cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A299794 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x or 2*y is a power of 4 (including 4^0 = 1) and x + 15*y is also a power of 4.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 4, 2, 1, 1, 2, 3, 1, 1, 2, 2, 1, 1, 6, 3, 1, 3, 3, 2, 2, 1, 3, 4, 2, 1, 5, 4, 4, 5, 1, 2, 3, 2, 5, 5, 2, 2, 8, 2, 2, 1, 5, 2, 4, 3, 4, 4, 4, 3, 6, 3, 2, 3, 3, 4, 3, 1, 3, 6, 4, 3, 11, 2, 2, 2, 4, 5, 1, 2, 3, 5, 3, 1
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 04 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0. Also, for any integer n > 1 we can write n^2 as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that 2*x or y is a power of 4 and also x + 15*y = 2^(2k+1) for some k = 0,1,2,....
Conjecture 2: Let d be 2 or 8, and let r be 0 or 1. Then any positive square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x or y is a power of 2 and x + d*y = 2^(2k+r) for some k = 0,1,2,....
We have verified Conjecture 1 for n up to 10^7.
See also A299537, A300219 and A300396 for similar conjectures.

Examples

			a(2) = 1 since 2^2 = 1^2 + 1^2 + 1^2 + 1^2 with 1 = 4^0 and 1 + 15*1 = 4^2.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4 = 4^1 and 4 + 15*0 = 4^1.
a(19) = 1 since 19^2 = 1^2 + 0^2 + 6^2 + 18^2 with 1 = 4^0 and 1 + 15*0 = 4^0.
a(159) = 1 since 159^2 = 34^2 + 2^2 + 75^2 + 136^2 with 2*2 = 4^1 and 34 + 15*2 = 4^3.
a(1998) = 1 since 1998^2 = 256^2 + 256^2 + 286^2 + 1944^2 with 256 = 4^4 and 256 + 15*256 = 4^6.
a(3742) = 1 since 3742^2 = 2176^2 + 128^2 + 98^2 + 3040^2 with 2*128 = 4^4 and 2176 + 15*128 = 4^6.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
    tab={};Do[r=0;Do[If[Pow[2y]||Pow[4^k-15y],Do[If[SQ[n^2-y^2-(4^k-15y)^2-z^2],r=r+1],{z,0,Sqrt[Max[0,(n^2-y^2-(4^k-15y)^2)/2]]}]],
    {k,0,Log[4,Sqrt[226]*n]},{y,0,Min[n,4^(k-2)]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A300362 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 2*y and (z + 2*w)/3 are squares and w is even.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 4, 1, 4, 2, 4, 1, 2, 2, 2, 3, 2, 1, 5, 1, 5, 2, 3, 3, 3, 1, 1, 2, 3, 1, 4, 3, 5, 1, 6, 6, 6, 1, 4, 6, 8, 2, 4, 4, 3, 1, 5, 3, 9, 1, 4, 4, 5, 3, 10, 4, 7, 3, 9, 2, 14, 2, 6, 2, 6, 1
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 04 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 7, 9, 14, 19, 22, 26, 34, 41, 4^k*m (k = 0,1,... and m = 1, 2, 3, 5, 10, 11, 13, 15).

Examples

			 a(9) = 1 since 9^2 = 9^2 + 0^2 + 0^2 + 0^2 with 9 + 2*0 = 3^2 and 0 + 2*0 = 3*0^2.
a(13) = 1 since 13^2 = 4^2 + 0^2 + 3^2 + 12^2 with 4 + 2*0 = 2^2 and 3 + 2*12 = 3*3^2.
a(14) = 1 since 14^2 = 4^2 + 6^2 + 12^2 + 0^2 with 4 + 2*6 = 4^2 and 12 + 2*0 = 3*2^2.
a(15) = 1 since 15^2 = 9^2 + 0^2 + 12^2 + 0^2 with 9 + 2*0 = 3^2 and 12 + 2*0 = 3*2^2.
a(41) = 1 since 41 = 38^2 + 13^2 + 8^2 + 2^2 with 38 + 2*13 = 8^2 and 8 + 2*2 = 3*2^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[x+2y]&&SQ[(n^2-x^2-y^2-z^2)/4]&&SQ[(z+2*Sqrt[n^2-x^2-y^2-z^2])/3],r=r+1],{x,0,n},{y,0,Sqrt[n^2-x^2]},{z,0,Sqrt[n^2-x^2-y^2]}];tab=Append[tab,r],{n,0,80}];Print[tab]

