A261029 -id:A261029 - cp's OEIS frontend

A272381 Primes p == 1 (mod 3) for which A261029(10*p) = 2.

Original entry on oeis.org

7, 13, 19, 31, 37, 73

Offset: 1

Vladimir Shevelev, Apr 28 2016

Peter J. C. Moses did not find any term > 73. The author proved that the sequence is full.

Subsequence of A030096.

Cf. A261029.

A272382 Primes p == 1 (mod 3) for which A261029(14*p) = 3.

Original entry on oeis.org

13, 19, 31, 37, 43, 61, 67, 97, 157

Offset: 1

Vladimir Shevelev, Apr 28 2016

Peter J. C. Moses did not find any term > 157. The author proved that the sequence is full. Moreover, he proved the following more general result.
Theorem. If p,q == 1 (mod 3) are prime and A261029(2*q*p) > 2, then sqrt(q)/2 < p < 4*q^2.
In this sequence q=7, so a(n) < 196.
Proof of Theorem is similar to proof of the theorem in A272384.

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A261029, A272381, A272384, A272404, A272406, A272407, A272409.

r[n_] := Reduce[0 <= x <= y <= z && z >= x + 1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a29[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Select[Select[Range[1, 1000, 3], PrimeQ], a29[14 #] == 3&] (* Jean-François Alcover, Nov 21 2018 *)

A272384 Primes p == 1 (mod 3) for which A261029(22*p) = 2.

Original entry on oeis.org

7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 181, 211, 229, 307, 313, 421

Offset: 1

Vladimir Shevelev, Apr 28 2016

Theorem. Let q==2 (mod 3) be a fixed prime and for a prime p==1 (mod 3), A261029(2*q*p) > 1. Then p < (2*q)^2.
In this sequence q=11, a(n) < 484.
Proof of Theorem. Recall that in A261029 we consider the number of representations of nonnegative numbers by the form: F(x,y,z) = x^3 + y^3 + z^3 - 3*x*y*z with the conditions 0 <= x <= y <= z, z >= x+1.
Note that F(x,y,z) = (x+y+z)*G(x,y,z), where G(x,y,z) = x^2 + y^2 + z^2 - x*y - x*z - y*z. It is easy to see that G==0 or 1(mod 3). Therefore, G differs from 2,2*p and q*p. In the case when G=1, x+y+z = 2*q*p, we have the representation [Shevelev] 2*p*q = F(k-1,k-1,k), where k=2(p*q + 1)/3.
Thus one can obtain more representations only in the case G=p, x+y+z = 2*q. Let a = z-x >= 1, then y = 2*q - 2*x - a. Then G=p yields 9*x^2 + (9a - 12q)*x + (3*a^2 - 6*a*q + 4q^2 - p) = 0. Solving this equation gives x = (4*q - 3*a +- sqrt(4*p - 3*a^2))/6. Since x >= 0, the choice of minus is possible only if p <= ((4*q - 3*a)^2 - 3*a^2)/4 or, since 1 <= a < 2*q, p < 4q^2. Now choose plus. Let b = sqrt(4*p - 3a^2). The condition y >= x yields b <= a. So p = (3*a^2 + b^2)/4 <= a^2 = (z-x)^2 < 4q^2.
It remains to show that the case G=2q, x+y+z = p is impossible. Indeed, in this case x = (2*p - 3*a +- sqrt(8*q - 3*a^2))/6. Let c = sqrt(8*q - 3*a^2), i.e., q = (3*a^2 + c*2)/8. Since q is prime, then 3*a^2 + c^2 should be divisible by 2^3. But it is easy to show that this is impossible for any integers a,c. QED

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A261029, A272381, A272382, A272404, A272406, A272407, A272409.

r[n_] := Reduce[0 <= x <= y <= z && z >= x + 1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a29[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Select[Select[Range[1, 1000, 3], PrimeQ], a29[22 #] == 2&] (* Jean-François Alcover, Nov 21 2018 *)

All terms (after author's first terms) were calculated by Peter J. C. Moses, Apr 28 2016

A272404 Primes p == 1 (mod 3) for which A261029(26*p) = 3.

