A300396 -id:A300396 - cp's OEIS frontend

A301579 Least nonnegative integer k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers, or -1 if no such k exists.

-1, 0, 0, 2, 0, 0, 1, 4, 1, 0, 0, 2, 0, 0, 1, 6, 1, 0, 0, 2, 0, 2, 1, 4, 1, 0, 0, 2, 0, 3, 3, 8, 1, 0, 3, 2, 0, 0, 3, 4, 1, 0, 1, 4, 0, 0, 1, 6, 3, 0, 0, 2, 1, 0, 1, 4, 3, 0, 1, 5, 0, 5, 1, 10, 1, 0, 0, 2, 3, 0, 4, 4, 1, 2, 0, 2, 0, 0, 3, 6

Offset: 1

Zhi-Wei Sun, Mar 23 2018

sign

The Square Conjecture in A301471 implies that a(n) >= 0 for all n > 1.
It is known that a positive integer n has the form x^2 + 2*y^2 with x and y integers if and only if the p-adic order of n is even for any prime p == 5 or 7 (mod 8).
Numbers t such that a(t) = 0 are 2, 3, 5, 6, 10, 11, 13, 14, 18, 19, 21, 26, 27, 29, 34, 37, ... - Altug Alkan, Mar 26 2018

			a(1) = -1 since 1^2 - 3*2^k < 0 for all k = 0,1,2,....
a(31) = 3 since 31^2 - 3*2^3 = 17^2 + 2*18^2.
a(2^k) = 2*k - 2 for all k = 1,2,3,..., because (2^k)^2 - 3*2^(2*k-2) = (2^(k-1))^2 + 2*0^2, and (2^k)^2 - 3*2^j = 2^j*(2^(2*k-j) - 3) with 0 <= j < 2*k-2 cannot be written as x^2 + 2*y^2 with x and y integers.

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. A000079, A000290, A002479, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301471, A301472, A301479.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n],i],1],8]==5||Mod[Part[Part[f[n],i],1],8]==7)&&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]);
tab={};Do[Do[If[QQ[n^2-3*2^k],tab=Append[tab,k];Goto[aa]],{k,0,Log[2,n^2/3]}];tab=Append[tab,-1];Label[aa],{n,1,80}];Print[tab]

A301640 Largest integer k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers, or -1 if no such k exists.

Original entry on oeis.org

-1, 0, 1, 2, 3, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 6, 7, 7, 7, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 8, 9, 9, 9, 9, 8, 9, 9, 7, 9, 7, 9, 9, 8, 9, 10, 10, 10, 10, 10, 10, 10, 9, 10, 10, 10, 9, 10, 10, 10

Offset: 1

Zhi-Wei Sun, Mar 25 2018

sign

Conjecture: a(n) > 0.6*log_2(log_2 n) for all n > 2, and also lim inf_{n->infinity} a(n)/(log n) = 0.
The author's Square Conjecture in A301471 would imply that a(n) >= 0 for all n > 1. We have verified that a(n) > 0.6*log_2(log_2 n) for all n = 3..4*10^9. For n = 2857932461, we have a(n) = 3 and 0.603 < a(n)/log_2(log_2 n) < 0.604.
It is known that a positive integer n has the form x^2 + 2*y^2 with x and y integers if and only if the p-adic order of n is even for any prime p == 5 or 7 (mod 8).

			a(2) = 0 since 2^2 - 3*2^0 = 1^2 + 2*0^2.
a(3) = 1 since 3^2 - 3*2^1 = 2^2 + 2*1^2.
a(5) = 3 since 5^2 - 3*2^3 = 1^2 + 2*0^2.
a(6434567) = 10 since 6434567^2 - 3*2^10 = 5921293^2 + 2*1780722^2.

