 62 0E AA 18 BE 1B

%e   B0 FC 56 3E 4B C6 D2 79 20 9A DB C0 FE 78 CD 5A F4

%e   C0 1F DD A8 33 88 07 C7 31 B1 12 10 59 27 80 EC 5F

%e   D0 60 51 7F A9 19 B5 4A 0D 2D E5 7A 9F 93 C9 9C EF

%e   E0 A0 E0 3B 4D AE 2A F5 B0 C8 EB BB 3C 83 53 99 61

%e   F0 17 2B 04 7E BA 77 D6 26 E1 69 14 63 55 21 0C 7D

%e The inverse Rijndael S-box written in decimal:

%e        +0  +1  +2  +3  +4  +5  +6  +7  +8  +9 +10 +11 +12 +13 +14 +15

%e     0  82   9 106 213  48  54 165  56 191  64 163 158 129 243 215 251

%e    16 124 227  57 130 155  47 255 135  52 142  67  68 196 222 233 203

%e    32  84 123 148  50 166 194  35  61 238  76 149  11  66 250 195  78

%e    48   8  46 161 102  40 217  36 178 118  91 162  73 109 139 209  37

%e    64 114 248 246 100 134 104 152  22 212 164  92 204  93 101 182 146

%e    80 108 112  72  80 253 237 185 218  94  21  70  87 167 141 157 132

%e    96 144 216 171   0 140 188 211  10 247 228  88   5 184 179  69   6

%e   112 208  44  30 143 202  63  15   2 193 175 189   3   1  19 138 107

%e   128  58 145  17  65  79 103 220 234 151 242 207 206 240 180 230 115

%e   144 150 172 116  34 231 173  53 133 226 249  55 232  28 117 223 110

%e   160  71 241  26 113  29  41 197 137 111 183  98  14 170  24 190  27

%e   176 252  86  62  75 198 210 121  32 154 219 192 254 120 205  90 244

%e   192  31 221 168  51 136   7 199  49 177  18  16  89  39 128 236  95

%e   208  96  81 127 169  25 181  74  13  45 229 122 159 147 201 156 239

%e   224 160 224  59  77 174  42 245 176 200 235 187  60 131  83 153  97

%e   240  23  43   4 126 186 119 214  38 225 105  20  99  85  33  12 125

%o (PARI) m(P) = Mod(P, 2);

%o A354501(n) = subst(lift(lift(Mod(lift(Mod(m(x^6+x^3+x)*Pol(binary(n))+m(x^2+1), m(x^8+1))), m(x^8+x^4+x^3+x+1))^254)), x, 2)

%Y Cf. A354500, A355891.

%K nonn,easy,fini,full

%O 0,1

%A _Jianing Song_, Aug 15 2022