A000378 -id:A000378 - cp's OEIS frontend

A347969 Numbers which are sum of three squares of positive numbers and also 5 times of the sum of their joint products.

1715, 6860, 12635, 15435, 27440, 42875, 47915, 50540, 53235, 61740, 84035, 109760, 113715, 138915, 171500, 191660, 202160, 207515, 212940, 218435, 246960, 289835, 302715, 315875, 329315, 336140, 385875, 415835, 431235, 439040, 454860, 479115, 495635, 555660, 582435, 619115, 686000

Offset: 1

Alexander Kritov, Sep 23 2021

nonn

The general problem is to find such numbers which can be represented as the sum of three squares of integers x, y, z, and additionally also satisfy: x^2 + y^2 + z^2 = k*(x*y + x*z + y*z).
For k=1 it is simply a(n) = 3*n^2 given by A033428.
For k=2 it is A347360.
The present sequence is for the next k=5.
All possible k-numbers are listed by A331605.

			    a(n)      ( x,  y,   z)
  ------      -------------
    1715      ( 3,  5,  41)
    6860      ( 6, 10,  82)
   12635      ( 5, 17, 111)
   15435      ( 9, 15, 123)
   27440      (12, 20, 164)
   42875      (15, 25, 205)
   47915      ( 3, 41, 215)
   50540      (10, 34, 222)
   53235      ( 5, 41, 227)
   61740      (18, 30, 246)
   84035      (21, 35, 287)
  109760      (24, 40, 328)

E. Grosswald, Representations of Integers as Sums of Squares, Springer-Verlag, NY, 1985.

Cf. A000378, A033428, A331605 (all possible k-numbers), A347360.

A076180 Indices of spheres mentioned in A071609.

Original entry on oeis.org

1, 2, 5, 8, 13, 23, 35, 63, 76, 86, 123, 163, 226, 264, 287, 374, 514, 523, 576, 664, 715, 787, 924, 927, 963, 1074, 1137, 1364, 1487, 1576, 1786, 2176, 2475, 2614, 2837, 2877, 2925, 3536, 3576, 4025, 4487, 4638, 5215, 6026, 6527, 6726, 6774

Offset: 1

Hugo Pfoertner, Nov 01 2002

nonn

			a(7)=35 because A000378(35)=A071609(7)=41

J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, Table 4.3, p. 107.

Hugo Pfoertner, Spheres containing record numbers of lattice points of Z^3

Cf. A000378, A005875, A071609.

a(n) = k for which A000378(k) = A071609(n)

A272189 Values of A004215(n) such that A004215(n+1) = A004215(n) + 8 = A004215(n-1) + 16.

Original entry on oeis.org

15, 39, 47, 71, 79, 103, 135, 143, 167, 175, 199, 207, 231, 263, 271, 295, 303, 327, 335, 359, 391, 399, 423, 431, 463, 487, 519, 527, 551, 559, 583, 591, 615, 647, 655, 679, 687, 711, 719, 743, 775, 783, 807, 815, 839, 847, 871, 903, 911, 935, 943, 975, 999, 1031, 1039, 1063, 1071, 1095

Offset: 1

Altug Alkan, Apr 22 2016

If there are k consecutive natural numbers and all of them are members of A000378, then the maximum value of k is 7. So if we randomly choose 2*7+1 consecutive natural numbers, at least one of them must be member of A004215. This sequence gives the average of 15 consecutive natural numbers in the case there is exactly one member from A004215 in these 15 consecutive natural numbers. In other words, this sequence gives the most isolated terms of A004215.

Numbers n which are 7 mod 16 such that n+5 and n-7 are sums of three squares, together with numbers n which are 15 mod 16 such that n+1 and n-3 are sums of three squares. - Charles R Greathouse IV, Apr 25 2016

			15 is a term because 8, 9, 10, 11, 12, 13, 14 and 16, 17, 18, 19, 20, 21, 22 are consecutive members of A000378.

Cf. A000378, A004215.

Take[#, {2}] & /@ Select[#, Union@ Differences@ # == {8} &] &@ Partition[#, 3, 1] &@ Select[Range@ 1200, Mod[#/4^IntegerExponent[#, 4], 8] == 7 &] // Flatten (* Michael De Vlieger, Apr 25 2016, after Ant King at A004215 *)

isA004215(n)=(n>>(2*valuation(n, 4)))%8==7
is(n)=my(m=n%16); n>9 && if(m==7, !isA004215(n+5) && !isA004215(n-7), m==15 && !isA004215(n+1) && !isA004215(n-3)) \\ Charles R Greathouse IV, Apr 25 2016

A287164 Primes having a unique partition into three squares.

