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}]

复制代码

王守恩

northwolves 发表于 2025-9-27 09:29

A178501——Zero followed by powers of ten——EXTENSIONS More terms from Elmo R. Oliveira, Jul 21 2025。

1, 10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000, 10000000000, 100000000000, 1000000000000, 10000000000000, 100000000000000, 1000000000000000, 10000000000000000, 100000000000000000,  

还不如——正整数,各个数位上的数码相加再相加再相加再相加, 直到最后是 ”9“。

这样的 1 位数有 1个。
这样的 2 位数有 10个。
这样的 3 位数有 100个。
这样的 4 位数有 1000个。
这样的 5 位数有 10000个。
这样的 6 位数有 100000个。
这样的 7 位数有 1000000个。
这样的 8 位数有 10000000个。
这样的 9 位数有 100000000个。

王守恩

Table[With[{t = n^3}, # + t/# &@ First@Nearest[Divisors[t], Sqrt[t]]], {n, 154}]——这通项公式蛮好的。
{2, 6, 12, 16, 30, 30, 56, 48, 54, 65, 132, 84, 182, 105, 120, 128, 306, 153, 380, 180, 210, 209, 552, 236, 250, 273, 324, 308, 870, 330, 992, 384, 418, 425, 420, 432, 1406, 513, 520, 506, 1722, 546, 1892, 594, 618, 713,
2256, 672, 686, 750, 748, 754, 2862, 810, 880, 840, 874, 1073, 3540, 930, 3782, 1209, 1008, 1024, 1170, 1078, 4556, 1122, 1150, 1186, 5112, 1224, 5402, 1665, 1300, 1330, 1386, 1378, 6320, 1440, 1458, 2009, 6972, 1540,
1870, 2193, 1624, 1672, 8010, 1710, 1820, 1794, 1798, 2585, 2280, 1888, 9506, 2058, 1980, 2000, 10302, 2074, 10712, 2184, 2154, 3233, 11556, 2268, 11990, 2310, 2368, 2396, 12882, 2451, 3220, 2610, 2574, 3953, 2856,
2630, 2662, 4209, 2788, 2914, 3750, 2830, 16256, 3072, 3010, 2977, 17292, 3036, 3458, 5025, 3240, 3332, 18906, 3243, 19460, 3315, 3478, 5609, 3432, 3456, 4466, 5913, 3724, 3737, 22350, 3675, 22952, 3876, 3978, 3850}

补充内容 (2025-10-14 08:03):
Table[With[{t = Fibonacci[n]}, # + t/# &@First@Nearest[Divisors[t], Sqrt[t]]], {n, 64}]

王守恩

A032742——a(1) = 1; for n > 1, a(n) = largest proper divisor of n (that is, for n>1, maximum divisor d of n in range 1 <= d < n).

{1, 1, 2, 1, 3, 1, 4, 3, 5, 1, 6, 1, 7, 5, 8, 1, 9, 1, 10, 7, 11, 1, 12, 5, 13, 9, 14, 1, 15, 1, 16, 11, 17, 7, 18, 1, 19, 13, 20, 1, 21, 1, 22, 15, 23, 1, 24, 7, 25}

Table[Divisors[n][[-2]], {n, 2, 50}]——好！！！

王守恩

A048611——Find smallest pair (x,y) such that x^2 - y^2 = 11...1 (n times) = (10^n-1)/9; sequence gives value of x.

{1, 6, 20, 56, 156, 340, 2444, 4440, 167000, 55556, 267444, 333400, 132687920, 5555556, 10731400, 40938800, 2682647040, 333334000, 555555555555555556, 3334367856}

Table[x /. Solve[{x^2 - y^2 == FromDigits[PadRight[{}, n, 1]]}, {x, y}, PositiveIntegers][[1]], {n, 2, 20}]

Table[Module[{d = Divisors[(10^n - 1)/9]}, d[[Length[d]/2 + 1]] + d[[Length[d]/2]]]/2, {n, 2, 20}]——Apr 05 2021——这个还是快一些。

A048612——Find smallest pair (x,y) such that x^2-y^2 = 11...1 (n times) = (10^n-1)/9; sequence gives value of y.

{0, 5, 17, 45, 115, 67, 2205, 2933, 166667, 44445, 245795, 6667, 132683733, 4444445, 2012917, 23767083, 2680575317, 666667, 555555555555555555, 83053525}

Table[y /. Solve[{x^2 - y^2 == FromDigits[PadRight[{}, n, 1]]}, {x, y}, PositiveIntegers][[1]], {n, 2, 20}]

Table[Module[{d = Divisors[(10^n - 1)/9]}, d[[Length[d]/2 + 1]] - d[[Length[d]/2]]]/2, {n, 2, 20}]——Apr 05 2021——这个还是快一些。

仿造2个(A,B)——OEIS就没有了。——这样的数字串不好找——你先得都有解——如果有“0”——就没味了。

{10, 100, 1000, 1432, 100000, 1432, 10000000, 65168, 32258080, 141832, 447380, 260894168, 8436340, 4347826086968, 9174311926660, 154374400, 455329520, 2526889600, 5816341160}——A
{09, 099, 0999, 1425, 099999, 0225, 09999999, 63615, 32258049, 010785, 012201, 260890335, 7153449, 4347826086945, 9174311926551, 061898751, 085586049, 2094080001, 3718847199}——B

Table[Module[{d = Divisors[10^n/5 - 1]}, d[[Length[d]/2 + 1]] + d[[Length[d]/2]]]/2, {n, 2, 20}]——A
Table[Module[{d = Divisors[10^n/5 - 1]}, d[[Length[d]/2 + 1]]  - d[[Length[d]/2]]]/2, {n, 2, 20}]——B

王守恩

A034797_——a(0) = 0; a(n+1) = a(n) + 2^a(n).——0, 1, 3, 11, 2059, ————长得太快了！！！！！！

a(0) = 0;

a(1) = 0 + 2^0 = 1;

a(2) = 1 + 2^1 = 3;

a(3) = 3 + 2^3 = 11;

a(4) = 11 + 2^11 = 2059;

a(5) = 2059 + 2^2059 = 6618522843404494295186406745839606161498952226757731129780294743557049372440144054926786849079892677363449438396804714392395685714020
54064027405360874460838310520368482324399959044049927980075147183260434105703798308704637800852606194444172051991971237512107049703527278337554258761027760
28267313405809429548880554782040765277562828362884238325465448520348307574943345990309941642666926723379729598185834735054732500415409883868361423159913770
8122187727119017722495531534022877597895171217443367553504659016552051849173709742024055869412110653955407655676631932971733672542303136122441829419995004023881954500530803855470;