A122141 Array: T(d,n) = number of ways of writing n as a sum of d squares, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 4, 0, 1, 6, 4, 0, 1, 8, 12, 0, 2, 1, 10, 24, 8, 4, 0, 1, 12, 40, 32, 6, 8, 0, 1, 14, 60, 80, 24, 24, 0, 0, 1, 16, 84, 160, 90, 48, 24, 0, 0, 1, 18, 112, 280, 252, 112, 96, 0, 4, 2, 1, 20, 144, 448, 574, 312, 240, 64, 12, 4, 0, 1, 22, 180, 672, 1136, 840, 544, 320, 24, 30, 8, 0
Offset: 1

Views

Author

R. J. Mathar, Oct 29 2006

Keywords

Comments

This is the transpose of the array in A286815.
T(d,n) is divisible by 2d for any n != 0 iff d is a power of 2. - Jianing Song, Sep 05 2018

Examples

			Array T(d,n) with rows d = 1,2,3,... and columns n = 0,1,2,3,... reads
  1  2   0   0    2    0     0     0     0     2      0 ...
  1  4   4   0    4    8     0     0     4     4      8 ...
  1  6  12   8    6   24    24     0    12    30     24 ...
  1  8  24  32   24   48    96    64    24   104    144 ...
  1 10  40  80   90  112   240   320   200   250    560 ...
  1 12  60 160  252  312   544   960  1020   876   1560 ...
  1 14  84 280  574  840  1288  2368  3444  3542   4424 ...
  1 16 112 448 1136 2016  3136  5504  9328 12112  14112 ...
  1 18 144 672 2034 4320  7392 12672 22608 34802  44640 ...
  1 20 180 960 3380 8424 16320 28800 52020 88660 129064 ...
		

Crossrefs

Cf. A000122 (1st row), A004018 (2nd row), A005875 (3rd row), A000118 (4th row), A000132 (5th row), A000141 (6th row), A008451 (7th row), A000143 (8th row), A008452 (9th row), A000144 (10th row), A008453 (11th row), A000145 (12th row), A276285 (13th row), A276286 (14th row), A276287 (15th row), A000152 (16th row).
Cf. A005843 (2nd column), A046092 (3rd column), A130809 (4th column).
Cf. A010052 (1st row divides 2), A002654 (2nd row divides 4), A046897 (4th row divides 8), A008457 (8th row divides 16), A302855 (16th row divides 32), A302857 (32nd row divides 64).

Programs

  • Maple
    A122141 := proc(d,n) local i,cnts ; cnts := 0 ; for i from -trunc(sqrt(n)) to trunc(sqrt(n)) do if n-i^2 >= 0 then if d > 1 then cnts := cnts+procname(d-1,n-i^2) ; elif n-i^2 = 0 then cnts := cnts+1 ; fi ; fi ; od ; cnts ;
    end:
    for diag from 1 to 14 do for n from 0 to diag-1 do d := diag-n ; printf("%d,",A122141(d,n)) ; od ; od;
    # second Maple program:
    A:= proc(d, n) option remember; `if`(n=0, 1, `if`(n<0 or d<1, 0,
          A(d-1, n) +2*add(A(d-1, n-j^2), j=1..isqrt(n))))
        end:
    seq(seq(A(h-n, n), n=0..h-1), h=1..14); # Alois P. Heinz, Jul 16 2014
  • Mathematica
    Table[ SquaresR[d - n, n], {d, 1, 12}, {n, 0, d - 1}] // Flatten (* Jean-François Alcover, Jun 13 2013 *)
    A[d_, n_] := A[d, n] = If[n==0, 1, If[n<0 || d<1, 0, A[d-1, n] + 2*Sum[A[d-1, n-j^2], {j, 1, Sqrt[n]}]]]; Table[A[h-n, n], {h, 1, 14}, {n, 0, h-1}] // Flatten (* Jean-François Alcover, Feb 28 2018, after Alois P. Heinz *)
  • Python
    from sympy.core.power import isqrt
    from functools import cache
    @cache
    def T(d, n):
      if n == 0: return 1
      if n < 0 or d < 1: return 0
      return T(d-1, n) + sum(T(d-1, n-(j**2)) for j in range(1, isqrt(n)+1)) * 2  # Darío Clavijo, Feb 06 2024

