切手問題
( A , B ) = { ( 22 9 , 0 ) ( n ≡ 0 ( mod 9 ) ) , ( 68 27 , 62 81 ) ( n ≡ 1 ( mod 9 ) ) , ( 71 27 , − 26 81 ) ( n ≡ 2 ( mod 9 ) ) , ( 23 9 , 0 ) ( n ≡ 3 ( mod 9 ) ) , ( 68 9 , − 154 81 ) ( n ≡ 4 ( mod 9 ) ) , ( 68 27 , 46 81 ) ( n ≡ 5 ( mod 9 ) ) , ( 23 9 , − 1 ) ( n ≡ 6 ( mod 9 ) ) , ( 71 27 , − 1 81 ) ( n ≡ 7 ( mod 9 ) ) , ( 68 27 , − 170 81 ) ( n ≡ 8 ( mod 9 ) ) {\displaystyle (A,B)={\begin{cases}\left({\frac {22}{9}},0\right)&(n\equiv 0{\pmod {9}}),\\\left({\frac {68}{27}},{\frac {62}{81}}\right)&(n\equiv 1{\pmod {9}}),\\\left({\frac {71}{27}},-{\frac {26}{81}}\right)&(n\equiv 2{\pmod {9}}),\\\left({\frac {23}{9}},0\right)&(n\equiv 3{\pmod {9}}),\\\left({\frac {68}{9}},-{\frac {154}{81}}\right)&(n\equiv 4{\pmod {9}}),\\\left({\frac {68}{27}},{\frac {46}{81}}\right)&(n\equiv 5{\pmod {9}}),\\\left({\frac {23}{9}},-1\right)&(n\equiv 6{\pmod {9}}),\\\left({\frac {71}{27}},-{\frac {1}{81}}\right)&(n\equiv 7{\pmod {9}}),\\\left({\frac {68}{27}},-{\frac {170}{81}}\right)&(n\equiv 8{\pmod {9}})\end{cases}}}
となる 。
3種類を適切に選ぶと、 h 枚の切手での解は最大で
3, 8, 15, 26, 35, 52, 69, 89, 112, 146, 172, 212, 259, 302, 354, 418, 476, 548, 633, 714, 805, 902, 1012, 1127, 1254, 1382,... (オンライン整数列大辞典の数列 A001208)
となる。
n ≥ 20 のとき
β = ⌊ 4 h + 4 9 ⌋ + 2 , γ = ⌊ 2 9 h ⌋ + 2 {\displaystyle \beta =\left\lfloor {\frac {4h+4}{9}}\right\rfloor +2,\gamma =\left\lfloor {\frac {2}{9}}h\right\rfloor +2}
とおくと
a 1 = 1 , a 2 = 2 β − γ + 1 , a 3 = γ a 2 − β {\displaystyle a_{1}=1,a_{2}=2\beta -\gamma +1,a_{3}=\gamma a_{2}-\beta }
が h 枚の切手での最大の解を与え、その最大の解は