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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A034170 Disjoint discriminants (one form per genus) of type 1.

Original entry on oeis.org

5, 13, 21, 33, 37, 57, 85, 93, 105, 133, 165, 177, 253, 273, 345, 357, 385, 1365
Offset: 1

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Author

Jonathan Borwein (jborwein(AT)cecm.sfu.ca), Stephen Choi (choi(AT)cecm.sfu.ca)

Keywords

Comments

A subsequence of A000926, A139826, and A232528. - Andrew Howroyd, Jun 09 2018

References

  • L. E. Dickson, Introduction to the theory of numbers, Dover, NY, 1929.
  • J. M. Borwein and P. B. Borwein, Pi and the AGM, page 293.

Crossrefs

Programs

  • PARI
    ok(n)={n<>1 && n%4==1 && issquarefree(n) && !select(t->t<>2, quadclassunit(-4*n).cyc)} \\ Andrew Howroyd, Jun 09 2018

Formula

Intersection of A004766 and A139826. - Andrew Howroyd, Jun 09 2018

A140768 Prime convenient numbers.

Original entry on oeis.org

2, 3, 5, 7, 13, 37
Offset: 1

Views

Author

Lekraj Beedassy, Jul 13 2008

Keywords

Comments

Prime terms of A000926. From Jud McCranie's comment in A000926, if an additional term exists it is > 100000000, and Weinberger showed that there is at most one further term. - Jonathan Vos Post, Jan 29 2011

References

  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 37, p. 15, Ellipses, Paris 2008.

A232527 Numbers n such that there are no positive integers m such that m^3 is of the form x^2 + n*y^2 and m is not of the form x^2 + n*y^2.

Original entry on oeis.org

1, 2, 3, 5, 6, 9, 10, 12, 13, 14, 17, 21, 22, 30, 33, 34, 37, 42, 46, 49, 57, 58, 65, 66, 69, 70, 73, 77, 78, 82, 85, 90, 93, 97, 102, 105, 114, 117, 130, 133, 138, 141, 142, 145, 154, 165, 177, 190, 193, 198, 205, 210, 213, 217, 238, 253, 258, 265, 273, 282, 285, 301, 310, 322, 330, 333, 345, 357, 385, 390
Offset: 1

Views

Author

V. Raman, Nov 25 2013

Keywords

Comments

Equivalently, numbers n such that there are no positive integers m such that m^r is of the form x^2 + n*y^2 and m is not of the form x^2 + n*y^2, for any odd integer r >= 3.
Because if m is not of the form x^2 + n*y^2 and m^(r-2) is not of the form x^2 + n*y^2, but m^r = a^2 + n*b^2, then m^(r+2) = (m*a)^2+n*(m*b)^2, m^(r+4) = (m^2*a)^2+n*(m^2*b)^2, etc. are also of the form x^2 + n*y^2. Then m^(3*r-6) = (m^(r-2))^3 = (m^(r-3)*a)^2 + n*(m^(r-3)*b)^2 is of the form x^2 + n*y^2 and m^(r-2) is not of the form x^2 + n*y^2.
All squarefree convenient numbers (A000926) congruent to {1, 2, 3, 5, 6} mod 8 are members of this sequence.
It appears that for a given value of n, if there is an m such that m^3 is of the form x^2 + n*y^2 and m is not of the form x^2 + n*y^2, then there exists some m < n that satisfies this condition.
Is this sequence finite?
Is 66045 the largest term of this sequence?

Examples

			n = 11 is not a member of this sequence because for m = 23, 23 is not of form x^2 + 11*y^2, but 23^3 = 12167 = 54^2 + 11*29^2.
		

Crossrefs

Cf. A000926.

Programs

  • PARI
    for(n=1,100000,flag=0;for(m=1,n,a=0;b=0;for(x=0,ceil(sqrt(m/n)),if(issquare(m-n*x^2),a=1; break));if(a==0,for(y=0,ceil(sqrt(m^3/n)),if(issquare(m^3-n*y^2),b=1; break)));if(a==0&&b==1,flag=1));if(flag==0,print1(n", ")))

A234001 Lowest common modulus to which the set of residue classes (mod 4n) that all the primes represented by a certain quadratic form of discriminant = -4n belong to, can be simplified to, for all quadratic forms of discriminant = -4n.