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. A000079, A000290, A002479, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301471, A301472, A301479, A301579.

f:= proc(n) local k,t;
    for k from floor(log[2](n^2/3)) by -1 to 0 do
       if g(n^2 - 3*2^k) then return k fi
    od;
    -1
end proc:
map(f, [$1..100]); # Robert Israel, Mar 26 2018

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n],i],1],8]==5||Mod[Part[Part[f[n],i],1],8]==7)&&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]);
tab={};Do[Do[If[QQ[n^2-3*2^(Floor[Log[2,n^2/3]]-k)],tab=Append[tab,Floor[Log[2,n^2/3]]-k];Goto[aa]],{k,0,Log[2,n^2/3]}];tab=Append[tab,-1];Label[aa],{n,1,70}];Print[tab]

A303235 Number of ordered pairs (x, y) with 0 <= x <= y such that n - 2^x - 2^y can be written as the sum of two triangular numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 4, 5, 6, 6, 6, 8, 7, 7, 8, 8, 8, 10, 10, 10, 10, 9, 9, 11, 9, 10, 11, 10, 9, 12, 10, 11, 14, 13, 11, 14, 12, 12, 13, 15, 12, 14, 12, 13, 14, 14, 14, 15, 13, 11, 14, 13, 11, 16, 13, 10, 11, 13, 11, 14

Offset: 1

Zhi-Wei Sun, Apr 20 2018

nonn

Conjecture: a(n) > 0 for all n > 1.
Note that a nonnegative integer m is the sum of two triangular numbers if and only if 4*m+1 (or 8*m+2) can be written as the sum of two squares.
We have verified a(n) > 0 for all n = 2..4*10^8. See also the related sequences A303233 and A303234.

			a(2) = 1 with 2 - 2^0 - 2^0 = 0*(0+1)/2 + 0*(0+1)/2.
a(3) = 2 with 3 - 2^0 - 2^0 = 0*(0+1)/2 + 1*(1+1)/2 and 3 - 2^0 - 2^1 = 0*(0+1)/2 + 0*(0+1)/2.

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, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000079, A000217, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&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]);
tab={};Do[r=0;Do[If[QQ[4(n-2^k-2^j)+1],r=r+1],{k,0,Log[2,n]-1},{j,k,Log[2,n-2^k]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A300448 Number of the integers 4^km with k >= 0 and m = 1, 2, 5 such that n^2 - 4^km can be written as the sum of two squares.

Original entry on oeis.org

1, 2, 4, 2, 4, 5, 4, 2, 7, 5, 6, 5, 6, 5, 5, 2, 5, 7, 6, 5, 10, 7, 5, 5, 7, 7, 6, 5, 6, 6, 7, 2, 10, 6, 8, 7, 7, 6, 6, 5, 7, 11, 5, 7, 10, 5, 6, 5, 8, 8, 9, 7, 8, 6, 6, 5, 10, 7, 5, 6, 5, 8, 7, 2, 5, 10, 8, 6, 10, 8, 7, 7, 11, 7, 7, 6, 10, 7, 5, 5

Offset: 1

Zhi-Wei Sun, Mar 06 2018

nonn

Conjecture 1: a(n) > 0 for all n > 0.
Conjecture 2: Each positive square n^2 can be written as 4^k*m + x^2 + y^2 with k,x,y nonnegative integers and m among 1, 5 17.
Conjecture 3: For each n = 1,2,3,... we can write n^2 as 4^k*m + x^2 + y^2 with k,x,y nonnegative integers and m among 1, 5, 65.
Conjecture 3 implies that A300441(n) > 0 for all n > 0 since  4*A001353(0)^2 + 1 = 1, 4*A001353(1)^2 + 1 = 5 and 4*A001353(2)^2 + 1 = 65.
We have verified Conjectures 1-3 for all n = 1..2*10^7.

			a(1) = 1 since 1^2 - 4^0*1 = 0 = 0^2 + 0^2.
a(2) = 2 since 2^2 - 4^0*2 = 2 = 1^2 + 1^2 and 2^2 - 4*1 = 0 = 0^2 + 0^2.
a(4) = 2 since 4^2 - 4^1*2 = 8 = 2^2 + 2^2 and 4^2 - 4^2*1 = 0 = 0^2 + 0^2.

