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.

User: ,11,

,11,'s wiki page.

,11, has authored 11878 sequences. Here are the ten most recent ones:

A386969 A variant of Recamán's sequence (A005132): a(0) = 0; for n > 0, a(n) = a(n-1) - n if nonnegative and not already in the sequence, otherwise a(n) = a(n-1) + n + 3.

Original entry on oeis.org

0, 4, 2, 8, 15, 10, 19, 12, 23, 14, 27, 16, 31, 18, 35, 20, 39, 22, 43, 24, 47, 26, 51, 28, 55, 30, 59, 32, 63, 34, 67, 36, 71, 38, 75, 40, 79, 42, 83, 44, 87, 46, 91, 48, 95, 50, 99, 52, 103, 54, 107, 56, 111, 58, 115, 60, 119, 62, 123, 64, 127, 66, 131, 68, 135, 70, 139, 72, 143, 74, 147, 76, 151
Offset: 0

Author

Jules Beauchamp, Aug 11 2025

Keywords

Comments

When visualized in connecting semicircular lines after a(n) = 8 it produces one increasingly large conical spiral.

Crossrefs

Programs

  • Mathematica
    a[n_] := a[n] = If[a[n-1] >= n && FreeQ[Table[a[k], {k, 0, n-1}], a[n-1] - n], a[n-1] - n, a[n-1] + n + 3]; a[0] = 0; Array[a, 100, 0] (* Amiram Eldar, Aug 22 2025 *)

A386962 Number of equivalence classes of connected 3-regular graphs on 2n unlabeled nodes up to local complementation.

Original entry on oeis.org

0, 1, 2, 4, 15, 60
Offset: 1

Author

Tristan Cam, Aug 11 2025

Keywords

Comments

Number of equivalences classes of 3-regular graphs on 2n nodes up to a sequence of local complementation or isomorphism, also called orbits for the local equivalence relation.
a(n) is necessarily less than:
A005638(n) (number of non-isomorphic, not necessarily connected 3-regular graphs);
A002851(n) (number of non-isomophic connected 3-regular graphs);
A090899(n) (number of local equivalence classes of connected graphs); and
A156800(n) (number of equivalence classes for connected graphs up to pivots and isomorphism).
This is relevant in the study of optimal quantum circuit synthesis for graph state preparation.

Examples

			There are only two 3-regular graphs with 6 nodes and they are not equivalent up to a sequence of local complementation, thus a(3) = 2.
		

Crossrefs

A386963 Gaps between positions of odd terms in A065090.

Original entry on oeis.org

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

Author

Aied Sulaiman, Aug 11 2025

Keywords

Comments

For n >= 2 we have a(n) in {2,3,4}:
a(n) = 2 if no prime lies between the two successive odd terms,
a(n) = 3 if a single prime lies between them,
a(n) = 4 if two primes lie between them.
The initial 5 comes from 3, 5, 7 between 1 and 9.
Conjecture: a(n) tends to 2 in frequency (i.e., {n : a(n) = 2} has natural density 1).
Conjecture is true because the primes have natural density 0. - Robert Israel, Aug 29 2025

Examples

			A065090: 1, 2, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, ...
Odd terms occur at positions: 1, 6, 10, 14, 17, 19, 23, 25, ...
Hence a(n): 5, 4, 4, 3, 2, 4, 2, ...
		

Crossrefs

Programs

  • Mathematica
    a065090=Select[Range[335],#==2||!PrimeQ[#]&];l=Length[a065090];p={};Do[If[OddQ[a065090[[i]] ],AppendTo[p,i]],{i,l}];Differences[p] (* James C. McMahon, Aug 29 2025 *)
  • PARI
    lista(nn) = my(vio = select(x->(x % 2), select(m->(!isprime(m) || m==2), [1..nn]), 1)); vector(#vio-1, k, vio[k+1] - vio[k]); \\ Michel Marcus, Aug 16 2025
  • Python
    from sympy import isprime
    def gaps_generator():
        pos = 0
        last = None
        k = 1
        while True:
            if not (k % 2 == 1 and isprime(k)):  # in A065090
                pos += 1
                if k % 2 == 1:  # odd term (A014076)
                    if last is None:
                        last = pos
                    else:
                        yield pos - last
                        last = pos
            k += 1
    def a(n: int) -> int:
        g = gaps_generator()
        for _ in range(n - 1):
            next(g)
        return next(g)
    

A385822 Numbers k such that phi(k) is not a perfect square.

Original entry on oeis.org

3, 4, 6, 7, 9, 11, 13, 14, 15, 16, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 33, 35, 36, 38, 39, 41, 42, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 54, 55, 56, 58, 59, 61, 62, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 75, 77, 78, 79, 80, 81, 82, 83, 84, 86, 87, 88, 89
Offset: 1

Author

Aidan Chen, Aug 11 2025

Keywords

Examples

			Since phi(35) = 24 and there is no integer n such that n^2 = 24.
		

Crossrefs

Cf. A000010. Complement of A039770.

