Displaying 1-10 of 32 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 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 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 as a*(3*a-1)/2 + b*(3*b-1)/2 + 3^c + 3^d with a,b,c,d nonnegative integers.
+10
27
0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 4, 1, 3, 2, 3, 2, 3, 3, 2, 1, 2, 3, 3, 2, 2, 2, 4, 4, 4, 3, 2, 3, 3, 3, 4, 3, 4, 2, 5, 4, 5, 1, 2, 3, 5, 2, 3, 2, 3, 2, 4, 5, 5, 3, 3, 3, 4, 4, 3, 2, 4, 4, 4, 3, 3, 3, 2, 3, 3, 2, 4, 2, 4, 5, 4, 5, 1, 3, 4
COMMENTS
Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two pentagonal numbers and two powers of 3.
a(n) > 0 for all n = 2..7*10^6. See A303434 for the numbers of the form x*(3*x-1)/2 + 3^y with x and y nonnegative integers. See also A303389 and A303432 for similar conjectures.
FORMULA
a(78) = 1 with 78 = 3*(3*3-1)/2 + 3*(3*3-1)/2 + 3^3 + 3^3.
a(285) = 1 with 285 = 3*(3*1-1)/2 + 11*(3*11-1)/2 + 3^3 + 3^4.
a(711) = 1 with 711 = 9*(3*9-1)/2 + 20*(3*20-1)/2 + 3^0 + 3^1.
a(775) = 1 with 775 = 7*(3*7-1)/2 + 21*(3*21-1)/2 + 3^3 + 3^3.
a(3200) = 1 with 12*(3*12-1)/2 + 44*(3*44-1)/2 + 3^3 + 3^4.
a(13372) = 1 with 13372 = 17*(3*17-1)/2 + 65*(3*65-1)/2 + 3^4 + 3^8.
a(16545) = 1 with 16545 = 0*(3*0-1)/2 + 98*(3*98-1)/2 + 3^0 + 3^7.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PenQ[n_]:=PenQ[n]=SQ[24n+1]&&(n==0||Mod[Sqrt[24n+1]+1, 6]==0);
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[12(n-3^j-3^k)+1], Do[If[PenQ[n-3^j-3^k-x(3x-1)/2], r=r+1], {x, 0, (Sqrt[12(n-3^j-3^k)+1]+1)/6}]], {j, 0, Log[3, n/2]}, {k, j, Log[3, n-3^j]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000244, A000326, A303233, A303338, A303363, A303389, A303393, A303399, A303428, A303432, A303434.
Number of ways to write n as a*(2*a-1) + b*(2*b-1) + 2^c + 2^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.
+10
27
0, 1, 2, 3, 3, 3, 2, 3, 4, 5, 4, 4, 2, 3, 3, 4, 5, 7, 5, 5, 4, 4, 4, 7, 5, 4, 3, 2, 2, 4, 5, 7, 8, 7, 5, 7, 5, 7, 7, 7, 4, 4, 2, 3, 5, 7, 6, 9, 7, 6, 5, 6, 5, 7, 7, 3, 3, 3, 3, 5, 7, 7, 8, 7, 6, 8, 5, 8, 8, 8, 5, 7, 4, 6, 7, 9, 8, 9, 7, 8
COMMENTS
Conjecture 1: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two hexagonal numbers and two powers of 2.
Conjecture 2: Any integer n > 1 can be written as a*(2*a+1) + b*(2*b+1) + 2^c + 2^d with a,b,c,d nonnegative integers.
Conjecture 3: Each integer n > 1 can be written as a*(2*a-1) + b*(2*b+1) + 2^c + 2^d with a,b,c,d nonnegative integers.
All the three conjectures hold for n = 2..2*10^6. Note that either of them is stronger than the conjecture in A303233.
EXAMPLE
a(2) = 1 with 2 = 0*(2*0-1) + 0*(2*0-1) + 2^0 + 2^0.
a(7) = 2 with 7 = 1*(2*1-1) + 1*(2*1-1) + 2^0 + 2^2 = 0*(2*0-1) + 1*(2*1-1) + 2^1 + 2^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
HexQ[n_]:=HexQ[n]=SQ[8n+1]&&(n==0||Mod[Sqrt[8n+1]+1, 4]==0);
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^j-2^k)+1], Do[If[HexQ[n-2^j-2^k-x(2x-1)], r=r+1], {x, 0, (Sqrt[4(n-2^j-2^k)+1]+1)/4}]], {j, 0, Log[2, n/2]}, {k, j, Log[2, n-2^j]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
Number of ways to write n as a^2 + b^2 + binomial(2*c,c) + binomial(2*d,d), where a,b,c,d are nonnegative integers with a <= b and c <= d.
+10
26
0, 1, 2, 3, 2, 2, 3, 4, 3, 2, 3, 6, 4, 2, 2, 4, 4, 2, 2, 5, 5, 5, 4, 4, 4, 4, 5, 6, 5, 5, 4, 5, 4, 4, 3, 4, 5, 5, 6, 5, 5, 5, 4, 7, 3, 4, 5, 6, 4, 2, 4, 6, 7, 4, 4, 5, 7, 6, 2, 5, 4, 6, 3, 2, 5, 5, 5, 4, 4, 3, 7, 9, 6, 5, 6, 11, 7, 3, 4, 8
COMMENTS
Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two squares and two central binomial coefficients.
It has been verified that a(n) > 0 for all n = 2..10^10.
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 10^11. - Zhi-Wei Sun, Jul 30 2022
EXAMPLE
a(2) = 1 since 2 = 0^2 + 0^2 + binomial(2*0,0) + binomial(2*0,0).
a(10) = 2 with 10 = 2^2 + 2^2 + binomial(2*0,0) + binomial(2*0,0) = 1^2 + 1^2 + binomial(2*1,1) + binomial(2*2,2).
a(2435) = 1 with 2435 = 32^2 + 33^2 + binomial(2*4,4) + binomial(2*5,5).