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, A000302, A001481, A271518, A281976, A299924, A299537, A299794, A300362, A300441, A300396.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[If[Mod[Part[Part[f[n],i],1]-3,4]==0&&Mod[Part[Part[f[n],i],2],2]==1,1,0],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=n==0||(n>0&&g[n]);
u[1]=1; u[2]=2; u[3]=5;
tab={};Do[r=0;Do[If[QQ[n^2-4^k*u[m]],r=r+1],{m,1,3},{k,0,Log[4,n^2/u[m]]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301534 Number of ways to write the n-th prime congruent to 7 modulo 12 as x^2 + 3y^2 + 152^z with x,y,z nonnegative integers.

Original entry on oeis.org

0, 2, 3, 4, 5, 5, 2, 6, 6, 4, 7, 4, 9, 6, 6, 6, 7, 9, 5, 10, 3, 9, 7, 9, 8, 11, 9, 8, 10, 5, 8, 9, 4, 10, 7, 7, 7, 8, 7, 13, 8, 6, 6, 14, 7, 15, 3, 11, 8, 10, 8, 7, 7, 9, 6, 9, 7, 7, 10, 12, 6, 9, 4, 7, 10, 12, 12, 7, 13, 9, 12, 6, 7, 10, 5, 8, 7, 12, 12, 10

Offset: 1

Zhi-Wei Sun, Apr 16 2018

nonn

Conjecture: a(n) > 0 for all n > 1. In other words, any prime p > 7 with p == 7 (mod 12) can be written as x^2 + 3*y^2 + 15*2^z with x,y,z nonnegative integers.

We have verified the conjecture for all primes p == 7 (mod 12) with 7 < p < 8*10^9.

			a(1) = 0 since 7 cannot be written as x^2 + 3*y^2 + 15*2^z with x,y,z nonnegative integers.
a(2) = 2 since the second prime congruent to 7 modulo 12 is 19 and 19 = 1^2 + 3*1^2 + 15*2^0 = 2^2 + 3*0^2 + 15*2^0.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Cf. A000040, A000079, A000290, A092572, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301471, A301472, A302920.

p[n_]:=p[n]=Prime[n];
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[MemberQ[{2},Mod[Part[Part[f[n],i],1],3]]&&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]);
n=0;Do[If[Mod[p[m],12]!=7,Goto[aa]];n=n+1;r=0;Do[If[QQ[p[m]-15*2^k],Do[If[SQ[p[m]-15*2^k-3x^2],r=r+1],{x,0,Sqrt[(p[m]-15*2^k)/3]}]],{k,0,Log[2,p[m]/15]}];Print[n," ",r];Label[aa],{m,1,315}]

A302641 Number of nonnegative integers k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, Apr 10 2018

nonn

The author's Square Conjecture in A301471 implies that a(n) > 0 for all n > 1.

We have a(2^n) = 1 for all n > 0. In fact, (2^n)^2 = (2^(n-1))^2 + 2*0^2 + 3*2^(2*n-2). If k > 2*n-2 then 3*2^k >= 6*2^(2*n-2) > (2^n)^2. If 0 <= k < 2*n-2, then 2*n-k is at least 3 and hence (2^n)^2 - 3*2^k = 2^k*(2^(2*n-k)-3) cannot be written as x^2 + 2*y^2 with x and y integers.

			a(2) = 1 with 2^2 = 1^2 + 2*0^2 + 3*2^0.
a(3) = 2 with 3^2 = 2^2 + 2*1^2 + 3*2^0 = 1^2 + 2*1^2 + 3*2^1.
a(2857932461) = 1 since 3 is the only nonnegative integer k such that 2857932461^2 - 3*2^k has the form x^2 + 2*y^2 with x and y integers.
a(4428524981) = 2 since 3 and 8 are the only nonnegative integers k such that 4428524981^2 - 3*2^k has the form x^2 + 2*y^2 with x and y integers.
a(4912451281) = 3 since 3, 6 and 7 are the only nonnegative integers k with 4428524981^2 - 3*2^k = x^2 + 2*y^2 for some integers x and y.

