A300365 -id:A300365 - cp's OEIS frontend

A300219 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and 4x - 3y are powers of 4 (including 4^0 = 1).

1, 1, 1, 1, 1, 3, 2, 1, 5, 2, 2, 1, 3, 3, 1, 1, 2, 2, 2, 1, 8, 3, 2, 3, 4, 3, 4, 2, 8, 5, 4, 1, 7, 6, 4, 5, 1, 3, 6, 2, 9, 6, 3, 2, 8, 4, 2, 1, 5, 3, 7, 3, 4, 6, 3, 3, 7, 4, 5, 1, 3, 5, 3, 1, 2, 9, 4, 2, 11, 3, 6, 2, 6, 7, 3, 2, 4, 5, 4, 1

Offset: 1

Zhi-Wei Sun, Feb 28 2018

nonn

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m (k = 0,1,2,... and m = 1, 2, 3, 5, 15, 37, 83, 263). Also, for each n = 2,3,... we can write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 4*x - 3*y lie in the set {2^(2k+1): k = 0,1,...}.
(ii) Let r be 0 or 1, and let n > r. Then n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and x + 3*y belong to the set {2^(2k+r): k = 0,1,2,...}, unless n has the form 2^(2k+r)*81503 with k a nonnegative integer and hence n^2 = (2^(2k+r)*28^2)^2 + (2^(2k+r)*80)^2 + (2^(2k+r)*55937)^2 + (2^(2k+r)*59272)^2 with 2^(2k+r)*28^2 = 2^r*(2^k*28)^2 and 2^(2k+r)*28^2 + 3*(2^(2k+r)*80) = 2^(2(k+5)+r). So we always can write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x/2^r is a square and (x+3*y)/2^r is a power of 4.
In arXiv:1701.05868 the author proved that for each r = 0,1 and n > r we can write n^2 as (2^(2k+r))^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers.
We have verified both parts of the conjecture for n up to 10^7.

			a(2) = 1 since 2^2 = 1^2 + 1^2 + 1^2 + 1^2 with 1 = 4^0 and 4*1 - 3*1 = 4^0.
a(3) = 1 since 3^2 = 1^2 + 0^2 + 2^2 + 2^2 with 1 = 4^0 and 4*1 - 3*0 = 4^1.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4 = 4^1 and 4*4 - 3*0 = 4^2.
a(15) = 1 since 15^2 = 4^2 + 4^2 + 7^2 + 12^2 with 4 = 4^1 and 4*4 - 3*4 = 4^1.
a(37) = 1 since 37^2 = 16^2 + 16^2 + 4^2 + 29^2 with 16 = 4^2 and 4*16 - 3*16 = 4^2.
a(83) = 1 since 83^2 = 4^2 + 4^2 + 56^2 + 61^2 with 4 = 4^1 and 4*4 - 3*4 = 4^1.
a(263) = 1 since 263^2 = 4^2 + 5^2 + 22^2 + 262^2 with 4 = 4^1 and 4*4 - 3*5 = 4^0.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000118, A271518, A279612, A281976, A299924, A300365, A300360.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0;Do[If[SQ[n^2-16^k-((4^(k+1)-4^m)/3)^2-z^2],r=r+1],{k,0,Log[4,n]},{m,Ceiling[Log[4,Max[1,4^(k+1)-3*Sqrt[n^2-16^k]]]],k+1},{z,0,Sqrt[(n^2-16^k-((4^(k+1)-4^m)/3)^2)/2]}];Print[n," ",r];Label[aa],{n,1,80}]

A300364 Balanced primes of order thirteen.

Original entry on oeis.org

353, 2351, 3673, 3863, 4759, 6271, 8539, 8821, 11261, 12073, 17839, 26711, 32797, 33769, 34679, 41357, 53269, 60217, 64879, 64891, 68713, 88493, 90121, 91811, 101347, 101411, 101641, 108139, 108203, 114659, 122029, 123637, 127843, 128237, 130447, 133279

Offset: 1

Muniru A Asiru, Mar 04 2018

nonn

			353 is a member because 353 = 271 + 277 + 281 + 283 + 293 + 307 + 311 + 313 + 317 + 331 + 337 + 347 + 349 + 353 + 359 + 367 + 373 + 379 + 383 + 389 + 397 + 401 + 409 + 419 + 421 + 431 + 433 = 9531/27.

Muniru A Asiru, Table of n, a(n) for n = 1..7200

Cf. Balanced primes of order b: A006562 (b=1), A082077 (b=2), A082078 (b=3), A082079 (b=4), A096697 (b=5), A096698 (b=6), A096699 (b=7), A096700 (b=8), A096701 (b=9), A096702 (b=10), A096703 (b=11), A096704 (b=12), this sequence (b=13), A300365 (b=14).

P:=Filtered([1..150000],IsPrime);;
a:=List(Filtered(List([0..13000],k->List([1..27],j->P[j+k])),i->Sum(i)/27=i[14]),m->m[14]);

A363168 Balanced primes of order 100.

Original entry on oeis.org

27947, 111337, 193283, 197341, 197621, 347063, 809821, 955193, 1029803, 1184269, 1292971, 1609163, 1630859, 1656019, 1752449, 1883381, 1935517, 1969661, 2120221, 2156383, 2238959, 2287133, 2548631, 2592089, 2750903, 2866403, 3165769, 3257941, 3590299, 3889423

Offset: 1

Harvey P. Dale, Jul 07 2023

nonn

A prime p is in this sequence if the sum of the 100 consecutive primes just less than p, plus p, plus the sum of the 100 consecutive primes just greater than p, divided by 201 equals p.

Harvey P. Dale, Table of n, a(n) for n = 1..350 (all terms up to the 4 millionth prime)

Module[{bal=100,nn=300000},Select[Partition[Prime[Range[nn]],2bal+1,1],Mean[#]== #[[bal+1]]&]] [[;;,101]]

cp's OEIS Frontend

A300219 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and 4x - 3y are powers of 4 (including 4^0 = 1).

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A300364 Balanced primes of order thirteen.

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A363168 Balanced primes of order 100.

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cp's OEIS Frontend

A300219 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and 4*x - 3*y are powers of 4 (including 4^0 = 1).

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Programs

Mathematica

A300364 Balanced primes of order thirteen.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

GAP

A363168 Balanced primes of order 100.

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Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

A300219 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z <= w such that both x and 4x - 3y are powers of 4 (including 4^0 = 1).