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: Miles Englezou

Miles Englezou's wiki page.

Miles Englezou has authored 36 sequences. Here are the ten most recent ones:

A386881 a(n) is the number of occurrences of n in A386838.

Original entry on oeis.org

1, 2, 2, 4, 2, 5, 2, 7, 4, 7, 3, 11, 4, 7, 6, 13, 3, 14, 5, 13, 7, 11, 3, 24, 7, 11, 8, 20, 4, 21, 8, 20, 10, 15, 7, 33, 7, 14, 12, 29, 4, 30, 8, 22, 15, 19, 7, 45, 10, 22, 14, 30, 8, 34, 12, 36, 16, 24, 7, 49, 12, 21, 20, 42, 10, 42, 12, 34, 13, 36, 10, 69, 15, 23, 23, 36
Offset: 1

Author

Miles Englezou, Aug 06 2025

Keywords

Comments

A386838(k) is the minimal area of the graph formed under the requirement that the straight line drawn from (0,0) to (x,y) (where x^2 + y^2 = k = A001481(n)) passes through an enclosed space on the square lattice and its edges are either vertical or horizontal. If A001481(n) = x^2 + y^2 for multiple x and y, then x and y are chosen such that A386838(A001481(n)) is minimal. a(n) is the number of graphs with area n, and equivalently the number of numbers of the form x^2 + y^2 = A001481(n) such that n = x + y - gcd(x,y) for such minimal x and y.
The offset is 1 since 0 occurs infinitely many times in A386838 (e.g., A386838(k) = 0 when A001481(k) is square).
The range in which n can occur in A386838 is bounded above by 2*n^2.
Does every integer n > 0 appear in this sequence?

Examples

			a(5) = 2 since 5 appears twice in A386838.
		

Crossrefs

Programs

  • PARI
    a(n) = my(f, A = []); (f(n) = my(g, S, T = []); (g(n) = my(P = []); for(x = 0, sqrtint(n), my(y2 = n - x^2); if(issquare(y2), my(y = sqrtint(y2)); if(x <= y, P = concat(P, [[x, y]])))); return(P)); S = g(n); if(#S == 0, return(0), for(k = 1, #S, T = concat(T, S[k][1] + S[k][2] - gcd(S[k][1], S[k][2]))); return(vecmin(T)))); for(k = 1, 2*n^2, if(f(k) == n, A = concat(A, f(k)))); return(#A)

A386838 Minimum number of unit squares that a straight line drawn from (0,0) to (x,y) passes through on the square lattice where x^2 + y^2 = A001481(n). If A001481(n) can be written as a sum of two squares in two or more ways, x and y are chosen such that a(n) is the least value.

Original entry on oeis.org

0, 0, 1, 0, 2, 2, 0, 3, 4, 0, 4, 3, 4, 0, 5, 6, 4, 7, 0, 6, 6, 8, 6, 0, 5, 8, 8, 9, 10, 0, 8, 8, 6, 10, 11, 8, 0, 9, 10, 12, 9, 12, 7, 0, 10, 10, 13, 12, 14, 12, 12, 0, 11, 10, 8, 13, 14, 14, 0, 12, 15, 12, 16, 12, 16, 12, 9, 16, 0, 13, 14, 15, 12, 18, 16, 18, 17, 0, 14
Offset: 1

Author

Miles Englezou, Aug 05 2025

Keywords

Comments

a(n) is the minimal area of the graph formed under the requirement that the straight line (0,0) to (x,y) passes through an enclosed space on the square lattice, with the graph drawn using only vertical and horizontal edges.
Every nonnegative integer n appears in this sequence. Proof: Since 2*n^2 = n^2 + n^2 then by the first formula in the formula section n + n - gcd(n,n) = n. To prove that a(m) = n when A001481(m) = 2*n^2, we have to prove that x = n and y = n is the choice such that a(m) is minimal. Let r and s be two other numbers such that 2*n^2 = r^2 + s^2. Let r > n: consequently s < n, 1 <= gcd(r,s) <= s, and s - gcd(r,s) >= 0. If r + s - gcd(r,s) <= n, then s - gcd(r,s) < 0. But s - gcd(r,s) >= 0. Therefore r + s - gcd(r,s) >= r > n, and a(m) = n.

