本帖最后由 王守恩 于 2022-12-13 06:13 编辑
谢谢 northwolves!整理一下,宝贵的资料。开了眼。谢谢 northwolves!
偶数分拆为 2 个素数之和, 恰好有 n 种分拆方式,我们取其中最小的那个数。
a(1)=04: 2+02,
a(2)=10: 3+07, 05+05,
a(3)=22: 3+19, 05+17, 11+11,
a(4)=34: 3+31, 05+29, 11+23, 17+17,
a(5)=48: 5+43, 07+41, 11+37, 17+31, 19+29,
a(6)=60: 7+53, 13+47, 17+43, 19+41, 23+37, 29+31,
a(7)=78: 5+73, 07+71, 11+67, 17+61, 19+59, 31+47, 33+43,
A023036 偶数分拆为 2 个素数之和, 恰好有 n 种分拆方式,我们取其中最小的那个数。
4, 10, 22, 34, 48, 60, 78, 84, 90, 114, 144, 120, 168, 180, 234, 246, 288, 240, 210, 324, 300, 360, 474, 330, 528, 576, 390, 462, 480, 420, 570, 510, 672, 792, 756, 876, 714,
798, 690, 1038, 630, 1008, 930, 780, 960, 870, 924, 900, 1134, 1434, 840, 990, 1302, 1080, 1230, 1518, 1050, 1140, 1386, 1290, 1380, 1554, 1896, 1596, 1620, 1320, 1500,
1260, 1530, 1848, 1590, 1560, 1470, 2508, 1800, 1650, 2244, 1710, 2160, 2652, 1920, 1980, 1680, 2010, 2040, 2250, 2478, 2688, 2838, 2220, 1890, 2280, 2460, 2340, 2580,
3534, 2100, 2610, 2550, 2700, 2772, 3066, 3498, 2790, 3366, 4002, 2640, 3588, 2760, 2850, 3978, 2520, 3060, 2310, 3390, 2970, 3120, 4032, 3420, 3300, 3948, 2940, ......
f[n_] :=Length@Select[2n - Prime@Range@PrimePi@n, PrimeQ]; nn=100; t=Table[0, {nn}]; k=1; cnt=0; While[cnt < nn, a=f@k; If[a <=nn && t[[a]]==0, t[[a]] =2k; cnt++]; k++]; t
A109679 偶数分拆为 2 个素数之和, 可以有 n 种分拆方式,我们取其中最小的那个数。
4, 10, 22, 34, 48, 60, 78, 84, 90, 114, 120, 168, 180, 210, 300, 330, 390, 420, 510, 630, 780, 840, 990, 1050, 1140, 1260, 1470, 1650, 1680, 1890, 2100, 2310, 2730, 3150, 3570,
3990, 4200, 4410, 4620, 5250, 5460, 6090, 6510, 6930, 7980, 8190, 9030, 9240, 10290, 10710, 10920, 11550, 13020, 13650, 13860, 15330, 15540, 15960, 16170, 17850, ......
f[n_] := Length[Select[n - Prime@Range@PrimePi[n/2], PrimeQ]]; t = {}; mxm = -1; Do[If[f[n] > mxm, AppendTo[t, n]; mxm = f[n]], {n, 4, 12000, 2}]; t
偶数分拆为 2 个素数之和, 可以有 A 种分拆方式。这里A=18
240, 366, 372, 560, 620, 640, 704, 740, 746, 772, 784, 812, 836, \842, 860, 872, 886, 916, 926, 928, 938, 944, 964, 1004, 1006, 1016, 1022, 1028, 1052, 1088, 1096, 1142, 1172, 1238, 1412,...
f[n_] := Count[ IntegerPartitions[n, {2}], _?(And @@ PrimeQ[#] &)]; Select[Range[1000], f[#] == A &]
a(3)=1: 3+03,
a(4)=1: 3+05,
a(5)=2: 3+07, 5+05,
a(6)=1: 5+07,
a(7)=2: 3+11, 7+07,
a(8)=2: 3+13, 5+11,
a(9)=2: 5+13, 7+11,
{1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 3, 2, 3, 2, 4, 4, 2, 3, 4, 3, 4, 5, 4, 3, 5, 3, 4, 6, 3, 5, 6, 2, 5, 6, 5, 5, 7, 4, 5, 8, 5, 4, 9, 4, 5, 7, 3, 6, 8, 5, 6, 8, 6, 7, 10, 6, 6, 12, 4, 5, 10, 3, 7, 9, 6, 5, 8, 7,
8, 11, 6, 5, 12, 4, 8, 11, 5, 8, 10, 5, 6, 13, 9, 6, 11, 7, 7, 14, 6, 8, 13, 5, 8, 11, 7, 9, 13, 8}
Table[Count[IntegerPartitions[2 n, {2}], _?(AllTrue[#, PrimeQ] && FreeQ[#, 2] &)], {n, 3, 100}] |