Displaying 1-10 of 24 results found.
Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c + 2^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.
+10
34
0, 1, 2, 3, 4, 5, 4, 6, 7, 7, 7, 9, 7, 8, 9, 9, 8, 12, 11, 11, 11, 11, 11, 14, 11, 13, 12, 11, 10, 14, 11, 12, 17, 15, 12, 16, 14, 15, 17, 19, 15, 16, 13, 15, 17, 17, 16, 20, 16, 14, 17, 17, 14, 22, 17, 14, 14, 17, 15, 19
COMMENTS
Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two triangular numbers and two powers of 2.
a(n) > 0 for all n = 2..10^9. See A303234 for numbers of the form x*(x+1)/2 + 2^y with x and y nonnegative integers. See also A303363 for a stronger conjecture. In contrast, Crocker proved in 2008 that there are infinitely many positive integers not representable as the sum of two squares and at most two powers of 2.
REFERENCES
R. C. Crocker, On the sum of two squares and two powers of k, Colloq. Math. 112(2008), 235-267.
EXAMPLE
a(2) = 1 with 2 = 0*(0+1)/2 + 0*(0+1)/2 + 2^0 + 2^0.
a(3) = 2 with 3 = 0*(0+1)/2 + 1*(1+1)/2 + 2^0 + 2^0 = 0*(0+1)/2 + 0*(0+1)/2 + 2^0 + 2^1.
a(4) = 3 with 4 = 1*(1+1)/2 + 1*(1+1)/2 + 2^0 + 2^0 = 0*(0+1)/2 + 1*(1+1)/2 + 2^0 + 2^1 = 0*(0+1)/2 + 0*(0+1)/2 + 2^1 + 2^1.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
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], Do[If[SQ[8(n-2^k-2^j-x(x+1)/2)+1], r=r+1], {x, 0, (Sqrt[4(n-2^k-2^j)+1]-1)/2}]], {k, 0, Log[2, n]-1}, {j, k, Log[2, n-2^k]}]; tab=Append[tab, r], {n, 1, 60}]; Print[tab]
CROSSREFS
Cf. A000079, A000217, A271518, A273812, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303234, A303338, A303363.
Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 2^c + 2^d, where a,b,c,d are nonnegative integers with a <= b, c <= d and 2|c.
+10
33
0, 1, 2, 2, 3, 3, 2, 4, 6, 3, 5, 6, 4, 6, 7, 4, 4, 9, 6, 6, 8, 4, 9, 9, 5, 7, 7, 5, 7, 9, 4, 8, 13, 7, 6, 11, 7, 10, 13, 8, 9, 10, 7, 9, 11, 7, 9, 15, 8, 8, 14, 6, 9, 16, 6, 8, 11, 11, 10, 12, 8, 7, 15, 10, 8, 11, 9, 14, 15, 9
COMMENTS
Conjecture: a(n) > 0 for all n > 1.
This is stronger than the author's conjecture in A303233. I have verified a(n) > 0 for all n = 2..10^9. In contrast, Corcker proved in 2008 that there are infinitely many positive integers not representable as the sum of two squares and at most two powers of 2.
EXAMPLE
a(2) = 1 with 2 = 0*(0+1)/2 + 0*(0+1)/2 + 2^0 + 2^0.
a(3) = 2 with 3 = 0*(0+1)/2 + 1*(1+1)/2 + 2^0 + 2^0 = 0*(0+1)/2 + 0*(0+1)/2 + 2^0 + 2^1.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
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-4^j-2^k)+1], Do[If[SQ[8(n-4^j-2^k-x(x+1)/2)+1], r=r+1], {x, 0, (Sqrt[4(n-4^j-2^k)+1]-1)/2}]], {j, 0, Log[4, n/2]}, {k, 2j, Log[2, n-4^j]}]; tab=Append[tab, r], {n, 1, 70}]; Print[tab]
CROSSREFS
Cf. A000079, A000217, A271518, A273812, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303389.
Number of ways to write n as x^2 + 2*y^2 + 3*2^z + 4^w with x,y,z,w nonnegative integers.
+10
32
0, 0, 0, 1, 1, 1, 3, 3, 2, 4, 3, 2, 6, 2, 4, 8, 2, 4, 7, 3, 4, 8, 5, 5, 10, 6, 4, 10, 8, 5, 12, 7, 3, 12, 4, 5, 12, 5, 5, 14, 7, 4, 12, 7, 6, 12, 6, 6, 10, 7, 7, 12, 7, 6, 14, 6, 8, 16, 4, 8, 18, 5, 6, 16, 5, 9, 13, 7, 7, 14
COMMENTS
Conjecture: a(n) > 0 for all n > 3.