Examples

			a(4) = 0 since A001481(4) = 4 and 4 = 2^2 + 0^2. A straight line from (0,0) to (2,0) stays on the x axis and therefore passes through no unit squares.
a(5) = 2 since A001481(5) = 5 and 5 = 2^2 + 1^2. A straight line from (0,0) to (2,1) passes through two unit squares. It looks like this:
      _ _ (2,1)
     |_|_|
(0,0)
a(6) = 2 since A001481(6) = 8 and 8 = 2^2 + 2^2. A straight line from (0,0) to (2,2) passes through two unit squares. It looks like this:
        _ (2,2)
      _|_|
     |_|
(0,0)
a(16) = 6 since A001481(16) = 29 and 29 = 5^2 + 2^2. A straight line from (0,0) to (5,2) passes through six unit squares. It looks like this:
          _ _ _ (5,2)
      _ _|_|_|_|
     |_|_|_|
(0,0)
a(14) = 0 since A001481(14) = 25 and 25 = 5^2 + 0^2 = 4^2 + 3^2. x + y - gcd(x,y) is minimal for x = 5 and y = 0 and is equal to zero, therefore a(14) = 0.
		

Crossrefs

Programs

  • PARI
    a(n) = my(f, S, T = []); (f(n) = my(P = []); for(x=0, sqrtint(n), my(y2 = n - x^2); if(issquare(y2), my(y = sqrtint(y2)); if(x <= y, P = concat(P, [[x, y]])))); return(P)); S = f(n); if(#S == 0, return(0), for(k = 1, #S, T = concat(T, S[k][1] + S[k][2] - gcd(S[k][1], S[k][2]))); return(vecmin(T))) \\ function will also return 0 for n not in A001481 so any loop of a(n) must filter n

Formula

For x and y defined in the title, a(n) = x + y - gcd(x,y).
a(n) = 0 when A001481(n) is square.
a(n) = k when A001481(n) = 2*k^2, for k >= 0.
a(n) = A328803(n) - gcd(x,y) for A001481(n) = x^2 + y^2 with exactly one decomposition into a sum of two squares.

A385999 Least k such that every group of order n embeds into a group of order k*n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 4, 3, 2, 1, 12, 1, 2, 1, 16, 1, 12, 1, 8, 3, 2, 1
Offset: 1

Author

Miles Englezou, Jul 14 2025

Keywords

Examples

			a(2) = 1 since there is one group of order 2 and therefore 2 is the least order such that all groups of order 2 are embedded, and 2/2 = 1.
a(4) = 2 since there are two groups of order 4 and both groups are embedded in a group of order 8, and 8/4 = 2.
a(12) = 12 since there are five groups of order 12 and 144 is the least order for which there is a group into which all five groups are embedded, and 144/12 = 12.
		

Crossrefs

Cf. A340514.

Programs

  • GAP
    # Checks for n within the range [u..v]. In general u should be made equal to 1 to avoid erroneous output. Choice in range given for efficiency in checking individual terms.
    a := function(n, u, v)
        local T, S, k, r, m;
        T := [];
        for k in [1..NrSmallGroups(n)] do
            T := Concatenation(T, [SmallGroup(n,k)]);
        od;
        for m in [u..v] do
            S := [];
            for r in [1..NrSmallGroups(m*n)] do
                S := Concatenation(S, [SmallGroup(m*n, r)]);
            od;
            if ForAny(S, H -> ForAll(T, G -> ForAny(AllSubgroups(H), K -> IsomorphismGroups(G, K) <> fail))) then
                return m;
                break;
            fi;
        od;
        return fail;
    end;

Formula

a(n) = A340514(n)/n.
a(p) = 1 for prime p.
a(p^2) = p.
a(p^3) = p^3 for p an odd prime.
If p < q are distinct primes, a(pq) = p if p divides (q-1), else a(pq) = 1.

A385480 Sum of the orders of every automorphism group of order n.