Formula

T(n,n) = A066535(n). - Alois P. Heinz, Jul 16 2014

A300396 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 2*x or y is a power of 4 (including 4^0 = 1) and x + 63*y = 2^(2k+1) for some k = 0,1,2,....

Original entry on oeis.org

0, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 3, 1, 1, 3, 1, 3, 4, 2, 2, 4, 1, 2, 1, 1, 2, 4, 3, 1, 1, 2, 1, 6, 2, 2, 2, 5, 1, 4, 1, 2, 6, 3, 3, 3, 1, 2, 3, 4, 3, 3, 2, 4, 2, 2, 1, 7, 3, 1, 4, 1, 2, 8, 1, 3, 7, 3, 4, 6, 3, 4, 4, 6, 4, 3, 2, 4, 3, 1, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 05 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 5, 13, 25, 29, 59, 61, 79, 91, 95, 101, 103, 1315, 2^k (k = 1,2,3,...), 2^(2k+1)*m (k = 0,1,2,... and m = 3, 5, 7, 11, 15, 19, 23, 887).
This is stronger than the conjecture that A300360(n) > 0 for all n > 1. Note that a(387) = 3 < A300360(387) = 4 and a(1774) = 1 < A300360(1774) = 2.
We have verified that a(n) > 0 for all n = 2..10^7.
See also A299537, A299794 and A300219 for similar conjectures.