Programs

  • Mathematica
    Select[Range[100], !IntegerQ[Sqrt[EulerPhi[#]]] &] (* Amiram Eldar, Aug 18 2025 *)
  • PARI
    isok(k) = !issquare(eulerphi(k)); \\ Michel Marcus, Aug 18 2025
  • Python
    from math import isqrt
    from sympy import totient as phi
    def ok(n): return isqrt(p:=phi(n))**2 != p
    print([k for k in range(1, 110) if ok(k)]) # Michael S. Branicky, Aug 17 2025
    

A386985 Smallest k > 0 such that the base-n number formed by concatenating k, k - 1, ..., 2, 1 (each written in base n) is prime, or -1 if no such k exists for the given n.

Original entry on oeis.org

2, 2, 4, 2, 2, 373, 2, 2, 82, 2, 3
Offset: 2

Author

Marco Ripà, Aug 11 2025

Keywords

Comments

If a(13) != -1, then the corresponding prime must have more than 4178 decimal digits.
Sequence continues n=13..36: ?, 2, 2, 28, 362, 2, ?, 2, 2, 4, 2, 3, ?, 2, 5, 4, 2, 2, 37, 3, 2, 4, 2, 2.
The increasing-order analog begins 15, 2, ?. See A048436.

Examples

			a(3) = 2 since 21_(base-3) = 7_(base-10), which is prime.
		

Crossrefs

Programs

  • PARI
    f(n, b) = my(p=1, L=1); sum(k=1, n, k*p*=L+(k==L&&!L*=b)); \\ adapted from A000422
    a(n) = my(k=1); while (!ispseudoprime(f(k, n)), k++); k; \\ Michel Marcus, Aug 16 2025

A386980 Number of acute Heronian triangles with integer inradius n.

Original entry on oeis.org

0, 0, 1, 1, 0, 4, 0, 2, 2, 2, 0, 6, 0, 1, 4, 3, 0, 8, 0, 6, 7, 2, 0, 17, 1, 0, 2, 8, 0, 14, 0, 3, 6, 1, 4, 17, 0, 0, 4, 12, 0, 27, 0, 4, 13, 1, 0, 27, 1, 4, 2, 4, 0, 13, 5, 14, 2, 0, 0, 32, 0, 0, 14, 4, 3, 18, 0, 5, 3, 15, 0, 41, 0, 0, 10, 4, 7, 16, 0, 18, 3, 0, 0, 60, 2, 0, 2, 18, 0, 39, 9
Offset: 1

Author

Frank M Jackson, Aug 11 2025

Keywords

Comments

If a Heronian triangle has an inradius n, and sides (x, y, z), where x <= y <= z, then the triangle is acute iff n < (x+y-z)/2.
The only Heronian triangle with inradius 1 is the right triangle (3, 4, 5). Also, it has been proved that other than n = 3, all acute Heronian triangles have no prime inradii. For n = 3, the Heronian triangle has sides (10, 10, 12).
Empirically, it appears that the remaining occurrences of zero counts (other than 1 and the primes excluding 3) are inradii of the form 2p where p is in the set 13, 19, 29 and all other primes > 29.
The number of right integer triangles with inradius n is given by A078644, the number of obtuse Heronian triangles with inradius n is given by A386981 and the total number of Heronian triangles with inradius n is given by A120062.

Examples

			a(6) = 4, and the 4 acute Heronian triangles with inradius 6 have sides (15, 34, 35), (17, 25, 28), (17, 25, 26), (20, 20, 24).
		

Crossrefs

Programs

  • Mathematica
    (* See link above. *)

A386981 Number of obtuse Heronian triangles with integer inradius n.

Original entry on oeis.org

0, 3, 9, 14, 12, 35, 21, 39, 44, 44, 23, 124, 28, 73, 97, 81, 30, 166, 31, 130, 169, 95, 39, 283, 59, 90, 131, 208, 33, 347, 43, 160, 196, 109, 160, 466, 35, 117, 197, 304, 41, 515, 57, 267, 354, 127, 61, 550, 110, 214, 219, 258, 44, 425, 215, 484, 265, 128, 51, 977, 41, 138, 582, 269, 169, 603, 48, 325, 252, 564, 47, 1058, 65, 133, 445, 341
Offset: 1

Author

Frank M Jackson, Aug 11 2025

Keywords

Comments

If a Heronian triangle has an inradius n, and sides (x, y, z), where x <= y <= z, then the triangle is obtuse iff n > (x+y-z)/2.
The only Heronian triangle with inradius 1 is the right triangle (3, 4, 5).
The number of right integer triangles with inradius n is given by A078644, the number of acute Heronian triangles with inradius n is given by A386980 and the total number of Heronian triangles with inradius n is given by A120062.

Examples

			a(2) = 3, and the 3 obtuse Heronian triangles with inradius 2 have sides (6, 25, 29), (7, 15, 20), (9, 10, 17).
		

Crossrefs

Programs

  • Mathematica
    (* See link above. *)

A386972 Numbers that are the product of a semiprime and the square of another semiprime.

