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.

Showing 1-4 of 4 results.

A278243 Filter-sequence for Stern polynomials: Least number with the same prime signature as A260443(n).

Original entry on oeis.org

1, 2, 2, 6, 2, 12, 6, 30, 2, 60, 12, 120, 6, 180, 30, 210, 2, 420, 60, 1080, 12, 2160, 120, 2520, 6, 2520, 180, 7560, 30, 6300, 210, 2310, 2, 4620, 420, 37800, 60, 90720, 1080, 75600, 12, 226800, 2160, 544320, 120, 453600, 2520, 138600, 6, 138600, 2520, 756000, 180, 2268000, 7560, 831600, 30, 415800, 6300, 2079000, 210, 485100, 2310, 30030, 2, 60060, 4620
Offset: 0

Views

Author

Antti Karttunen, Nov 16 2016

Keywords

Comments

This sequence can be used for filtering certain Stern polynomial (see A125184, A260443) related sequences, because it matches only with any such sequence b that can be computed as b(n) = f(A260443(n)), where f(n) is any function that depends only on the prime signature of n (some of these are listed under the index entry for "sequences computed from exponents in ...").
Matching in this context means that the sequence a matches with the sequence b iff for all i, j: a(i) = a(j) => b(i) = b(j). In other words, iff the sequence b partitions the natural numbers to the same or coarser equivalence classes (as/than the sequence a) by the distinct values it obtains.
Some of these are listed on the last line ("Sequences that partition N into ...") of Crossrefs section.

Crossrefs

Sequences that partition or seem to partition N into same or coarser equivalence classes: A002487, A126606, A277314, A277315, A277325, A277326, A277700, A277705.
The following are less certain: A007302 (not proved, but the first 10000 terms match), A072453, A110955 (uncertain, but related to A007302), A218799, A218800.
Note that the base-2 related sequences A069010 and A277561 (= 2^A069010(n)) do not match, although at first it seems so, up to for at least 139 initial terms. Also A028928 belongs to a different family.

Programs

  • Mathematica
    a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[Times @@ MapIndexed[Prime[First@ #2]^#1 &, Sort[FactorInteger[#][[All, -1]], Greater]] - Boole[# == 1] &@ a@ n, {n, 0, 66}] (* Michael De Vlieger, May 12 2017 *)
  • Scheme
    (define (A278243 n) (A046523 (A260443 n)))

Formula

a(n) = A046523(A260443(n)).

A218800 Number of nonnegative integer solutions to x^2 + 2y^2 = (3n)^2.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 2, 2, 4, 2, 5, 3, 2, 2, 3, 2, 5, 4, 5, 2, 3, 5, 2, 3, 2, 2, 5, 2, 2, 3, 2, 2, 8, 5, 2, 4, 2, 5, 3, 2, 5, 3, 5, 5, 4, 2, 2, 3, 2, 2, 8, 2, 2, 5, 5, 2, 8, 2, 5, 3, 2, 2, 4, 2, 2, 8, 5, 5, 3, 2, 2, 4, 5, 2, 3, 5, 5, 3, 2, 2, 6, 5, 5, 3, 5, 5
Offset: 0

Views

Author

Jon Perry, Nov 06 2012

Keywords

Comments

For n > 0, a(n) > 1 since n^2 + 2(2n)^2 = (3n)^2 and (3n)^2 + 2*0^2 = (3n)^2.
a(3k) > 2 as we also have (7k)^2 + 2*(4k)^2 = 81k^2 =
(9k)^2 = (3*3k)^2.

Examples

			a(2) = 2 because we have 6^2 + 2*0^2 = 6^2 and 2^2 + 2*4^2 = 6^2 and no others.
		

Crossrefs

Cf. A218799.

Programs

  • JavaScript
    for (i=0; i<200; i+=3) {
    d=0; e=0;
    for (a=0; a<=i; a++)
    for (b=0; b<=i; b++) {
    t1=Math.pow(a, 2)+2*Math.pow(b, 2);
    t2=Math.pow(i, 2);
    if (t1
    				

A218706 Number of nonnegative integer solutions to x^2 + 2y^2 <= n^2.

