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对给定的正整数 n ,满足 a^2-b^2=n^2 的正整数 a,b 有几组?

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发表于 2022-11-24 08:38 | 显示全部楼层 |阅读模式
\(a,b是正整数,满足 a^2-b^2=n^2,n=1,2,3,4,5,6,7,8,9,...\)

\(a(1)=0\)
\(a(2)=0\)
\(a(3)=1: 05^2-04^2\)
\(a(4)=1: 05^2-03^2\)
\(a(5)=1: 13^2-12^2\)
\(a(6)=1: 10^2-08^2\)
\(a(7)=1: 25^2-24^2\)
\(a(8)=2: 10^2-06^2, 17^2-15^2\)
\(a(9)=2: 15^2-12^2, 41^2-40^2\)

得到这样一串数: 0, 0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4, 4, 1, 1, 7, 2, 1,
3, 4, 1, 4, 1, 4, 4, 1, 4, 7, 1, 1, 4, 7, 1, 4, 1, 4, 7, 1, 1, 10, 2, 2, 4, 4, 1, 3, 4, 7, ......

\(\displaystyle a(n)=\sum_{i=1}^{n-1}\bigg(1-\bigg\lceil\frac{i(2  n - i)}{2(n-i)}\bigg\rceil+\bigg\lfloor\frac{i(2  n - i)}{2(n-i)}\bigg\rfloor\bigg)\)

参考 OEIS--A046079,这通项公式可是2020年才有的。


\(a,b是正整数,满足 a^2-b^2=n^3,n=1,2,3,4,5,6,7,8,9,...\)

\(a(1)=0\)
\(a(2)=1: 03^2-01^2\)
\(a(3)=2: 06^2-03^2, 014^2-013^2\)
\(a(4)=2: 10^2-06^2, 017^2-015^2\)
\(a(5)=2: 15^2-10^2, 063^2-062^2\)
\(a(6)=4: 15^2-03^2, 021^2-015^2, 029^2-025^2, 055^3-053^2\)
\(a(7)=2: 28^2-21^2, 172^2-171^2\)
\(a(8)=4: 24^2-08^2, 036^2-028^2, 066^2-062^2, 129^3-127^2\)
\(a(9)=3: 45^2-36^2, 123^2-120^2, 365^2-364^2\)

得到这样一串数: 0, 1, 2, 2, 2, 4, 2, 4, 3, 4, 2, 10, 2, 4, 8, 5, 2, 7, 2, 10, 8, 4, 2, 16, 3, 4,
5, 10, 2, 16, 2, 7, 8, 4, 8, 17, 2, 4, 8, 16, 2, 16, 2, 10, 14, 4, 2, 22, 3, 7, 8, 10, 2, ......

OEIS--还没有这串数,好奇的网友!你来试试?
 楼主| 发表于 2022-11-24 20:07 | 显示全部楼层
\(a,b是正整数,满足 a^2-b^2=n^3,n=1,2,3,4,5,6,7,8,9,...\)

谢谢 mathe 给出\(a^2-b^2=n^3\) 解数目的通项公式。谢谢 mathe!

Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^3]/2], {n, 1, 130}]

{0, 1, 2, 2, 2, 4, 2, 4, 3, 4, 2, 10, 2, 4, 8, 5, 2, 7, 2, 10, 8, 4, 2, 16, 3, 4, 5, 10, 2, 16,  2, 7, 8, 4, 8,
17, 2, 4, 8, 16, 2, 16, 2, 10, 14, 4, 2, 22, 3, 7, 8, 10, 2, 10, 8, 16, 8, 4, 2, 40, 2, 4, 14, 8,  8, 16, 2,
10, 8, 16, 2, 28, 2, 4, 14, 10, 8, 16, 2, 22, 6, 4, 2, 40, 8, 4, 8, 16, 2, 28, 8, 10, 8, 4, 8, 28, 2, 7, 14,
17, 2, 16, 2, 16, 32, 4, 2, 25, 2, 16, 8, 22, 2, 16, 8, 10, 14, 4, 8, 64, 3, 4, 8, 10, 5, 28, 2, 10, 8, 16}
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 楼主| 发表于 2022-11-25 10:10 | 显示全部楼层
\(接a^2-b^2=n^3,在相同方式中,我们取最小的那个数。\)

\(a(1)=02: 0003^2-001^2,\)
\(a(2)=03: 0006^2-003^2, 0014^2-0013^2,\)
\(a(3)=09: 0045^2-036^2, 0123^2-0120^2, 0365^2-0364^2,\)
\(a(4)=06: 0015^2-003^2, 0021^2-0015^2, 0029^2-0025^2, 00055^2-00053^2,\)
\(a(5)=16: 0080^2-048^2, 0136^2-0120^2, 0260^2-0252^2, 00514^2-00510^2, 01025^2-01023^2,\)
\(a(6)=81: 1215^2-972^2, 3321^2-3240^2, 9855^2-9828^2, 29529^2-29520^2, 88575^2-88572^2, 265721^2-265720^2,\)
\(a(7)=18: 0081^2-027^2, 0099^2-0063^2, 0171^2-0153^2, 00249^2-00237^2, 00489^2-00483^2, 000731^2-000727^2, 1459^2-1457^2,\)
\(a(8)=15: 0060^2-015^2, 0076^2-0049^2, 0080^2-0055^2, 00120^2-00105^2, 00192^2-00183^2, 000340^2-000335^2, 0564^2-0561^2, 1688^2-1687^2,\)

2, 3, 9, 6, 16, 81, 18, 15,729, 12, ?, 6561,162, 45,59049, 24, 36,531441, ?, 135,4782969, 48, ?, 225,108, ....