Examples

			a(29) = 1 since 29^2 = 2^2 + 2^2 + 7^2 + 28^2 with 2*2 = 4^1 and 2 + 63*2 = 2^7.
a(86) = 2 since 65^2 + 1^2 + 19^2 + 53^2 = 65^2 + 1^2 + 31^2 + 47^2 with 1 = 4^0 and 65 + 63*1 = 2^7.
a(1774) = 1 since 1774^2 = 8^2 + 520^2 + 14^2 + 1696^2 with 2*8 = 4^2 and 8 + 63*520 = 2^15.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
    tab={};Do[r=0;Do[If[Pow[y]||Pow[(2*4^k-63y)/2],Do[If[SQ[n^2-y^2-(2*4^k-63y)^2-z^2],r=r+1],{z,0,Sqrt[Max[0,(n^2-y^2-(2*4^k-63y)^2)/2]]}]],{k,0,Log[4,Sqrt[63^2+1]*n/2]},{y,0,Min[n,2*4^k/63]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A272084 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 4*x^2 + 5*y^2 + 20*z*w a square, where x,y,z,w are nonnegative integers with z < w.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 3, 1, 2, 1, 2, 1, 3, 3, 2, 2, 1, 2, 3, 1, 3, 4, 4, 1, 4, 1, 2, 1, 3, 3, 3, 2, 2, 1, 2, 3, 5, 4, 2, 3, 3, 3, 2, 1, 2, 6, 6, 2, 3, 2, 2, 1, 3, 4, 4, 2, 3, 1, 6, 1, 5, 3, 4, 3, 4, 1, 4, 3, 4, 8, 4, 2, 1, 3, 2, 2, 5, 4, 4, 1, 6, 3, 6, 2, 5, 6, 7, 3, 2, 2, 2, 1, 3, 5, 9, 3
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 19 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k*q (k = 0,1,2,... and q = 1, 3, 7), 4^k*m (k = 0,1,2,... and m = 21, 30, 38, 62, 70, 77, 142, 217, 237, 302, 382, 406, 453, 670).
(ii) For each triple (a,b,c) = (1,8,8), (7,9,-12), (9,40,-24), (9,40,-60), any positive integer can be written as x^2 + y^2 + z^2 + w^2 with a*x^2 + b*y^2 + c*z*w a square, where w is a positive integer and x,y,z are nonnegative integers.
(iii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with (3*x+5*y)^2 -24*z*w a square, where x,y,z,w are nonnegative integers. Also, for each ordered pair (a,b) = (1,4), (1,8), (1,12), (1,24), (1,32), (1,48), (25,24), (1,-4), (9,-4), (121,-20), every natural number can be written as x^2 + y^2 + z^2 + w^2 with a*x^2 + b*y*z a square, where x,y,z,w are nonnegative integers.
(iv) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with (x^2-y^2)*(w^2-2*z^2) (or (x^2-y^2)*(2*w^2-z^2) or (x^2-y^2)*(w^2-5*z^2)) a square, where w,x,y,z are integers.
See also A262357, A268507, A269400, A271510, A271513, A271518, A271665, A271714, A271721, A271724, A271775, A271778 and A271824 for other conjectures refining Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 < 1 and 4*0^2 + 5*0^2 + 20*0*1 = 0^2.
a(2) = 1 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 0 < 1 and 4*1^2 + 5*0^2 + 20*0*1 = 2^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 0 < 1 and 4*1^2 + 5*1^2 + 20*0*1 = 3^2.
a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 0 < 2 and 4*1^2 + 5*1^2 + 20*0*2 = 3^2.
a(14) = 1 since 14 = 1^2 + 3^2 + 0^2 + 2^2 with 0 < 2 and 4*1^2 + 5*3^2 + 20*0*2 = 7^2.
a(21) = 1 since 21 = 0^2 + 2^2 + 1^2 + 4^2 with 1 < 4 and 4*0^2 + 5*2^2 + 20*1*4 = 10^2.
a(30) = 1 since 30 = 4^2 + 2^2 + 1^2 + 3^2 with 1 < 3 and 4*4^2 + 5*2^2 + 20*1*3 = 12^2.
a(38) = 1 since 38 = 1^2 + 1^2 + 0^2 + 6^2 with 0 < 6 and 4*1^2 + 5*1^2 + 20*0*6 = 3^2.
a(62) = 1 since 62 = 1^2 + 3^2 + 4^2 + 6^2 with 4 < 6 and 4*1^2 + 5*3^2 + 20*4*6 = 23^2.
a(70) = 1 since 70 = 7^2 + 1^2 + 2^2 + 4^2 with 2 < 4 and 4*7^2 + 5*1^2 + 20*2*4 = 19^2.
a(77) = 1 since 77 = 4^2 + 6^2 + 3^2 + 4^2 with 3 < 4 and 4*4^2 + 5*6^2 + 20*3*4 = 22^2.
a(142) = 1 since 142 = 4^2 + 6^2 + 3^2 + 9^2 with 3 < 9 and 4*4^2 + 5*6^2 + 20*3*9 = 28^2.
a(217) = 1 since 217 = 6^2 + 6^2 + 8^2 + 9^2 with 8 < 9 and 4*6^2 + 5*6^2 + 20*8*9 = 42^2.
a(237) = 1 since 237 = 5^2 + 8^2 + 2^2 + 12^2 with 2 < 12 and 4*5^2 + 5*8^2 + 20*2*12 = 30^2.
a(302) = 1 since 302 = 11^2 + 9^2 + 6^2 + 8^2 with 6 < 8 and 4*11^2 + 5*9^2 + 20*6*8 = 43^2.
a(382) = 1 since 382 = 11^2 + 7^2 + 4^2 + 14^2 with 4 < 14 and 4*11^2 + 5*4^2 + 20*4*14 = 43^2.
a(406) = 1 since 406 = 8^2 + 6^2 + 9^2 + 15^2 with 9 < 15 and 4*8^2 + 5*6^2 + 20*9*15 = 56^2.
a(453) = 1 since 453 = 8^2 + 14^2 + 7^2 + 12^2 with 7 < 12 and 4*8^2 + 5*14^2 + 20*7*12 = 54^2.
a(670) = 1 since 670 = 17^2 + 11^2 + 2^2 + 16^2 with 2 < 16 and 4*17^2 + 5*11^2 + 20*2*16 = 49^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[4x^2+5y^2+20*z*Sqrt[n-x^2-y^2-z^2]],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[(n-1-x^2-y^2)/2]}];Print[n," ",r];Continue,{n,1,100}]