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.
Zhi-Wei Sun, My square conjecture with prize, a message to Number Theory List, April 7, 2018.

Cf. A000079, A000290, A002479, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301472, A301479, A301579, A301640.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[(Mod[Part[Part[f[n],i],1],8]==5||Mod[Part[Part[f[n],i],1],8]==7)&&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]);
tab={};Do[r=0;Do[If[QQ[n^2-3*2^k],r=r+1],{k,0,Log[2,n^2/3]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301805 Number of ways to write 3n^2 as x^2 + 10y^2 + 2^z, where x, y and z are nonnegative integers with z > 3.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 3, 3, 3, 2, 4, 4, 4, 4, 4, 4, 6, 4, 4, 3, 5, 4, 4, 5, 7, 5, 4, 4, 6, 4, 7, 5, 5, 7, 7, 5, 5, 4, 8, 5, 7, 6, 11, 6, 6, 5, 8, 5, 6, 7, 5, 7, 6, 5, 5, 5, 7, 7, 4, 4, 8, 8, 8, 6, 6, 6, 9, 8, 8, 7, 8, 6, 10, 6, 10, 6, 8, 8, 8, 5

Offset: 1

Zhi-Wei Sun, Mar 27 2018

nonn

It might seem that a(n) > 0 for all n > 2. However, we find that a(323525083) = 0, moreover 3*323525083^2 cannot be written as x^2 + 10*y^2 + 2^z with x,y,z nonnegative integers. We note also that a(270035155) = 0 but 3*270035155^2 - 2^0 has the form x^2 + 10*y^2 with x and y integers.
My way to check whether 3*n^2 can be written as x^2 + 10*y^2 + 2^z is to find z such that 3*n^2 - 2^z can be written as x^2 + 10*y^2. I observe that a positive integer n has the form x^2 + 10*y^2 with x and y integers if and only if the p-adic order ord_p(n) of n is even for any prime p == 3, 17, 21, 27, 29, 31, 33, 39 (mod 40) and the sum of those ord_p(n) with p prime and p == 2, 5, 7, 13, 23, 37 (mod 40) is even.
From David A. Corneth, Mar 27 2018: (Start)
If a(n) > 0 then a(2*n) > 0; 3*n^2 = x^2 + 10*y^2 + 2^z <=> 3*(2*n)^2 = 4 * 3*n^2 = 4 * (x^2 + 10*y^2 + 2^z) = (2*x)^2 + 10 * (2*y)^2 + 2^(z + 2).
So we just need to check odd n and as z > 0, 2 | 2^z and furthermore 2 | 10 * y^2 so 3*x^2 must be odd, i.e., x must be odd for 3*n^2 to be odd. Also, y must be odd. For odd n, 3*n^2 == 3 (mod 4), for odd x, x^2 == (1 mod 4), for z >= 3, 2^z == 0 (mod 4) so 10 * y^2 must be == 2 (mod 4) which happens if and only if y is odd. (End)

			a(1) = a(2) = 0 since 3*1^2 < 3*2^2 < 2^4.
a(3) = 1 since 3*3^2 = 1^2 + 10*1^2 + 2^4.
a(4) = 1 since 3*4^2 = 4^2 + 10*0^2 + 2^5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..800
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. A000079, A000290, A020673, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301471, A301472, A301479, A301579, A301640.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[SQ[3*n^2-2^k-10x^2],r=r+1],{k,4,Log[2,3n^2]},{x,0,(3*n^2-2^k)/10}];tab=Append[tab,r],{n,1,80}];Print[tab]

A301891 Number of ways to write 2n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3y + 5z + 15w is a power of 4.

Original entry on oeis.org

2, 1, 1, 2, 3, 1, 2, 1, 3, 1, 2, 1, 6, 3, 8, 2, 5, 4, 6, 3, 4, 3, 6, 1, 3, 3, 7, 2, 8, 9, 8, 1, 15, 5, 8, 3, 11, 1, 5, 1, 4, 4, 2, 2, 10, 7, 17, 1, 18, 11, 14, 6, 16, 6, 17, 3, 21, 10, 16, 8, 19, 8, 30, 2, 15, 9, 18, 5, 28, 5, 27, 4, 13, 11, 24, 6, 28, 17, 20, 3

Offset: 1

Zhi-Wei Sun, Mar 28 2018

nonn

Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 4^k*m with k = 0,1,2,... and m = 2, 3, 6, 10, 38.