这些乱七八糟的数,好像是没有规律,
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 楼主| 发表于 2022-11-25 16:42 | 显示全部楼层
     谢谢 mathe!       谢谢 northwolves!         一并给出!

(2): a^2-b^2=n^2,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^2]/2], {n, 1,130}]
{0, 0, 1, 1, 1, 1, 1, 2, 2, 1, 1, 4, 1, 1, 4, 3, 1, 2, 1, 4, 4, 1, 1, 7, 2, 1, 3, 4, 1, 4, 1, 4, 4, 1, 4, 7, 1, 1, 4,7, 1, 4, 1, 4,

(3): a^2-b^2=n^3,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^3]/2], {n, 1, 130}]
{0, 1, 2, 2, 2, 4, 2, 4, 3, 4, 2, 10, 2, 4, 8, 5, 2, 7, 2, 10, 8, 4, 2, 16, 3, 4, 5, 10, 2, 16, 2, 7, 8, 4, 8, 17, 2, 4, 8,16,

(4): a^2-b^2=n^4,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^4]/2], {n, 1, 130}]
{0, 1, 2, 3, 2, 7, 2, 5, 4, 7, 2, 17, 2, 7, 12, 7, 2, 13, 2, 17, 12, 7, 2, 27, 4, 7, 6, 17, 2, 37, 2, 9, 12, 7, 12, 31, 2, 7,

(5): a^2-b^2=n^5,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^5]/2], {n, 1, 130}]
{0, 2, 3, 4, 3, 12, 3, 7, 5, 12, 3, 27, 3, 12, 18, 9, 3, 22, 3, 27, 18, 12, 3, 42, 5, 12, 8, 27, 3, 72, 3, 12, 18, 12,

(6): a^2-b^2=n^6,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^6]/2], {n, 1, 130}]
{0, 2, 3, 5, 3, 17, 3, 8, 6, 17, 3, 38, 3, 17, 24, 11, 3, 32, 3, 38, 24, 17, 3, 59, 6, 17, 9, 38, 3, 122, 3, 14, 24, 17,

(7): a^2-b^2=n^7,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^7]/2], {n, 1, 130}]
{0, 3, 4, 6, 4, 24, 4, 10, 7, 24, 4, 52, 4, 24, 32, 13, 4, 45, 4, 52, 32, 24, 4, 80, 7, 24, 11, 52, 4, 192, 4, 17, 32,

(8): a^2-b^2=n^8,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^8]/2], {n, 1, 130}]
{0, 3, 4, 7, 4, 31, 4, 11, 8, 31, 4, 67, 4, 31, 40, 15, 4, 59, 4, 67, 40, 31, 4, 103, 8, 31, 12, 67, 4, 283, 4, 19, 40,

(9): a^2-b^2=n^9,Table[Floor[DivisorSigma[0, GCD[((n - 1)/2)^2, ((n + 1)/2)^2] n^9]/2], {n, 1, 130}]
{0, 4, 5, 8, 5, 40, 5, 13, 9, 40, 5, 85, 5, 40, 50, 17, 5, 76, 5, 85, 50, 40, 5, 130, 9, 40, 14, 85, 5, 400, 5, 22, 50,
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 楼主| 发表于 2022-11-26 13:02 | 显示全部楼层
         A100073           \(a^2-b^2=n\) 补一个通项公式

Table[Floor[DivisorSigma[0, n/4]/2] (Floor[n/4] - Floor[(n - 1)/4])
  + Floor[DivisorSigma[0, n]/2] (Floor[(n - 1)/2] - Floor[(n - 2)/2]), {n, 1, 200}]

{0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 2, 1, 1, 0, 1, 1, 2, 0, 1, 2, 1, 0, 2, 1, 1, 0, 1, 2, 2, 0, 2, 1, 1, 0, 2,
  2, 1, 0, 1, 1, 3, 0, 1, 3, 1, 0, 2, 1, 1, 0, 2, 2, 2, 0, 1, 2, 1, 0, 3, 2, 2, 0, 1, 1, 2, 0, 1, 3, 1, 0, 3, 1, 2, 0,
  1, 3, 2, 0, 1, 2, 2, 0, 2, 2, 1, 0, 2, 1, 2, 0, 2, 4, 1, 0, 3, 1, 1, 0, 1, 2, 4, 0, 1, 2, 1, 0, 2, 3, 1, 0, 2, 1, 3,
  0, 2, 4, 1, 0, 2, 1, 2, 0, 1, 3, 2, 0, 1, 2, 2, 0, 4, 2, 1, 0, 1, 2, 2, 0, 2, 4, 2, 0, 3, 1, 1, 0, 1, 2, 3, 0, 2, 2,
  1, 0, 2, 4, 2, 0, 1, 1, 4, 0, 1, 4, 1, 0, 3, 1, 1, 0, 3, 3, 2, 0, 1, 3, 1, 0, 2, 2, 2, 0, 2, 1, 4, 0, 1, 5, 1, 0, ...}

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