Original entry on oeis.org

2, 3, 5, 11, 13, 19, 37, 43, 67, 163

Offset: 1

Ilya Gutkovskiy, May 20 2017

D. H. Lehmer conjectures that there are no more terms (see A094739 and A094942).

			-------------------------------
|  n | a(n) | representation  |
|-----------------------------|
|  1 |   2  | 0^2 + 1^2 + 1^2 |
|  2 |   3  | 1^2 + 1^2 + 1^2 |
|  3 |   5  | 0^2 + 1^2 + 2^2 |
|  4 |  11  | 1^2 + 1^2 + 3^2 |
|  5 |  13  | 0^2 + 2^2 + 3^2 |
|  6 |  19  | 1^2 + 3^2 + 3^2 |
|  7 |  37  | 0^2 + 1^2 + 6^2 |
|  8 |  43  | 3^2 + 3^2 + 5^2 |
|  9 |  67  | 3^2 + 3^2 + 7^2 |
| 10 | 163  | 1^2 + 9^2 + 9^2 |
-------------------------------
157 is the prime of the form x^2 + y^2 + z^2 with x, y, z >= 0, but is not in the sequence because 157 = 0^2 + 6^2 + 11^2 = 2^2 + 3^2 + 12^2.

Index entries for sequences related to sums of squares

Cf. A000164, A000378, A002313, A042998, A094739, A094942.

Select[Range[200], Length[PowersRepresentations[#, 3, 2]] == 1 && PrimeQ[#] &]

A307219 a(n) is the number of partitions of (prime(n)^2 + 2)/3 into 3 squares.

Original entry on oeis.org

1, 1, 2, 2, 1, 2, 2, 5, 6, 2, 6, 3, 6, 5, 14, 8, 6, 5, 6, 15, 10, 6, 14, 24, 14, 6, 12, 12, 6, 16, 19, 18, 18, 36, 18, 10, 16, 20, 20, 12, 28, 18, 8, 24, 38, 27, 40, 20, 17, 30, 52, 18, 22, 26, 29, 21, 42, 31, 26, 26, 18, 44, 38, 40, 46, 26, 30, 44, 38, 36, 52, 28, 27, 38, 103, 22, 38, 78

Offset: 3

Marius A. Burtea, Mar 29 2019

nonn

If p >= 5 is a prime number it can be written as p = 6m-1 or p = 6m+1. The identities ((6m-1)^2 + 2)/3 = (2m)^2 + (2m)^2 + (2m-1)^2 and ((6m+1)^2 + 2)/3 = (2m)^2 + (2m)^2 + (2m+1)^2 show that the number (p^2 + 2)/3 can be written as a sum of 3 squares of integers in at least one way.

			For n = 3, p = prime(3) = 5, (5^2+2)/3 = 9 = 2^2 + 2^2 + 1^2, so a(3) = 1.
For n = 9, p = prime(9) = 23, (23^2+2)/3 = 177 = 13^2 + 2^2 + 2^2 = 8^2 + 8^2 + 7^2, so a(9) = 2.
For n = 17, p = prime(17) = 59, (59^2+2)/3 = 1161 = 34^2 + 2^2 + 1^2 = 33^2 + 6^2 + 6^2 = 24^2 + 11^2 + 4^2 = 31^2 + 14^2 + 2^2 = 31^2 + 10^2 + 10^2 = 30^2 + 15^2 + 6^2 = 29^2 + 16^2 + 8^2 = 28^2 + 19^2 + 4^2 = 28^2 + 16^2 + 11^2 = 26^2 + 22^2 + 1^2 = 26^2 + 17^2 + 14^2 = 24^2 + 24^2 + 3^2 = 24^2 + 21^2 + 12^2 = 20^2 + 20^2 + 19^2, so a(17) = 14.

Ion Cucurezeanu, Pătrate și cuburi perfecte de numere întregi (Squares and perfect cubes of integer numbers), Ed. Gil., Zalău, 2007, ch. 1, p. 21, pr. 166.
Laurențiu Panaitopol, Dinu Șerbănescu, Number theory and combinatorial problems for juniors, Ed. Gil, Zalău, (2003), ch. 1, p. 5, pr. 4. (in Romanian).

Cf. A000040, A000378, A024795, A025427.

[#RestrictedPartitions(Floor((p*p+2)/3),3,{d*d:d in [1..p]}): p in PrimesInInterval(5,500) ];

a(n)={k=(prime(n+2)^2+2)/3;sum(a=1, k, sum(b=1, a, sum(c=1, b, a^2+b^2+c^2==k)));} \\ Jinyuan Wang, Mar 30 2019

cp's OEIS Frontend

A347969 Numbers which are sum of three squares of positive numbers and also 5 times of the sum of their joint products.

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A076180 Indices of spheres mentioned in A071609.

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A272189 Values of A004215(n) such that A004215(n+1) = A004215(n) + 8 = A004215(n-1) + 16.

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A287164 Primes having a unique partition into three squares.

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A307219 a(n) is the number of partitions of (prime(n)^2 + 2)/3 into 3 squares.

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