Original entry on oeis.org

4, 8, 3, 4, 20, 24, 14, 8, 12, 40, 11, 12, 52, 56, 30, 8, 68, 24, 19, 20, 84, 88, 46, 24, 20, 104, 3, 28, 116, 120, 62, 8, 132, 136, 35, 12, 148, 152, 78, 40, 164, 168, 43, 44, 60, 184, 94, 24, 28, 40, 51, 52, 212, 24, 110, 56, 228, 232, 59, 60, 244, 248, 42, 8, 260, 264, 67, 68, 276, 280
Offset: 1

Views

Author

V. Raman, Dec 18 2013

Keywords

Comments

If n is a convenient number (A000926), the set of residue classes (mod 4n) that a prime p represented by x^2+n*y^2 belong to are those for which p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n), assuming that p^2 does not divide n. For non-convenient numbers n, some of the primes in these set of residue classes (mod 4n) can be represented by x^2+n*y^2, but not all.
A prime p such that p^2 does not divide n, can be represented by some primitive quadratic form of discriminant = -4n, if and only if -n is a quadratic residue (mod p).
A prime p can be represented by some quadratic form of discriminant = -4n if and only if there is a multiple of p that can be written in the x^2+n*y^2 form, in which prime factor of p appears raised to an odd power or if p = 2 and n == 3 (mod 4).
a(n) is always a divisor of 4n.
If n is squarefree and n == 1 (mod 4) or n == 2 (mod 4), then a(n) = 4n.
If p^2 divides n for some prime p, a(n) is a divisor of (4n)/p.
If n == 3 (mod 8), then a(n) is a divisor of n because numbers of the form x^2+n*y^2 cannot have any prime factors that are congruent to 2+n (mod 2n) raised to an odd power.
If n == 7 (mod 8), then a(n) is a divisor of 2n because numbers of the form x^2+n*y^2 can have prime factors that are congruent to 2+n (mod 2n) raised to an odd power, but they cannot be congruent to 2 (mod 4). So, we need to characterize the prime factor of 2 from the remaining prime factors that are congruent to 2+n (mod 2n) separately.

Examples

			For n = 7, consider the set of all residue classes to which a prime represented by the quadratic form x^2+7*y^2 belong to, {1, 9, 11, 15, 23, 25} mod 28. This can be simplified to {1, 9, 11} mod 14 and this is the lowest modulo this set of residue classes can be simplified to. So, a(7) = 14. x^2+7*y^2 is the only primitive quadratic form of discriminant = -28.
For n = 15, there are two quadratic forms of discriminant = -60, x^2+15*y^2 and 3*x^2+5*y^2. x^2+15*y^2 can be used to represent all primes in set of residue classes {1, 4} mod 15. 3*x^2+5*y^2 can be used to represent all primes in set of residue classes {3, 5, 17, 23} mod 30. The lowest common modulo is 30, because {1, 4} mod 15 can also be written as {1, 4, 16, 19} mod 30, and so a(15) = 30.
		

Crossrefs

A234002 4n/A234001(n).

Original entry on oeis.org

1, 1, 4, 4, 1, 1, 2, 4, 3, 1, 4, 4, 1, 1, 2, 8, 1, 3, 4, 4, 1, 1, 2, 4, 5, 1, 36, 4, 1, 1, 2, 16, 1, 1, 4, 12, 1, 1, 2, 4, 1, 1, 4, 4, 3, 1, 2, 8, 7, 5, 4, 4, 1, 9, 2, 4, 1, 1, 4, 4, 1, 1, 6, 32, 1, 1, 4, 4, 1, 1, 2, 12, 1, 1, 20, 4, 1, 1, 2, 8, 27, 1, 4, 4, 1, 1, 2, 4, 1, 3, 4, 4, 1, 1, 2
Offset: 1

Views

Author

V. Raman, Dec 18 2013

Keywords

Comments

Please look into A234001 for a more detailed description.
If n is squarefree and n == 1 (mod 4) or n == 2 (mod 4), then a(n) = 1.
If p^2 divides n for some prime p, a(n) is a multiple of p.
If n == 3 (mod 8), then a(n) is a multiple of 4 because numbers of the form x^2+n*y^2 cannot have any prime factors that are congruent to 2+n (mod 2n) raised to an odd power.
If n == 7 (mod 8), then a(n) is a multiple of 2 because numbers of the form x^2+n*y^2 can have prime factors that are congruent to 2+n (mod 2n) raised to an odd power, but they cannot be congruent to 2 (mod 4). So, we need to characterize the prime factor of 2 from the remaining prime factors that are congruent to 2+n (mod 2n) separately.

Crossrefs

A290870 a(n) is the number of ways to represent n as n = x*y + y*z + z*x where 0 < x < y < z.

Original entry on oeis.org

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

Views

Author

Joerg Arndt, Aug 13 2017

Keywords

Comments

a(n) = 0 if and only if n is a term of A000926.
a(n) = 1 if and only if n is a term of A093669.

Examples

			For (x, y, z) = (1, 3, 5), we have x * y + y * z + z * x = 1 * 3 + 3 * 5 + 5 * 1 = 23 and similarily for (x, y, z) = (1, 2, 7), we have x * y + y * z + z * x = 23. Those 2 triples are all for n=23, so a(23) = 2. - _David A. Corneth_, Oct 01 2017
		

Crossrefs

Cf. A066955 (ways to represent n as n = x*y + y*z + z*x where 0 <= x <= y <= z).
Cf. A094377 (greatest number having exactly n representations).
Cf. A094376 (indices of records).