Original entry on oeis.org

7, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 157, 163, 181, 193, 199, 223, 241, 277, 349, 463, 601

Offset: 1

Vladimir Shevelev, Apr 29 2016

By theorem in A272382, case q=13, the sequence is finite with a(n) <= 676.

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A261029, A272381, A272382, A272384.

r[n_] := Reduce[0 <= x <= y <= z && z >= x + 1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a29[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Select[Range[1, 997, 3], PrimeQ[#] && a29[26#] == 3&] (* Jean-François Alcover, Nov 06 2018 *)

All terms (after first author's ones) were calculated by Peter J. C. Moses, Apr 29 2016

A272406 Primes p == 1 (mod 3) for which A261029(34*p) = 2.

Original entry on oeis.org

7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 157, 163, 181, 193, 199, 211, 223, 229, 241, 271, 277, 283, 307, 313, 331, 337, 367, 373, 397, 409, 421, 439, 457, 487, 571, 709, 787, 877

Offset: 1

Vladimir Shevelev, Apr 29 2016

By theorem in A272384, case q=17, the sequence is finite with a(n)<1156.

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A261029, A272381, A272382, A272384, A272404.

r[n_] := Reduce[0 <= x <= y <= z && z >= x+1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a29[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Select[Select[Range[7, 997, 3], PrimeQ], a29[34 #] == 2&] (* Jean-François Alcover, Dec 01 2018 *)

All terms (after first author's ones) were calculated by Peter J. C. Moses, Apr 29 2016

A272407 Primes p == 1 (mod 3) for which A261029(38*p) = 3.

Original entry on oeis.org

7, 13, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 157, 163, 181, 193, 199, 211, 223, 229, 241, 271, 277, 283, 307, 313, 331, 337, 349, 367, 373, 379, 397, 409, 433, 463, 499, 523, 541, 577, 601, 607, 619, 661, 757, 853, 937, 1123, 1129

Offset: 1

Vladimir Shevelev, Apr 29 2016

By theorem in A272382, case q=19, the sequence is finite with a(n)<1444.

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A261029, A272381, A272382, A272384, A272404, A272406.

r[n_] := Reduce[0 <= x <= y <= z && z >= x+1 && n == x^3+y^3+z^3 - 3 x y z, {x, y, z}, Integers];
a261029[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Select[Select[Range[1, 1171, 3], PrimeQ], a261029[38 #] == 3&] (* Jean-François Alcover, Dec 04 2018 *)

All terms (after first author's ones) were calculated by Peter J. C. Moses, Apr 29 2016

A272409 Primes p == 1 (mod 3) for which A261029(46*p) = 2.

Original entry on oeis.org

7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, 109, 127, 139, 151, 157, 163, 181, 193, 199, 211, 223, 229, 241, 271, 277, 283, 307, 313, 331, 337, 349, 367, 373, 379, 397, 409, 421, 433, 439, 457, 463, 487, 499, 523, 541, 547, 571, 577, 613, 631, 643, 673, 709, 733, 739, 787, 811, 829, 859, 877, 907, 1009, 1063, 1093, 1117, 1279, 1297, 1381, 1483, 1489, 1723

Offset: 1

Vladimir Shevelev, Apr 29 2016

By theorem in A272384, case q=23, the sequence is finite with a(n)<2116.

Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.

Cf. A261029, A272381, A272382, A272384, A272404, A272406, A272407.

r[n_] := Reduce[0 <= x <= y <= z && z >= x+1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a29[n_] := Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
Select[Select[Range[1, 2002, 3], PrimeQ], a29[ 46 # ] == 2&] (* Jean-François Alcover, Dec 06 2018 *)

All terms (after first author's ones) were calculated by Peter J. C. Moses, Apr 29 2016

A260935 Smallest k such that A261029(k) = n.