Original entry on oeis.org

1, 1, 2, 8, 4, 8, 6, 212, 54, 24, 10, 64, 12, 48, 8, 20936, 16, 552, 18, 132, 54, 120, 22, 856, 500, 168, 11844, 216, 28, 192, 30, 10047248, 20, 288, 24, 2856, 36, 360, 180, 2128, 40, 468, 42, 520, 216, 528, 46, 61696, 2058, 13080, 32, 876, 52, 320400, 150, 3960
Offset: 1

Author

Miles Englezou, Jun 30 2025

Keywords

Comments

Let G be a group of order n, let N = {1, 2, ..., n}, and let f: G -> N be a bijection whereby f(G) = I is an index set of G. An automorphism phi of G is a permutation of N via f(phi(G)). It is tempting to ask the question 'how many permutations of N obey the group laws?'. However this question is not well-defined since it would require there being a natural single choice of bijection for every group of order n, which in general does not exist. Enumerating permutations of N which are automorphisms for every isomorphism class G will therefore depend on the choice of bijection for G. a(n) is the upper bound for all such enumerations of permutations of size n since a(n) is either: the maximum enumeration when the choice of bijections ensures that all permutations are distinct; or a(n) is the enumeration including all multiplicities when the choice of bijections leads to permutations which are not distinct.

Examples

			a(3) = 2 since for the one group of order 3, C_3, |Aut(C_3)| = 2.
a(4) = 8 since for the two groups of order 4, C_4 and C_2 x C_2, |Aut(C_4)| + |Aut(C_2 x C_2)| = 2 + 6 = 8.
a(8) = 212 since for the five groups of order 8, the sum of the orders of their automorphism groups is 4 + 8 + 8 + 24 + 168 = 212.
		

Crossrefs

Cf. A137316.

Programs

  • GAP
    a := function(n)
    local T, k;
    T := [];
    for k in [1..NrSmallGroups(n)] do
        T := Concatenation(T, [Order(AutomorphismGroup(SmallGroup(n,k)))]);
    od;
    return Sum(T);
    end;

Formula

a(n) is the sum of the k-th row in A137316.

A385209 Least k such that A086369(k) = n.

Original entry on oeis.org

1, 2, 3, 4, 9, 6, 15, 16, 81, 12, 45, 64, 729, 24, 105, 36, 225, 48, 405, 1024, 59049, 60, 315, 4096, 531441, 192, 3645, 144, 2025, 120, 945, 65536, 43046721, 180, 1575, 262144, 387420489, 240, 2835, 576, 18225, 3072, 295245, 4194304, 31381059609, 360, 3465, 1296
Offset: 1

Author

Miles Englezou, Jun 21 2025

Keywords

Comments

A086369(n) is also the cardinality of the set containing the divisors d of n and those 0 < m < n satisfying m + d = n.

Examples

			a(4) = 4 since 4 is the least k such that A086369(k) = 4.
a(13) = 729 since 729 is the least k such that A086369(k) = 13.
		

Crossrefs

Programs

  • PARI
    a(n,k) = my(f); (f(n) = if(n%2==1, 1+2*sumdiv(n,d,d
    				

Formula

a(2*p-1) = 3^(p-1) for prime p. - Jinyuan Wang, Jun 30 2025

Extensions

a(45)-a(48) from Jinyuan Wang, Jun 30 2025

A385202 Irregular triangle read by rows: let S be an ordered set of nondivisors of n such that a and b belong to S if a + b = n. T(n,k) is the k-th member of S. If S is empty, T(n,k) = 0.

Original entry on oeis.org

0, 0, 0, 0, 2, 3, 0, 2, 3, 4, 5, 3, 5, 2, 4, 5, 7, 3, 4, 6, 7, 2, 3, 4, 5, 6, 7, 8, 9, 5, 7, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 3, 4, 5, 6, 8, 9, 10, 11, 2, 4, 6, 7, 8, 9, 11, 13, 3, 5, 6, 7, 9, 10, 11, 13, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 4, 5, 7, 8, 10, 11, 13, 14
Offset: 1