A272332 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 36*x^2*y + 12*y^2*z + z^2*x a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 2, 2, 6, 4, 3, 3, 3, 8, 5, 2, 6, 6, 4, 1, 7, 10, 6, 8, 8, 5, 2, 2, 7, 16, 8, 3, 12, 6, 4, 3, 6, 13, 8, 8, 8, 6, 5, 7, 15, 14, 4, 2, 12, 7, 3, 2, 5, 18, 8, 12, 14, 8, 7, 4, 6, 8, 7, 5, 14, 8, 5, 2, 12, 18, 8, 12, 10, 6, 3, 5, 10, 19, 10, 3, 8, 3, 1, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 26 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 79, 591, 599, 1752, 1839, 10264).
We have verified that a(n) > 0 for all n = 1,...,400000.
For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0 and 36*0^2*0 + 12*0^2*0 + 0^2*0 = 0^2.
a(79) = 1 since 79 = 7^2 + 1^2 + 5^2 + 2^2 with 7 > 0 and 36*1^2*5 + 12*5^2*2 + 2^2*1 = 28^2.
a(591) = 1 since 591 = 23^2 + 1^2 + 6^2 + 5^2 with 23 > 0 and 36*1^2*6 + 12*6^2*5 + 5^2*1 = 49^2.
a(599) = 1 since 599 = 6^2 + 1^2 + 11^2 + 21^2 with 6 > 0 and 36*1^2*11 + 12*11^2*21 + 21^2*1 = 177^2.
a(1752) = 1 since 1752 = 10^2 + 4^2 + 40^2 + 6^2 with 10 > 0 and 36*4^2*40 + 12*40^2*6 + 6^2*10 = 372^2.
a(1839) = 1 since 1839 = 17^2 + 37^2 + 9^2 + 10^2 with 17 > 0 and 36*37^2*9 + 12*9^2*10 + 10^2*37 = 676^2.
a(10264) = 1 since 10264 = 96^2 + 30^2 + 2^2 + 12^2 with 96 > 0 and 36*30^2*2 + 12*2^2*12 + 12^2*30 = 264^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[36*x^2*y+12*y^2*z+z^2*x],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]

A272351 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with x^4 + 8*y*z*(y^2+z^2) a fourth power, where w,x,y,z are nonnegative integers with w >= x and y > z.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 2, 1, 2, 3, 2, 1, 4, 5, 2, 2, 3, 3, 2, 2, 3, 5, 4, 1, 5, 6, 1, 1, 5, 4, 5, 3, 2, 5, 2, 3, 7, 7, 3, 2, 5, 4, 2, 1, 5, 8, 7, 2, 5, 9, 1, 3, 4, 4, 5, 2, 5, 8, 6, 1, 8, 8, 4, 4, 6, 5, 1, 5, 5, 10, 6, 2, 6, 8, 1, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 27 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k, 4^k*m (k = 0,1,2,... and m = 3, 7, 31, 55, 71, 79, 151, 191).
(ii) For (b,c) = (8,8), (16,64), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with x^4 + b*y^3*z + c*y*z^3 a fourth power, where w is a positive integer and x,y,z are nonnegative integers.
(iii) For each triple (a,b,c) = (1,20,60), (1,24,56), (9,20,60), (9,32,96), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with a*x^4 + b*y^3*z + c*y*z^3 a square, where w is a positive integer and x,y,z are nonnegative integers.
The author has proved part (ii) of the conjecture in arXiv:1604.06723. - Zhi-Wei Sun, May 09 2016