MAPLE
N:= 100: # for a(1)..a(N)
A:= Vector(N):
for b from 0 to floor(sqrt(N)) do
for a from 0 to min(b, floor(sqrt(N-b^2))) do
t:= a^2+b^2;
for d from 0 do
s:= t + binomial(2*d, d);
if s > N then break fi;
for c from 0 to d do
u:= s + binomial(2*c, c);
if u > N then break fi;
A[u]:= A[u]+1;
od od od od:
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
c[n_]:=c[n]=Binomial[2n, 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; k=0; Label[bb]; If[c[k]>n, Goto[aa]]; Do[If[QQ[n-c[k]-c[j]], Do[If[SQ[n-c[k]-c[j]-x^2], r=r+1], {x, 0, Sqrt[(n-c[k]-c[j])/2]}]], {j, 0, k}]; k=k+1; Goto[bb]; Label[aa]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000290, A000984, A001481, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303541, A303543, A303601, A303639.
Number of ways to write n as x*(3*x-2) + y*(3*y-2) + 3^u + 3^v, where x,y,u,v are integers with x <= y and 0 <= u <= v.
+10
25
0, 1, 1, 2, 1, 2, 2, 2, 1, 3, 3, 5, 2, 3, 3, 2, 2, 5, 4, 5, 2, 3, 5, 2, 3, 5, 4, 7, 2, 4, 5, 3, 4, 6, 4, 7, 3, 6, 6, 4, 4, 5, 5, 9, 5, 6, 6, 2, 5, 5, 7, 8, 4, 5, 4, 4, 4, 6, 6, 8, 3, 6, 6, 3, 4, 6, 7, 8, 5, 8, 6, 5, 4, 6, 7, 8, 6, 6, 6, 2
COMMENTS
Conjecture: a(n) > 0 for all n > 1. Moreover, any integer n > 1 can be written as x*(3*x+2) + y*(3*y+2) + 3^z + 3^w, where x is an integer and y,z,w are nonnegative integers.
a(n) > 0 for all n = 2..3*10^8. Those x*(3*x-2) with x integral are called generalized octagonal numbers ( A001082). 76683391 is the least integer n > 1 not representable as the sum of two generalized octagonal numbers and two powers of 2.
EXAMPLE
a(2) = 1 with 2 = 0*(3*0-2) + 0*(3*0-2) + 3^0 + 3^0.
a(3) = 1 with 3 = 0*(3*0-2) + 1*(3*1-2) + 3^0 + 3^0.
a(4) = 2 with 4 = 1*(3*1-2) + 1*(3*1-2) + 3^0 + 3^0 = 0*(3*0-2) + 0*(3*0-2) + 3^0 + 3^1.
a(5) = 1 with 5 = 0*(3*0-2) + 1*(3*1-2) + 3^0 + 3^1.
a(9) = 1 with 9 = (-1)*(3*(-1)-2) + 0*(3*0-2) + 3^0 + 3^1.
a(4360) = 4 with 4360 = (-35)*(3*(-35)-2) + (-13)*(3*(-13)-2) + 3^0 + 3^4 = (-37)*(3*(-37)-2) + (-7)*(3*(-7)-2) + 3^2 + 3^2 = (-27)*(3*(-27)-2) + (-23)*(3*(-23)-2) + 3^5 + 3^5 = (-25)*(3*(-25)-2) + (-1)*(3*(-1)-2) + 3^5 + 3^7.
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[3(n-3^j-3^k)+2], Do[If[SQ[3(n-3^j-3^k-x(3x-2))+1], r=r+1], {x, -Floor[(Sqrt[3(n-3^j-3^k)/2+1]-1)/3], (Sqrt[3(n-3^j-3^k)/2+1]+1)/3}]],
{j, 0, Log[3, n/2]}, {k, j, Log[3, n-3^j]}]; tab=Append[tab, r], {n, 1, 80}]; Print[tab]
CROSSREFS
Cf. A000244, A000567, A001082, A045944, A280472, A303233, A303338, A303363, A303389, A303393, A303399, A303401, A303432.
Numbers of the form x*(3*x-1)/2 + 3^y with x and y nonnegative integers.
+10
24
1, 2, 3, 4, 6, 8, 9, 10, 13, 14, 15, 21, 23, 25, 27, 28, 31, 32, 36, 38, 39, 44, 49, 52, 54, 60, 62, 71, 73, 78, 79, 81, 82, 86, 93, 95, 97, 101, 103, 116, 118, 119, 120, 126, 132, 144, 146, 148, 151, 154, 172, 173, 177, 179, 185
COMMENTS
The author's conjecture in A303401 has the following equivalent version: Each integer n > 1 can be written as the sum of two terms of the current sequence. This has been verified for all n = 2..7*10^6.
EXAMPLE
a(1) = 1 with 1 = 0*(3*0-1)/2 + 3^0.
a(2) = 2 with 2 = 1*(3*1-1)/2 + 3^0.
a(5) = 6 with 6 = 2*(3*2-1)/2 + 3^0.
a(6) = 8 with 8 = 2*(3*2-1)/2 + 3^1.
MATHEMATICA
PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1, 6]==0);
tab={}; Do[Do[If[PenQ[m-3^k], n=n+1; tab=Append[tab, m]; Goto[aa]], {k, 0, Log[3, m]}]; Label[aa], {m, 1, 185}]; Print[tab]
CROSSREFS
Cf. A000244, A000326, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432.
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