Author

Miles Englezou, Jun 21 2025

Keywords

Examples

			 n | Triangle begins:
---+-----------------
 1 | {}
 2 | {}
 3 | {}
 4 | {}
 5 | {2, 3}
 6 | {}
 7 | {2, 3, 4, 5}
 8 | {3, 5}
 9 | {2, 4, 5, 7}
10 | {3, 4, 6, 7}
11 | {2, 3, 4, 5, 6, 7, 8, 9}
12 | {5, 7}
13 | {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
14 | {3, 4, 5, 6, 8, 9, 10, 11}
15 | {2, 4, 6, 7, 8, 9, 11, 13}
16 | {3, 5, 6, 7, 9, 10, 11, 13}
17 | {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
18 | {4, 5, 7, 8, 10, 11, 13, 14}
19 | {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17}
20 | {3, 6, 7, 8, 9, 11, 12, 13, 14, 17}
21 | {2, 4, 5, 6, 8, 9, 10, 11, 12, 13, 15, 16, 17, 19}
22 | {3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19}
23 | {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21}
24 | {5, 7, 9, 10, 11, 13, 14, 15, 17, 19}
25 | {2, 3, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 22, 23}
		

Crossrefs

Cf. A086369.

Programs

  • PARI
    T(n,k) = my(S); S = select(x -> setsearch(divisors(n), x)==0 && setsearch(divisors(n), n-x)==0, [1..n]); if(k <= #S, S[k], 0) \\ function made to output 0 if k exceeds the size of S to avoid breaking

Formula

n = T(n, m) + T(n, k-(m-1)), 1 <= m <= k, for every row of length k.
S defined as in the name, n - |S| = A086369(n).

A385030 Orders of characteristically simple groups.

Original entry on oeis.org

2, 3, 4, 5, 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, 37, 41, 43, 47, 49, 53, 59, 60, 61, 64, 67, 71, 73, 79, 81, 83, 89, 97, 101, 103, 107, 109, 113, 121, 125, 127, 128, 131, 137, 139, 149, 151, 157, 163, 167, 168, 169, 173, 179, 181, 191, 193, 197, 199, 211
Offset: 1

Author

Miles Englezou, Jun 15 2025

Keywords

Comments

Equivalently, orders k of groups G where a G exists as a direct product of isomorphic simple groups.
A group G is characteristically simple if it contains no characteristic proper subgroups (a subgroup which is invariant under every automorphism of G). Since a finite group is characteristically simple if and only if it is a direct product of isomorphic simple groups, G is characteristically simple if and only if it is an elementary abelian group or a direct product of isomorphic nonabelian simple groups.

Examples

			5 is a term since C_5 is prime cyclic and contains no proper subgroups. Therefore it contains no characteristic proper subgroups.
60 is a term since the alternating group A_5 is simple and contains no normal subgroups. Therefore it contains no characteristic proper subgroups.
3600 is a term since the direct product A_5 x A_5, though it contains A_5 twice as a normal subgroup and is therefore not simple, it contains no characteristic proper subgroups.
		

Crossrefs

Programs

  • GAP
    isok := function(G)
        if Order(G) = 1 then
            return false;
        elif IsElementaryAbelian(G) then
            return true;
        elif IsSimpleGroup(G) then
            return true;
        else
            for K in AllSubgroups(G) do
                if IsCharacteristicSubgroup(G, K) then
                    return false;
                fi;
            od;
            return true;
        fi;
    end;

Formula

Union of A246655 and the nonzero powers of every term in A001034.

A384310 Numbers k such that A383844(k) and A383844(k+1) are nonzero.

Original entry on oeis.org

0, 3, 6, 7, 12, 20, 26, 27, 28, 53, 56, 61, 74, 88, 145, 146, 252, 289, 299, 308, 320, 323, 340, 471, 577, 578, 739, 1240, 1517, 1568, 1579, 1857, 2638, 3042, 3043, 3133, 3455, 3565, 4910, 8683, 8684, 8857, 8858, 9291, 14549, 17913, 18117, 20005, 21989, 32552, 37902, 42514, 44869, 47877, 49942
Offset: 1

Author

Miles Englezou, Jun 04 2025

Keywords

Comments

a(n) is the lesser term of a pair of consecutive nonzero terms in A383844.
Triplets of consecutive nonzero terms can also be found in A383844 and are represented here as pairs. Up to n = 83354 there are 8 such triplets, the least terms of each being 6, 26, 27, 145, 577, 3042, 8683, 8857.

Examples

			26 is a term since A383844(26) and A383844(27) are nonzero.
27 is a term since A383844(27) and A383844(28) are nonzero.
252 is a term since A383844(252) and A383844(253) are nonzero.
61890 is a term since A383844(61890) and A383844(61891) are nonzero.
		