Programs

  • PARI
    N=10^3; V=vector(N);
    { for (x=1, N,
        for (y=x+1, N, t=x*y; if( t > N, break() );
          for (z=y+1, N,
            tt = t + y*z + z*x;  if( tt > N, break() );
            V[tt]+=1;
    ); ); ); }
    V \\ Joerg Arndt, Oct 01 2017
    
  • PARI
    a(n) = {my(res = 0);
    for(x = 1, sqrtint(n\3), for(y = x + 1, (n - x^2) \ (2 * x), z = (n - x*y) / (x + y); if(z > y && z == z\1, res++))); res} \\ David A. Corneth, Oct 01 2017

Formula

For the triples (x,y,z) we have x < sqrt(n / 3), y < (n - x^2) / (2 * x), z = (n - x*y) / (x + y) which must be integer. - David A. Corneth, Oct 01 2017

A380807 Numbers not of form a*(b+1) + b*(c+1) + c*(a+1) for 0

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 19, 20, 22, 23, 24, 26, 28, 30, 31, 34, 36, 38, 40, 43, 44, 46, 48, 50, 54, 55, 58, 64, 66, 68, 76, 78, 79, 80, 90, 94, 100, 106, 108, 118, 120, 124, 134, 138, 148, 156, 160, 170, 178, 198, 220, 271, 274, 288, 358, 376, 498, 750, 828
Offset: 1

Views

Author

Seiichi Manyama, Feb 04 2025

Keywords

Crossrefs

A025064 Position of numbers of form 3*n^2 in A025060 (numbers of form j*k + k*i + i*j, where 1 <=i < j < k).

Original entry on oeis.org

8, 43, 70, 105, 146, 194, 248, 307, 374, 448, 528, 615, 707, 805, 910, 1021, 1138, 1260, 1388, 1523, 1664, 1810, 1963, 2122, 2287, 2458, 2635, 2818, 3007, 3202, 3403, 3610, 3823, 4042, 4267, 4498, 4735, 4978
Offset: 1

Views

Author

Keywords

Programs

  • Maple
    N:= 10000: # to get positions of all 3*n^2 <= N
    B:= sort(convert({seq(seq(seq(i*j + j*k + i*k, i=1..min(j-1, (N-j*k)/(j+k))),j=2..min(k-1,(N-k)/(1+k))),k=3..(N-2)/3)},list)):
    count:= 1:
    for n from 1 to floor(sqrt(N/3)) do
      if member(3*n^2,B,A[count]) then count:= count+1 fi
    od:
    seq(A[i],i=1..count-1); # Robert Israel, Sep 06 2016

Formula

It is conjectured that A000926 ends at 1848, in which case a(n) = 3*n^2+18*n-38 for all n >= 22. - Robert Israel, Sep 06 2016

Extensions

More terms and a(4)-a(7) corrected by Gionata Neri, Sep 06 2016

A140411 Conjectured complete list of squarefree numbers that can be written as a sum of at most two positive squares, but not as a sum of three positive squares.

Original entry on oeis.org

1, 2, 5, 10, 13, 37, 58, 85, 130
Offset: 1

Views

Author

Jonathan Vos Post, Jun 25 2008

Keywords

Comments

Conjecture 1,9, p. 4, of Goswick et al. "The squarefree numbers in question form a subset of Euler's numeri idonei [A000926], therefore at most one number can be absent from the list above. If such a number does exist, it must exceed 2 * 10^11 and if it is even the Generalized Riemann Hypothesis is false."

Crossrefs

Programs

  • Mathematica
    Join[{1},Select[Range[500], Abs[MoebiusMu[#]] == 1 && Length[Select[PowersRepresentations[#,2,2], Not[MemberQ[#, 0, 2]] &]] > 0 && Length[Select[PowersRepresentations[#,3,2], Not[MemberQ[#, 0, 2]] &]] == 0 &]] (* Alonso del Arte, Sep 12 2019 *)
    Select[Range[500], SquareFreeQ[#] && (p = IntegerPartitions[#, {1, 3}, Range[Sqrt@#]^2]; p != {} && ! MemberQ[Length /@ p, 3]) &] (* Giovanni Resta, Sep 12 2019 *)

Formula

a(n) in A005117 and a(n) in {i^2 + j^2 for i,j > 1} and a(n) not in {i^2 + j^2 + k^2 for i,j,k > 1}.

A181564 Semiprime convenient numbers.

Original entry on oeis.org

4, 6, 9, 10, 15, 21, 22, 25, 33, 57, 58, 85, 93, 133, 177, 253
Offset: 1

Views

Author

Jonathan Vos Post, Jan 29 2011

Keywords

Comments

This is to semiprimes A001358 as A140768 is to primes A000040. Because Euler's "numerus idoneus" (idoneal, or suitable, or convenient numbers) are finite, this subset is finite. Conjecture: this list is complete (see Weinberger's result and comment by Jud McCranie in A140768).

Examples

			a(16) = A000926(50) = 253 = 11 * 23.
		

Crossrefs

Formula

A000926 INTERSECTION A001358.
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