Original entry on oeis.org

0, 1, 8, 28, 108, 189, 324, 648, 972, 756, 1701, 2457, 1512, 3888, 2268, 4536, 6048, 13104, 10584, 15120, 6804, 16848, 9072, 14364, 9828, 28728, 19656, 21168, 36288, 31752, 50544, 27216, 46683, 70308, 29484, 57456, 39312, 81648, 111132, 63504, 58968, 108864

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Aug 22 2015

nonn

Theorem. For every n>=0, a(n) exists.
Are all terms from a(4)=108 onward divisible by 9?
a(139) = 12006176 is not divisible by 9. - Chai Wah Wu, Aug 25 2015

			By condition z>=x+1>=1. By induction one can prove that F(x,y,z)>=3*z-2 (cf.[Shevelev]).
Since F>=1, then A261029(0)=0 and a(0)=0;
Further,
x y z F
0 0 1 1
0 1 1 2
Since F(x,y,2)>=4>1, A261029(1)=1 and a(1)=1.
0 0 2 8
0 1 2 9
0 2 2 16
1 1 2 4
1 2 2 5
0 0 3 27
0 1 3 28
0 2 3 35
0 3 3 54
1 1 3 20
1 2 3 18
1 3 3 28
2 2 3 7
2 3 3 8
Since F(x,y,4)>=10>8, A261029(8)=2 and a(2)=8,
etc.

Chai Wah Wu, Table of n, a(n) for n = 0..698
Vladimir Shevelev, Representation of positive integers by the form x^3+y^3+z^3-3xyz, arXiv:1508.05748 [math.NT], 2015.
Robert G. Wilson v, For n: solutions of A261029(k).

Cf. A261029.

r[n_] := Reduce[0 <= x <= y <= z && z >= x + 1 && n == x^3 + y^3 + z^3 - 3 x y z, {x, y, z}, Integers];
a29[n_] := a29[n] = Which[rn = r[n]; rn === False, 0, rn[[0]] === And, 1, rn[[0]] === Or, Length[rn], True, Print["error ", rn]];
a[n_] := For[k=0, True, k++, If[a29[k] == n, Print[n, " ", k]; Return[k]]];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 15 2018 *)

A261029(a(n)) = n.

For n>=1, a(n) <= 8^(n-1).

a(11)-a(41) from Chai Wah Wu, Aug 25 2015

A268665 Number of primes p==1 (mod 3) for which A261029(2prime(n)p) is 4-i if prime(n)==i (mod 3), where i=1 or 2.

Original entry on oeis.org

6, 9, 22, 26, 44, 52, 73, 111, 122, 164, 201, 214, 254, 311, 374, 398, 465, 521, 542, 617, 684, 774, 899, 969, 1005, 1064, 1100, 1181, 1441, 1548, 1658, 1694, 1918, 1977, 2114, 2255, 2376, 2537, 2684, 2727, 3019, 3068, 3181, 3238, 3611, 3985, 4114, 4182, 4313

Offset: 3

Vladimir Shevelev and Peter J. C. Moses, May 07 2016

nonn

			Let n=3, then prime(n)=5. Since 5==2(mod 3), then i=2. So a(3) is the number of primes p==1(mod 3) for which A261029(10*p)=4-2=2. So it is number of terms in A272381, i.e., a(3)=6.
Let n=4, then prime(n)=7. Since 7==1(mod 3), then i=1. So a(4) is the number of primes p==1(mod 3) for which A261029(14*p)=4-1=3. So it is number of terms in A272382, i.e., a(4)=9.

Cf. A261029, A272381, A272382, A272384, A272404, A272406, A272407, A272409.

A272856 Greatest length of a chain of consecutive primes p==1 (mod 3) for which A261029 (2prime(n)p) is 4-i if prime(n) == i (mod 3), where i=1,2.