Crossrefs

Programs

  • PARI
    isok(n) = (count(n) = my(f, S=[], b);(f(m)=my(r=0); forprime(p=2, m, r+=m%p); return(r)); if(n<=21, b=26); if(n>21, b=n); if(n>=250, b=n^0.8); if(n>=6000, b=n^0.7); if(n>=21000, b=n^0.68); if(n>=43000, b=n^0.67); for(k=0, b, if(f(k)==n, S=concat(S,k))); return(S)); if(n==0 || (n>1 && count(n)<>[] && count(n+1)<>[]), return(1), return(0))

A384584 Numbers k such that A383844(k) = 2.

Original entry on oeis.org

4, 46, 62, 119, 145, 180, 200, 247, 305, 522, 707, 900, 1235, 1504, 1532, 1540, 2396, 3140, 4181, 4231, 6419, 9066, 9885, 14292, 17914, 22696, 33924, 35933, 38951, 80602
Offset: 1

Author

Miles Englezou, Jun 04 2025

Keywords

Comments

Numbers k such that there are exactly two m such that Sum_{i=1..t} m mod prime(i) for prime(t) <= m < prime(t+1) is equal to k (see A024934).
A383844(s) <= 3 for s <= 82000, with A383844(s) = 3 only for s = 0, 1, 8, 37, 781.

Examples

			4 is a term since 7 and 10 are the only numbers r such that A024934(r) = 4.
46 is a term since 29 and 30 are the only numbers r such that A024934(r) = 46.
9066 is a term since 552 and 566 are the only numbers r such that A024934(r) = 9066.
		

Crossrefs

Programs

  • PARI
    isok(n) = (count(n) = my(f, S=[], b); (f(m) = my(r=0); forprime(p=2, m, r+=m%p); return(r)); if(n<=21, b=26, b=n); for(k=0, b, if(f(k)==n, S=concat(S, k))); return(#S)); if(count(n)==2, return(1), return(0))

A384544 Numbers k such that A383327(k) = 1.

Original entry on oeis.org

1, 3, 5, 9, 15, 23, 35, 63, 65, 69, 113, 125, 141, 149, 173, 209, 231, 275, 279, 299, 321, 353, 365, 383, 419, 465, 509, 519, 555, 575, 603, 653, 695, 749, 765, 875, 945, 951, 959, 983
Offset: 1

Author

Miles Englezou, Jun 02 2025

Keywords

Comments

Numbers k such that there is only one m such that Sum_{i=1..t} m mod 2^i for 2^t <= m < 2^(t+1) is equal to k (see A049802). Every m = 2^k+1 (see A383327).
All terms are odd since for every even r there are at least two numbers u, v where the congruences sum to r (for u = 2^r+1 and v = 2^(r/2+1)+2).

Examples

			3 is a term since 2^3+1 = 9 is the only number such that the congruences sum to 3.
15 is a term since 2^15+1 = 32769 is the only number such that the congruences sum to 15.
63 is a term since 2^63+1 = 9223372036854775809 is the only number such that the congruences sum to 63.
		

Crossrefs

Programs

  • PARI
    isok(n) = (count(n) = local(tuple_sum, section, expansion, T=[], breakout, S, K); (tuple_sum(m) = sum(k=1, logint(m, 2), m % 2^k)); (section(r) = my(S=[]); for(n=1, 2^(r+1), if(logint(n, 2)==r, S=concat(S, n))); return(S[#S/2+1..#S])); (expansion(a, l) = my(k=a, K=[]); K=concat(K, a); for(n=1, l-1, K=concat(K, k+2^(logint(a, 2)-1+n)); k=k+2^(logint(a, 2)-1+n)); return(K)); for(k=1, n, for(i=1, #section(k), breakout=0; if(tuple_sum(section(k)[1]) > n, breakout=1); K=expansion(section(k)[i], n); for(j=1, #K, if(tuple_sum(K[j]) > n, break, if(tuple_sum(K[j])==n, T=concat(T, K[j]); break)))); if(breakout==1, break)); return(#T)); if(count(n)==1, return(1), return(0))