数字串的通项公式

王守恩

northwolves 发表于 2025-7-27 08:14

A029113——仿造1个。 ——Hoang Xuan Thanh, Jul 09 2025。

前n个自然数，有a(n)个5,7倍数的数。
a(0)=1,
a(1)=1,
a(2)=1,
a(3)=1,
a(4)=1,
a(5)=2,
a(6)=2,
a(7)=3,
a(8)=3,
a(9)=3,
a(10)=4,
a(11)=4,
a(12)=4,
a(13)=4,
a(14)=5,
a(15)=6,
a(16)=6,
a(17)=6,
a(18)=6,
a(19)=6,
a(20)=6,
a(21)=7,
a(22)=7,
a(23)=7,
a(24)=7,
a(25)=8,
a(26)=8,
a(27)=8,
a(28)=9,
a(29)=9,
a(30)=10,
a(31)=10,
a(32)=10,
a(33)=10,
a(34)=10,
a(35)=11,
a(36)=11,
a(37)=11,
a(38)=11,
a(39)=11,
a(40)=12,

{1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 10, 10,
11, 11, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 19, 19, 20, 20, 20, 20, 20,
21, 21, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 24, 24, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 28, 28, 28, 29, 29, 30, 30, 30, 30, 30,
31, 31, 31, 31, 31, 32, 32, 33, 33, 33, 34, 34, 34, 34, 35, 36, 36, 36, 36, 36, 36, 37, 37, 37, 37, 38, 38, 38, 39, 39, 40, 40, 40, 40, 40}
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 10, 10, 11}, 140]

northwolves

1. 
   
2. t = Table[Count[Range[0, n], _?(Divisible[#, 5] || Divisible[#, 7] &)], {n, 0, 139}]
   
3. 
   
4. v = Table[1 + Floor[n/5] + Floor[n/7] - Floor[n/35], {n, 0, 139}]

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