Original entry on oeis.org

1, 2, 5, 9, 12, 19, 27, 33, 42, 54, 66, 77, 91, 105, 120, 138, 156, 175, 197, 218, 240, 263, 287, 314, 339, 367, 398, 430, 459, 493, 526, 556, 595, 637, 670, 709, 752, 794, 833, 878, 921, 965, 1018, 1065, 1112, 1163, 1215, 1266, 1317, 1370, 1433, 1492, 1544
Offset: 0

Views

Author

Jon Perry, Nov 04 2012

Keywords

Examples

			There are 5 solutions for n=2, namely (0,0), (0,1), (1,1), (1,0) and (2,0), so a(2) = 5.
		

Crossrefs

Programs

  • JavaScript
    for (i=0;i<50;i++) {
    c=0;
    for (a=0;a<=i;a++)
    for (b=0;b<=i;b++)
    if (Math.pow(a,2)+2*Math.pow(b,2)<=Math.pow(i,2)) c++;
    document.write(c+", ");
    }
  • Mathematica
    nn = 50; t = Sort[Select[Flatten[Table[x^2 + 2*y^2, {x, 0, nn}, {y, 0, nn}]], # <= nn^2 &]]; Table[Count[t, ?(# <= n^2 &)], {n, 0, nn}] (* _T. D. Noe, Nov 06 2012 *)

A339377 Number of triples (x, y, z) of natural numbers satisfying x+y = n and 2*x*y = z^2.

Original entry on oeis.org

1, 2, 2, 4, 2, 2, 4, 2, 2, 6, 2, 4, 4, 2, 2, 4, 2, 4, 6, 4, 2, 4, 4, 2, 4, 2, 2, 8, 2, 2, 4, 2, 2, 10, 4, 2, 6, 2, 4, 4, 2, 4, 4, 4, 4, 6, 2, 2, 4, 2, 2, 10, 2, 2, 8, 4, 2, 10, 2, 4, 4, 2, 2, 6, 2, 2, 10, 4, 4, 4, 2, 2, 6, 4, 2, 4, 4, 4, 4, 2, 2, 10, 4, 4, 4, 4, 4, 4
Offset: 0

Views

Author

Bernard Schott, Dec 02 2020

Keywords

Comments

This sequence is inspired by the 4th problem proposed during the second day of the final round of the 18th Austrian Mathematical Olympiad in 1987. The problem asked to find all triples solutions (x, y, z) only for n = 1987 (see Link, Reference and last example).
Some properties:
-> Inequalities, 0 <= x, y <= n; 0 <= z <= floor(n*sqrt(2)/2)
-> z is even and (x,y) are not together even.
-> a(n) = 1 iff n = 0, and the only solution is (0,0,0).
-> for n >= 1, a(n) >= 2 because (0,n,0) and (n,0,0) are always solutions.
-> a(n) is even for n >= 1.
-> If n = 3k, then (k,2k,2k) and (2k,k,2k) are solutions.
-> If 2*(n-1) = m^2, then (1,n-1,m) and (n-1,1,m) are solutions (with n in A058331).
-> The formula for n>0 comes from (x+y=n and 2*x*y=z^2) <==> n^2 = |x-y|^2 + 2*z^2.

Examples

			a(9) = 6 and these 6 solutions are: (0, 9, 0), (1, 8, 4), (3, 6, 6), (6, 3, 6), (8, 1, 4), (9, 0, 0).
a(1987) = 4 and these 4 solutions are: (0, 1987, 0), (529, 1458, 1242), (1458, 529, 1242), (1987, 0, 0); this is the answer to the Olympiad problem in link.
		

References

  • Steve Dinh, The Hard Mathematical Olympiad Problems And Their Solutions, AuthorHouse, 2011, Problem 4 of Austrian Mathematical Olympiad 1987, page 29 [Warning: solution proposed in this book has a mistake with (x, y, z) = ([0, 1987], 1987-x, sqrt(2xy))].

Crossrefs

Cf. A058331, A218799, A339378 (variant with x+y = n and x*y = z^2).

Formula

a(0)=A218799(0); then for n>=1, a(n)=2*A218799(n) (remark from Hugo Pfoertner, Dec 02 2020).
Showing 1-4 of 4 results.