A379744 -id:A379744 - cp's OEIS frontend

A380729 Smallest n-digit number e such that there exists a primitive Pythagorean n-digit quintuple (a,b,c,d,e) with 10^(n-1) <= a < b < c < d < e < 10^n.

9, 27, 215, 2035, 20095, 200287, 2000851, 20002663, 200008317, 2000025997, 20000082213, 200000259021, 2000000817463, 20000002584459, 200000008167303, 2000000025828219, 20000000081661683, 200000000258208463, 2000000000816541333

Offset: 1

Jean-Marc Rebert, Jan 31 2025

From David A. Corneth, Feb 01 2025: (Start)
Let s1, s2, s3, and s4 be primitive positive distinct integers such that s1^2 + s2^2 + s3^2 + s4^2 = S^2. As squares are 0 or 1 (mod 4) and the quintuple (s1, s2, s3, s4, S) is primitive they cannot all be even. Hence at least one of s1, s2, s3, s4 must be odd. Without loss of generality let s4 be odd. Then s1, s2 and s3 all have the same parity (even or odd).
We may write s1^2 + s2^2 + s3^2 + s4^2 = S^2 as s1^2 + s2^2 + s3^2 = S^2 - s4^2 = (S - s4)*(S + s4) and so look at divisor pairs of s1^2 + s2^2 + s3^2 that multiply to (S - s4)*(S + s4), solve for S and s4 to see if the quintuple (s1, s2, s3, s4, S) meets the criteria for a(n). (End)
[10000005, 10000018, 10000098, 10005204, 20002663] is a Pythagorean 8-digit quintuple, so a(8) <= 20002663.
From David Consiglio, Jr., Mar 05 2025: (Start)
a(9) <= 200008317 [100000000, 100000008, 100000220, 100016405, 200008317];
a(10) <= 2000026127 [1000000000, 1000000004, 1000000457, 1000051792, 2000026127];
a(11) <= 20000082345 [10000000008, 10000000030, 10000001006, 10000163645, 20000082345]. (End)
2e-(a+b+c+d) >= 1 for all quintuples, with equality if e is close to the lower bound. See C++ program for details. - Martin Fuller, Mar 18 2025

			Pythagorean n-digit quintuples in strictly increasing order:
  [2, 4, 5, 6, 9];
  [10, 12, 14, 17, 27];
  [100, 101, 110, 118, 215];
  [1000, 1005, 1008, 1056, 2035];
  [10005, 10006, 10008, 10170, 20095];
  [100000, 100005, 100038, 100530, 200287];
  [1000001, 1000010, 1000040, 1001650, 2000851];
  [10000005, 10000018, 10000098, 10005204, 20002663];
  [100000000, 100000008, 100000220, 100016405, 200008317];
  [1000000005, 1000000020, 1000000240, 1000051728, 2000025997];
  [10000000001, 10000000102, 10000000742, 10000163580, 20000082213];
  [100000000010, 100000000054, 100000001169, 100000516808, 200000259021];
  [1000000000005, 1000000000062, 1000000001382, 1000001633476, 2000000817463];
  [10000000000006, 10000000000050, 10000000003649, 10000005165212, 20000002584459];
  [100000000000037, 100000000000142, 100000000003326, 100000016331100, 200000008167303];
  [1000000000000041, 1000000000000150, 1000000000012304, 1000000051643942, 2000000025828219];
  [10000000000000018, 10000000000000210, 10000000000017809, 10000000163305328, 20000000081661683];
  [100000000000000146, 100000000000000309, 100000000000013904, 100000000516402566, 200000000258208463];
  [1000000000000000210, 1000000000000000482, 1000000000000066436, 1000000001633015537, 2000000000816541333]

Martin Fuller, C++ program
Sean A. Irvine, Java program (github)
Eric Weisstein's World of Mathematics, Pythagorean Quadruple.

Cf. A096910, A379744.

Java
```
// See Links.
```

From Martin Fuller, Mar 16 2025: (Start)
a(n) > 2*10^(n-1) + ((2/3)*10^(n-1))^0.5.
  n even: a(n) > 2*10^(n-1) + 10^(n/2-1) * 2.5819888...
  n odd:  a(n) > 2*10^(n-1) + 10^((n-1)/2-1) * 8.1649658...
See proof in the C++ program. (End)

a(5) corrected by Jinyuan Wang, Feb 25 2025
a(8)-a(9) confirmed by Sean A. Irvine, Mar 06 2025
a(10)-a(19) from Martin Fuller, Mar 16 2025

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A380729 Smallest n-digit number e such that there exists a primitive Pythagorean n-digit quintuple (a,b,c,d,e) with 10^(n-1) <= a < b < c < d < e < 10^n.

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