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0 = 0, 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4.
a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4.
a(3) = 1 since 3 = 1^2 + 1^2 + 1^2 + 0^2 with 1 = 1, 1 > 0 and 1^4 + 8*1*0*(1^2+0^2) = 1^4.
a(7) = 1 since 7 = 1^2 + 1^2 + 2^2 + 1^2 with 1 = 1, 3 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.
a(31) = 1 since 31 = 5^2 + 1^2 + 2^2 + 1^2 with 5 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.
a(55) = 1 since 55 = 7^2 + 1^2 + 2^2 + 1^2 with 7 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.
a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 3 > 1, 6 > 5 and 1^4 + 8*6*5*(6^2+5^2) = 11^4.
a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5 > 3, 6 > 3 and 3^4 + 8*6*3*(6^2+3^2) = 9^4.
a(151) = 1 since 151 = 5^2 + 3^2 + 9^2 + 6^2 with 5 > 3, 9 > 6 and 3^4 + 8*9*6*(9^2+6^2) = 15^4.
a(191) = 1 since 191 = 3^2 + 1^2 + 10^2 + 9^2 with 3 > 1, 10 > 9 and 1^4 + 8*10*9*(10^2+9^2) = 19^4.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    QQ[n_]:=QQ[n]=IntegerQ[n^(1/4)]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&QQ[x^4+8y*z*(y^2+z^2)],r=r+1],{z,0,(Sqrt[2n-1]-1)/2},{y,z+1,Sqrt[n-z^2]},{x,0,Sqrt[(n-y^2-z^2)/2]}];Print[n," ",r];Continue,{n,1,80}]

A286815 Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of (Product_{j>=1} (1 - x^(2*j))^5/((1 - x^j)*(1 - x^(4*j)))^2)^k.

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 0, 0, 1, 6, 4, 0, 0, 1, 8, 12, 0, 2, 0, 1, 10, 24, 8, 4, 0, 0, 1, 12, 40, 32, 6, 8, 0, 0, 1, 14, 60, 80, 24, 24, 0, 0, 0, 1, 16, 84, 160, 90, 48, 24, 0, 0, 0, 1, 18, 112, 280, 252, 112, 96, 0, 4, 2, 0, 1, 20, 144, 448, 574, 312, 240, 64, 12
Offset: 0

Views

Author

Seiichi Manyama, May 27 2017

Keywords

Comments

A(n,k) is the number of ways of writing n as a sum of k squares.
This is the transpose of the array in A122141.

Examples

			Square array begins:
   1, 1, 1,  1,  1, ...
   0, 2, 4,  6,  8, ...
   0, 0, 4, 12, 24, ...
   0, 0, 0,  8, 32, ...
   0, 2, 4,  6, 24, ...
		

Crossrefs

Programs

  • Maple
    A:= proc(n, k) option remember; `if`(n=0, 1, `if`(n<0 or k<1, 0,
          A(n, k-1) +2*add(A(n-j^2, k-1), j=1..isqrt(n))))
        end:
    seq(seq(A(n, d-n), n=0..d), d=0..12);  # Alois P. Heinz, May 27 2017
  • Mathematica
    A[n_, k_] := A[n, k] = If[n == 0, 1, If[n < 0 || k < 1, 0, A[n, k-1] + 2*Sum[A[n-j^2, k-1], {j, 1, Sqrt[n]}]]];
    Table[A[n, d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Feb 28 2018, after Alois P. Heinz *)

Formula

G.f. of column k: (Product_{j>=1} (1 - x^(2*j))^5/((1 - x^j)*(1 - x^(4*j)))^2)^k.

A066535 Number of ways of writing n as a sum of n squares.