Original entry on oeis.org

5, 7, 16, 19, 32, 35, 50, 76, 81, 108, 140, 139, 171, 206, 254, 259, 305, 346, 349, 404, 449, 504, 582, 634, 645, 699, 707, 772, 930, 1006, 1078, 1097, 1258, 1271, 1362, 1448, 1529, 1633, 1737, 1752, 1951, 1970, 2064, 2082, 2310, 2550, 2659, 2672, 2783, 2917

Offset: 3

Vladimir Shevelev and Peter J. C. Moses, May 08 2016

nonn

			Let n=3; then prime(n)=5. Since 5 == 2 (mod 3), i=2. So a(3) is the greatest length of a chain of consecutive primes p == 1 (mod 3) for which A261029(10*p) = 4 - 2 = 2. So these primes are in A272381. The first term is 7, and we have the chain of consecutive primes == 1 (mod 3): {7, 13, 19, 31, 37}. Since the following prime 43 == 1 (mod 3) is not in A272381, the chain ends and its length is 5. The second chain is the singleton {71}. So a(3)=5.

Cf. A261029, A268665, A272381, A272382, A272384, A272404, A272406, A272407, A272409.

a261029[n_]:=a261029[n]={x,y,z}/.{ToRules[Reduce[x^3+y^3+z^3-3 x y z==n&&0<=x<=y<=z&&z>=x+1,Integers]]}/.{x,y,z}->{};
data={};
Do[p=Prime[n];
primes=Select[Prime[Range[1+PrimePi[(2p)^2]]],Mod[#,3]==1&];
tmp=Map[{#,Length[a261029[2 # p]]}&,primes];
AppendTo[data,{{n,2p,1+Mod[2p,3]},{{Length[#],Max[Map[Length,Select[Split[Differences[Flatten[Map[Position[primes,#,1,1]&,#]]]],#[[1]]==1&]]+1]},#}&[Map[#[[1]]&,Select[tmp,#[[2]]==(1+Mod[2p,3])&]]]}];Print[Last[data]],{n,3,10}]
Map[Length[a261029[#]]&,Range[0,20]] (* A261029 *)
Last[Last[data[[1]]]] (* A272381 *)
Last[Last[data[[2]]]] (* A272382 *)
Last[Last[data[[3]]]] (* A272384 *)
Last[Last[data[[4]]]] (* A272404 *)
Last[Last[data[[5]]]] (* A272406 *)
Last[Last[data[[6]]]] (* A272407 *)
Last[Last[data[[7]]]] (* A272409 *)
Map[#[[2]][[1]][[1]]&,data] (* A268665 *)
Map[#[[2]][[1]][[2]]&,data] (* A272856 *)

cp's OEIS Frontend

A272381 Primes p == 1 (mod 3) for which A261029(10*p) = 2.

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A272382 Primes p == 1 (mod 3) for which A261029(14*p) = 3.

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A272384 Primes p == 1 (mod 3) for which A261029(22*p) = 2.

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A272404 Primes p == 1 (mod 3) for which A261029(26*p) = 3.

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A272406 Primes p == 1 (mod 3) for which A261029(34*p) = 2.

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A272407 Primes p == 1 (mod 3) for which A261029(38*p) = 3.

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A272409 Primes p == 1 (mod 3) for which A261029(46*p) = 2.

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A260935 Smallest k such that A261029(k) = n.

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A268665 Number of primes p==1 (mod 3) for which A261029(2*prime(n)*p) is 4-i if prime(n)==i (mod 3), where i=1 or 2.

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A268665 Number of primes p==1 (mod 3) for which A261029(2prime(n)p) is 4-i if prime(n)==i (mod 3), where i=1 or 2.

A272856 Greatest length of a chain of consecutive primes p==1 (mod 3) for which A261029 (2prime(n)p) is 4-i if prime(n) == i (mod 3), where i=1,2.