This is stronger than the author's previous conjecture in A302983. It has been verified that a(n) > 0 for all n = 4..10^9. Jiao-Min Lin (a student at Nanjing University) has found a counterexample to the conjecture: a(12558941213) = 0. - Zhi-Wei Sun, Jul 30 2022
EXAMPLE
a(4) = 1 with 4 = 0^2 + 2*0^2 + 3*2^0 + 4^0.
a(5) = 1 with 5 = 1^2 + 2*0^2 + 3*2^0 + 4^0.
a(6) = 1 with 6 = 0^2 + 2*1^2 + 3*2^0 + 4^0.
a(9) = 2 with 9 = 0^2 + 2*1^2 + 3*2^0 + 4^1 = 0^2 + 2*1^2 + 3*2^1 + 4^0.
MATHEMATICA
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[MemberQ[{5, 7}, Mod[Part[Part[f[n], i], 1], 8]]&&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[QQ[n-3*2^k-4^j], Do[If[SQ[n-3*2^k-4^j-2x^2], r=r+1], {x, 0, Sqrt[(n-3*2^k-4^j)/2]}]], {k, 0, Log[2, n/3]}, {j, 0, If[3*2^k==n, -1, Log[4, n-3*2^k]]}]; tab=Append[tab, r], {n, 1, 70}]; Print[tab]
CROSSREFS
Cf. A000079, A000290, A002479, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303363.
Number of ways to write n as x^2 + 5*y^2 + 2^z + 3*2^w with x,y,z,w nonnegative integers.
+10
30
0, 0, 0, 1, 2, 1, 2, 4, 3, 3, 5, 4, 6, 7, 4, 7, 5, 4, 7, 8, 5, 5, 8, 5, 9, 7, 6, 13, 10, 7, 9, 10, 7, 12, 11, 8, 11, 7, 7, 11, 11, 6, 11, 13, 6, 10, 7, 7, 17, 13, 6, 13, 14, 9, 11, 18, 10, 13, 14, 11
COMMENTS
Conjecture: a(n) > 0 for all n > 3.
Clearly, a(4*n) > 0 if a(n) > 0. We have verified a(n) > 0 for all n = 4..2*10^8.
EXAMPLE
a(4) = 1 with 4 = 0^2 + 5*0^2 + 2^0 + 3*2^0.
a(5) = 2 with 5 = 1^2 + 5*0^2 + 2^0 + 3*2^0 = 0^2 + 5*0^2 + 2^1 + 3*2^0.
a(6) = 1 with 6 = 1^2 + 3*0^2 + 2^1 + 3*2^0.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[r=0; Do[If[SQ[n-3*2^k-2^j-5x^2], r=r+1], {k, 0, Log[2, n/3]}, {j, 0, If[n==3*2^k, -1, Log[2, n-3*2^k]]}, {x, 0, Sqrt[(n-3*2^k-2^j)/5]}]; tab=Append[tab, r], {n, 1, 60}]; Print[tab]
CROSSREFS
Cf. A000079, A000290, A020669, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A301534, A302920, A302981, A302983, A302984.
Number of ways to write n as x^2 + 2*y^2 + 2^z + 5*2^w with x,y,z,w nonnegative integers.
+10
30
0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 3, 4, 5, 5, 8, 5, 5, 7, 4, 6, 7, 9, 9, 10, 10, 7, 9, 8, 10, 15, 10, 9, 10, 8, 6, 10, 10, 11, 14, 14, 8, 12, 13, 13, 20, 15, 12, 16, 10, 15, 12, 10, 15, 17, 16, 12, 16, 14, 14, 21
COMMENTS
Conjecture: a(n) > 0 for all n > 5.
Clearly, a(2*n) > 0 if a(n) > 0. We have verified a(n) > 0 for all n = 6...10^9.