Original entry on oeis.org

1, 2, 4, 8, 24, 112, 544, 2368, 9328, 34802, 129064, 491768, 1938336, 7801744, 31553344, 127083328, 509145568, 2035437440, 8148505828, 32728127192, 131880275664, 532597541344, 2153312518240, 8710505815360, 35250721087168, 142743029326162, 578472382307304
Offset: 0

Views

Author

Peter Bertok (peter(AT)bertok.com), Jan 07 2002

Keywords

Examples

			There are a(3) = 8 solutions (x,y,z) of 3 = x^2 + y^2 + z^2: (1,1,1), (-1,-1,-1), 3 permutations of (1,1,-1) and 3 permutations of (1,-1,-1).
		

Crossrefs

Cf. A122141, A166952. - Paul D. Hanna, Oct 25 2009
a(n^2) gives A361431.

Programs

  • Maple
    b:= proc(n, t) option remember; `if`(n=0, 1, `if`(n<0 or t<1, 0,
          b(n, t-1) +2*add(b(n-j^2, t-1), j=1..isqrt(n))))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..30);  # Alois P. Heinz, Jul 16 2014
  • Mathematica
    Join[{1}, Table[SquaresR[n, n], {n, 24}]]
  • PARI
    {a(n)=local(THETA3=1+2*sum(k=1,sqrtint(n),x^(k^2))+x*O(x^n)); polcoeff(THETA3^n, n)} /* Paul D. Hanna, Oct 25 2009 */

Formula

a(n) equals the coefficient of x^n in the n-th power of Jacobi theta_3(x) where theta_3(x) = 1 + 2*Sum_{n>=1} x^(n^2). - Paul D. Hanna, Oct 25 2009
a(n) ~ c * d^n / sqrt(n), where d = 4.13273137623493996302796465... (= 1/radius of convergence A166952), c = 0.2820942036723951157919967... . - Vaclav Kotesovec, Sep 12 2014

Extensions

Edited by Dean Hickerson, Jan 12 2002
a(0) added by Paul D. Hanna, Oct 25 2009
Edited by R. J. Mathar, Oct 29 2009

A301376 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 x^2-(3*y)^2 = 4^k for some k = 0,1,2,....

Original entry on oeis.org

1, 1, 2, 1, 1, 3, 1, 1, 4, 2, 2, 3, 3, 3, 3, 1, 5, 6, 2, 2, 10, 5, 4, 3, 2, 7, 7, 3, 5, 4, 3, 1, 12, 8, 2, 6, 4, 5, 10, 2, 7, 13, 8, 5, 10, 6, 6, 3, 8, 4, 7, 7, 8, 11, 4, 3, 17, 9, 5, 4, 8, 5, 9, 1, 8, 14, 8, 8, 13, 5, 8, 6, 11, 10, 7, 5, 13, 15, 7, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 19 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0. Moreover, any positive square n^2 can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers and y even such that x^2 - (3*y)^2 = 4^k for some k = 0,1,2,....
We have verifed this for all n = 1..10^7.
Compare this conjecture with the conjectures in A299537.
As 3*A001353(n)^2 + 1 = A001075(n)^2, the conjecture in A300441 implies that any positive square can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x^2 - 3*y^2 = 4^k for some k = 0,1,2,....
See also A301391 for a similar conjecture.

Examples

			a(1) = 1 since 1^2 = 1^2 + 0^2 + 0^2 + 0^2 with 1^2 - (3*0)^2 = 4^0.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (3*0)^2 = 4^2.
a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (3*0)^2 = 4^1.
a(31) = 3 since 31^2 = 10^2 + 2^2 + 4^2 + 29^2 with 10^2 - (3*2)^2 = 4^3, and 31^2 = 20^2 + 4^2 + 4^2 + 23^2 = 20^2 + 4^2 + 16^2 + 17^2 with 20^2 - (3*4)^2 = 4^4.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[4^k+9y^2]&&QQ[n^2-4^k-10y^2],Do[If[SQ[n^2-(4^k+10y^2)-z^2],r=r+1],{z,0,Sqrt[(n^2-4^k-10y^2)/2]}]],{k,0,Log[2,n]},{y,0,Sqrt[(n^2-4^k)/10]}];tab=Append[tab,r],{n,1,80}];Print[tab]
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