Conjecture 2: For any positive integer n, we can write 2*n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3*y + 5*z + 15*w is twice a power of 4.

			a(1) = 2 since 2*1^2 = 1^2 + 1^2 + 0^2 + 0^2 with 1 + 3*1 + 5*0 + 15*0 = 4, and 2*1^2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 + 3*0 + 5*0 + 15*1 = 4^2.
a(2) = 1 since 2*2^2 = 0^2 + 2^2 + 2^2 + 0^2 with 0 + 3*2 + 5*2 + 15*0 = 4^2.
a(3) = 1 since 2*3^2 = 1^2 + 1^2 + 0^2 + 4^2 with 1 + 3*1 + 5*0 + 15*4 = 4^3.
a(6) = 1 since 2*6^2 = 0^2 + 8^2 + 2^2 + 2^2 with 0 + 3*8 + 5*2 + 15*2 = 4^3.
a(10) = 1 since 2*10^2 = 10^2 + 8^2 + 6^2 + 0^2 with 10 + 3*8 + 5*6 + 15*0 = 4^3.
a(38) = 1 since 2*38^2 = 34^2 + 34^2 + 24^2 + 0^2 with 34 + 3*34 + 5*24 + 15*0 = 4^4.

Zhi-Wei Sun, Table of n, a(n) for n = 1..225
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. A000290, A000302, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300510, A301376, A301391, A301452, A301471, A301579, A301640.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]]
tab={};Do[r=0;Do[If[SQ[2n^2-x^2-y^2-z^2]&&Pow[x+3y+5z+15*Sqrt[2n^2-x^2-y^2-z^2]],r=r+1],{x,0,Sqrt[2]n},{y,0,Sqrt[2n^2-x^2]},{z,0,Sqrt[2n^2-x^2-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

cp's OEIS Frontend

A301579 Least nonnegative integer k such that n^2 - 3*2^k can be written as x^2 + 2*y^2 with x and y integers, or -1 if no such k exists.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A301640 Largest integer k such that n^2 - 3*2^k can be written as x^2 + 2*y^2 with x and y integers, or -1 if no such k exists.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

A303235 Number of ordered pairs (x, y) with 0 <= x <= y such that n - 2^x - 2^y can be written as the sum of two triangular numbers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A300448 Number of the integers 4^k*m with k >= 0 and m = 1, 2, 5 such that n^2 - 4^k*m can be written as the sum of two squares.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A301534 Number of ways to write the n-th prime congruent to 7 modulo 12 as x^2 + 3*y^2 + 15*2^z with x,y,z nonnegative integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A302641 Number of nonnegative integers k such that n^2 - 3*2^k can be written as x^2 + 2*y^2 with x and y integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A301805 Number of ways to write 3*n^2 as x^2 + 10*y^2 + 2^z, where x, y and z are nonnegative integers with z > 3.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A301891 Number of ways to write 2*n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3*y + 5*z + 15*w is a power of 4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A301579 Least nonnegative integer k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers, or -1 if no such k exists.

A301640 Largest integer k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers, or -1 if no such k exists.

A300448 Number of the integers 4^km with k >= 0 and m = 1, 2, 5 such that n^2 - 4^km can be written as the sum of two squares.

A301534 Number of ways to write the n-th prime congruent to 7 modulo 12 as x^2 + 3y^2 + 152^z with x,y,z nonnegative integers.

A302641 Number of nonnegative integers k such that n^2 - 32^k can be written as x^2 + 2y^2 with x and y integers.

A301805 Number of ways to write 3n^2 as x^2 + 10y^2 + 2^z, where x, y and z are nonnegative integers with z > 3.

A301891 Number of ways to write 2n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3y + 5z + 15w is a power of 4.