EXAMPLE
a(6) = 1 with 6 = 0^2 + 2*0^2 + 2^0 + 5*2^0.
a(7) = 2 with 7 = 1^2 + 2*0^2 + 2^0 + 5*2^0 = 0^2 + 2*0^2 + 2^1 + 5*2^0.
a(8) = 2 with 8 = 0^2 + 2*1^2 + 2^0 + 5*2^0 = 1^2 + 2*0^2 + 2^1 + 5*2^0.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[MemberQ[{5, 7}, Mod[Part[Part[f[n], i], 1], 8]]&&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-5*2^k-2^j], Do[If[SQ[n-5*2^k-2^j-2x^2], r=r+1], {x, 0, Sqrt[(n-5*2^k-2^j)/2]}]], {k, 0, Log[2, n/5]}, {j, 0, Log[2, Max[1, n-5*2^k]]}]; tab=Append[tab, r], {n, 1, 60}]; Print[tab]
CROSSREFS
Cf. A000079, A000290, A002479, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302985.
Numbers of the form x*(x+1)/2 + 2^y with x and y nonnegative integers.
+10
30
1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 22, 23, 25, 26, 29, 30, 31, 32, 33, 35, 36, 37, 38, 40, 42, 44, 46, 47, 49, 52, 53, 56, 57, 59, 60, 61, 63, 64, 65, 67, 68, 70, 71, 74, 77, 79, 80
COMMENTS
Conjecture: Any integer n > 1 can be written as the sum of two terms of the current sequence.
This is equivalent to the author's conjecture in A303233.
EXAMPLE
a(1) = 1 with 1 = 0*(0+1)/2 + 2^0.
a(2) = 2 with 2 = 1*(1+1)/2 + 2^0 = 0*(0+1)/2 + 2^1.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={}; Do[Do[If[SQ[8(n-2^k)+1], tab=Append[tab, n]; Goto[aa]], {k, 0, Log[2, n]}]; Label[aa], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000079, A000217, A271518, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233.
Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + 5^c + 5^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.
+10
30
0, 1, 1, 1, 1, 2, 1, 3, 2, 2, 2, 4, 3, 2, 2, 3, 3, 3, 2, 2, 2, 4, 3, 2, 1, 5, 4, 3, 2, 5, 5, 5, 5, 3, 3, 5, 5, 4, 4, 4, 5, 5, 2, 5, 3, 5, 4, 7, 2, 4, 6, 6, 5, 4, 4, 5, 8, 4, 4, 4, 7, 6, 4, 3, 4, 8, 4, 7, 3, 3, 6, 8, 2, 5, 6, 5, 4, 6, 4, 3
COMMENTS
Conjecture: a(n) > 0 for all n > 1. In other words, any integers n > 1 can be written as the sum of two triangular numbers and two powers of 5.
This has been verified for all n = 2..10^10.
See A303393 for the numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.
EXAMPLE
a(4) = 1 with 4 = 1*(1+1)/2 + 1*(1+1)/2 + 5^0 + 5^0.
a(5) = 1 with 5 = 0*(0+1)/2 + 2*(2+1)/2 + 5^0 + 5^0.
a(7) = 1 with 7 = 0*(0+1)/2 + 1*(1+1)/2 + 5^0 + 5^1.
a(25) = 1 with 25 = 0*(0+1)/2 + 5*(5+1)/2 + 5^1 + 5^1.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
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-5^j-5^k)+1], Do[If[SQ[8(n-5^j-5^k-x(x+1)/2)+1], r=r+1], {x, 0, (Sqrt[4(n-5^j-5^k)+1]-1)/2}]], {j, 0, Log[5, n/2]}, {k, j, Log[5, n-5^j]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000217, A000351, A271518, A273812, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303363, A303393, A303401, A303432, A303540.
Numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.
+10
29
1, 2, 4, 5, 6, 7, 8, 11, 15, 16, 20, 22, 25, 26, 28, 29, 31, 33, 35, 37, 40, 41, 46, 50, 53, 56, 60, 61, 67, 70, 71, 79, 80, 83, 91, 92, 96, 103, 106, 110, 116, 121, 125, 126, 128, 130, 131, 135, 137, 140, 141, 145, 146, 153, 154, 158, 161, 170, 172, 176
COMMENTS
The author's conjecture in A303389 has the following equivalent version: Each integer n > 1 can be expressed as the sum of two terms of the current sequence. This has been verified for all n = 2..2*10^8.
EXAMPLE
a(1) = 1 with 1 = 0*(0+1)/2 + 5^0.
a(2) = 2 with 2 = 1*(1+1)/2 + 5^0.
a(3) = 4 with 4 = 2*(2+1)/2 + 5^0.
MATHEMATICA
TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={}; Do[Do[If[TQ[m-5^k], tab=Append[tab, m]; Goto[aa]], {k, 0, Log[5, m]}]; Label[aa], {m, 1, 176}]; Print[tab]
CROSSREFS
Cf. A000217, A000351, A271518, A273812, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303363, A303389.