Original entry on oeis.org

96, 144, 160, 224, 240, 324, 336, 352, 360, 400, 416, 486, 504, 528, 540, 544, 560, 600, 608, 624, 736, 756, 784, 792, 810, 816, 880, 900, 912, 928, 936, 992, 1040, 1104, 1134, 1176, 1184, 1188, 1215, 1224, 1232, 1260, 1312, 1350, 1360, 1368, 1376, 1392, 1400, 1404, 1456, 1488, 1500
Offset: 1

Author

Ian Hahus, Aug 11 2025

Keywords

Comments

Numbers with prime signature [5, 1], [4, 2], [4, 1, 1], [3, 2, 1], [2, 2, 2] or [2, 2, 1, 1]. So, necessarily but not sufficiently, terms t have bigomega(t) = 6. - David A. Corneth, Aug 11 2025

Examples

			96 = 6 * 4^2;
144 = 9 * 4^2 or 4 * 6^2.
		

Crossrefs

Cf. A001358 (semiprimes), A046306, A054753, A386977.

Programs

  • Maple
    M:= 2000: # for terms <= M
    P:= select(isprime, [2,seq(i,i=3..M/8,2)]): nP:= nops(P):
    S:= {}:
    for i1 from 1 to nP do
      p1:= P[i1];
      if p1^2*4^2 > M then break fi;
      for i2 from i1 to nP do
        p2:= P[i2];
        if p1*p2*4^2 > M then break fi;
        for i3 from 1 to nP do
          p3:= P[i3];
          if p1*p2*p3^4 > M then break fi;
          for i4 from i3 to nP do
            p4:= P[i4];
            v:= p1*p2*(p3*p4)^2;
            if v > M then break fi;
            if p1*p2 = p3*p4 then next fi;
            S:= S union {v}
    od od od od:
    sort(convert(S,list)); # Robert Israel, Aug 11 2025
  • Mathematica
    Select[Range@ 1500, MemberQ[{{1,5}, {2,4}, {1,1,4}, {1,2,3}, {2,2,2}, {1,1,2,2}}, Sort[ Last /@ FactorInteger[#]]] &] (* Giovanni Resta, Aug 12 2025 *)
  • PARI
    is(n) = {my(f = factor(n), b = bigomega(f)); if(b != 6, return(0)); f = vecsort(f[,2]~); #setminus(Set([f]), Set([[1, 5], [2, 4], [1, 1, 4], [1, 2, 3], [2, 2, 2], [1, 1, 2, 2]])) == 0} \\ David A. Corneth, Aug 12 2025

A386216 Values of v in the quartets (2, u, v, w) of type 2; i.e., values of v for solutions to 2(2 + u) = v(v - w), in distinct positive integers, with v > 1, sorted by nondecreasing values of u; see A385884.

Original entry on oeis.org

6, 10, 12, 14, 16, 6, 18, 5, 20, 22, 8, 24, 26, 7, 28, 6, 10, 30, 8, 32, 34, 9, 12, 36, 38, 8, 10, 40, 7, 14, 42, 11, 44, 46, 12, 16, 48, 10, 50, 13, 52, 9, 18, 54, 8, 14, 56, 58, 10, 12, 15, 20, 60, 62, 16, 64, 11, 22, 66, 17, 68, 10, 14, 70, 9, 12, 18, 24
Offset: 1

Author

Clark Kimberling, Aug 11 2025

Keywords

Crossrefs

Cf. A385884.

A386984 Number of 2-dense sublists of divisors of the n-th hexagonal number.

Original entry on oeis.org

1, 1, 1, 3, 1, 3, 1, 3, 1, 5, 3, 5, 1, 5, 1, 5, 1, 5, 1, 3, 1, 7, 3, 3, 1, 5, 3, 7, 1, 5, 1, 3, 1, 7, 3, 5, 1, 3, 1, 5, 1, 7, 1, 7, 1, 5, 3, 5, 1, 5, 1, 7, 3, 5, 1, 7, 1, 7, 5, 7, 1, 5, 3, 3, 1, 7, 1, 7, 1, 7, 5, 5, 1, 7, 1, 7, 3, 5, 1, 3, 1, 9, 3, 7, 1, 7, 1, 7, 1, 5, 1, 5, 1, 9, 3, 3, 1, 3, 1, 9, 1
Offset: 0

Author

Omar E. Pol, Aug 11 2025

Keywords

Comments

In a sublist of divisors of k the terms are in increasing order and two adjacent terms are the same two adjacent terms in the list of divisors of k.
The 2-dense sublists of divisors of k are the maximal sublists whose terms increase by a factor of at most 2.
Conjecture: all terms are odd.

Examples

			For n = 3 the third positive hexagonal number is 15. The list of divisors of 15 is [1, 3, 5, 15]. There are three 2-dense sublists of divisors of 15, they are [1], [3, 5], [15], so a(3) = 3.
		

Crossrefs

Bisection of A384928.

Programs

  • Mathematica
    A386984[n_] := Length[Split[Divisors[PolygonalNumber[6, n]], #2 <= 2*# &]];
    Array[A386984, 100, 0] (* Paolo Xausa, Aug 29 2025 *)

Formula

a(n) = A237271(A000384(n)) for n >= 1 (conjectured).