Number of ordered pairs (a, b) with 0 <= a <= b such that n - 5^a - 5^b can be written as the sum of two triangular numbers.
+10
27
0, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 1, 4, 3, 3, 2, 5, 4, 4, 4, 3, 3, 4, 4, 3, 4, 4, 4, 3, 2, 4, 3, 3, 3, 5, 2, 4, 5, 4, 4, 4, 4, 3, 5, 3, 4, 4, 4, 4, 4, 3, 3, 5, 4, 5, 3, 3, 5, 5, 2, 4, 6, 3, 3, 4, 4, 3
COMMENTS
Conjecture: a(n) > 0 for all n > 1.
This is equivalent to the author's conjecture in A303389. It has been verified that a(n) > 0 for all n = 2..6*10^9. Note that a nonnegative integer m is the sum of two triangular numbers if and only if 4*m + 1 can be written as the sum of two squares.
EXAMPLE
a(6) = 2 with 6 - 5^0 - 5^0 = 1*(1+1)/2 + 2*(2+1)/2 and 6 - 5^0 - 5^1 = 0*(0+1)/2 + 0*(0+1)/2.
a(7) = 1 with 7 - 5^0 - 5^1 = 0*(0+1)/2 + 1*(1+1)/2.
a(25) = 1 with 25 - 5^1 - 5^1 = 0*(0+1)/2 + 5*(5+1)/2.
MATHEMATICA
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-5^j-5^k)+1], r=r+1], {j, 0, Log[5, n/2]}, {k, j, Log[5, n-5^j]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000217, A000351, A271518, A273812, A281976, A299924, A300219, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303363, A303389, A303393.
Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y even such that x^2 - (6*y)^2 = 4^k for some k = 0,1,2,....
+10
25
1, 1, 2, 1, 1, 2, 1, 1, 4, 1, 1, 2, 1, 1, 2, 1, 3, 4, 1, 1, 8, 2, 2, 2, 3, 2, 6, 1, 2, 2, 1, 1, 11, 3, 2, 4, 4, 3, 3, 1, 6, 10, 6, 2, 7, 2, 3, 2, 6, 3, 8, 2, 7, 7, 2, 1, 11, 4, 4, 2, 2, 1, 6, 1, 3, 11, 3, 3, 16, 3, 5, 4, 8, 5, 2, 3, 11, 5, 8, 1
COMMENTS
Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 11, 13, 19, 2^k*m (k = 0,1,2,... and m = 1, 5, 7, 31).
We have verified a(n) > 0 for all n = 1..10^7.
See also A301376 for a similar conjecture.
EXAMPLE
a(2) = 1 since 2^2 = 2^2 + 0^2 + 0^2 + 0^2 with 2^2 - (6*0)^2 = 4^1.
a(5) = 1 since 5^2 = 4^2 + 0^2 + 0^2 + 3^2 with 4^2 - (6*0)^2 = 4^2.
a(7) = 1 since 7^2 = 2^2 + 0^2 + 3^2 + 6^2 with 2^2 - (6*0)^2 = 4^1.
a(11) = 1 since 11 = 2^2 + 0^2 + 6^2 + 9^2 with 2^2 - (6*0)^2 = 4^1.
a(13) = 1 since 13 = 4^2 + 0^2 + 3^2 + 12^2 with 4^2 - (6*0)^2 = 4^2.
a(19) = 1 since 19 = 1^2 + 0^2 + 6^2 + 18^2 with 1^2 - (6*0)^2 = 4^0.
a(31) = 1 since 31^2 = 20^2 + 2^2 + 14^2 + 19^2 with 20^2 - (6*2)^2 = 4^4.
a(75) = 2 since 75^2 = 68^2 + 10^2 + 1^2 + 30^2 = 68^2 + 10^2 + 15^2 + 26^2 with 68^2 - (6*10)^2 = 4^5.
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+144m^2]&&QQ[n^2-4^k-148m^2], Do[If[SQ[n^2-(4^k+148m^2)-z^2], r=r+1], {z, 0, Sqrt[(n^2-4^k-148m^2)/2]}]], {k, 0, Log[2, n]}, {m, 0, Sqrt[(n^2-4^k)/148]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000118, A000290, A000302, A299537, A299794, A299924, A300219, A300396, A300441, A300510, A301376.
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