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 1-10 of 12 results. Next

A085311 Number of distinct 8th powers modulo n.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 4, 2, 4, 4, 6, 4, 4, 8, 4, 2, 3, 8, 10, 4, 8, 12, 12, 4, 6, 8, 10, 8, 8, 8, 16, 2, 12, 6, 8, 8, 10, 20, 8, 4, 6, 16, 22, 12, 8, 24, 24, 4, 22, 12, 6, 8, 14, 20, 12, 8, 20, 16, 30, 8, 16, 32, 16, 3, 8, 24, 34, 6, 24, 16, 36, 8, 10, 20, 12, 20, 24, 16, 40, 4, 28, 12, 42, 16, 6
Offset: 1

Views

Author

Labos Elemer, Jun 27 2003

Keywords

Comments

This sequence is multiplicative. - Leon P Smith, Apr 16 2005

Crossrefs

Cf. A000224[k=2], A046530[k=3], A052273[k=4], A052274[k=5], A052275[k=6], A085310[k=7], A085312[k=9], A085313[k=10], A085314[k=11], A228849[k=12], A055653.

Programs

  • Maple
    A085311 := proc(m)
        {seq( modp(b^8,m),b=0..m-1) };
        nops(%) ;
    end proc:
    seq(A085311(m),m=1..100) ; # R. J. Mathar, Sep 22 2017
  • Mathematica
    a[n_] := Table[PowerMod[i, 8, n], {i, 0, n - 1}] // Union // Length;
    Array[a, 100] (* Jean-François Alcover, Mar 24 2020 *)
  • PARI
    a(n)=my(f=factor(n)); prod(i=1, #f[, 1], my(k=f[i, 1]^f[i, 2]); #vecsort(vector(k, i, i^8%k), , 8)) \\ Charles R Greathouse IV, Sep 05 2013

A319099 Number of solutions to x^5 == 1 (mod n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 5, 1, 1, 1, 1, 1, 5, 1, 5, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 5, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 5, 1, 1, 1, 1, 5, 1, 1, 1, 5, 1, 5, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Jianing Song, Sep 10 2018

Keywords

Comments

All terms are powers of 5. Those n such that a(n) > 1 are in A066500.

Examples

			Solutions to x^5 == 1 (mod 11): x == 1, 3, 4, 5, 9 (mod 11).
Solutions to x^5 == 1 (mod 25): x == 1, 6, 11, 16, 21 (mod 25) (x == 1 (mod 5)).
Solutions to x^5 == 1 (mod 31): x == 1, 2, 4, 8, 16 (mod 31).
		

Crossrefs

Number of solutions to x^k == 1 (mod n): A060594 (k=2), A060839 (k=3), A073103 (k=4), this sequence (k=5), A319100 (k=6), A319101 (k=7), A247257 (k=8).
Mobius transform gives A307380.

Programs

  • Mathematica
    f[p_, e_] := If[Mod[p, 5] == 1, 5, 1]; f[5, 1] = 1; f[5, e_] := 5; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)
  • PARI
    a(n)=my(Z=znstar(n)[2]); prod(i=1,#Z,gcd(5,Z[i]));

Formula

Multiplicative with a(5) = 1, a(5^e) = 5 if e >= 2; for other primes p, a(p^e) = 5 if p == 1 (mod 5), a(p^e) = 1 otherwise.
If the multiplicative group of integers modulo n is isomorphic to C_{k_1} x C_{k_2} x ... x C_{k_m}, where k_i divides k_j for i < j; then a(n) = Product_{i=1..m} gcd(5, k_i).
a(n) = A000010(n)/A293482(n). - Jianing Song, Nov 10 2019

A288341 Expansion of 1 / ((1-x)^2*(1-x^2)*(1-x^3)*...*(1-x^6)).

Original entry on oeis.org

1, 2, 4, 7, 12, 19, 30, 44, 64, 90, 125, 169, 227, 298, 388, 498, 634, 797, 996, 1231, 1513, 1844, 2235, 2689, 3221, 3833, 4542, 5353, 6284, 7341, 8547, 9907, 11447, 13176, 15121, 17293, 19725, 22427, 25436, 28767, 32459, 36529, 41023, 45958, 51385, 57327
Offset: 0

Views

Author

Seiichi Manyama, Jun 08 2017

Keywords

Comments

Number of partitions of at most n into at most 6 parts.

Crossrefs

Number of partitions of at most n into at most k parts: A002621 (k=4), A002622 (k=5), this sequence (k=6), A288342 (k=7), A288343 (k=8), A288344 (k=9), A288345 (k=10).
Cf. A288253. Column 6 of A092905. A001402 (first differences).

Programs

  • PARI
    x='x+O('x^99); Vec(1/((1-x)*prod(i=1, 6, (1-x^i)))) \\ Altug Alkan, Mar 28 2018

A092905 Triangle, read by rows, such that the partial sums of the n-th row form the n-th diagonal, for n>=0, where each row begins with 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 6, 4, 2, 1, 1, 6, 9, 7, 4, 2, 1, 1, 7, 12, 11, 7, 4, 2, 1, 1, 8, 16, 16, 12, 7, 4, 2, 1, 1, 9, 20, 23, 18, 12, 7, 4, 2, 1, 1, 10, 25, 31, 27, 19, 12, 7, 4, 2, 1, 1, 11, 30, 41, 38, 29, 19, 12, 7, 4, 2, 1, 1, 12, 36, 53, 53, 42, 30, 19, 12, 7, 4, 2, 1
Offset: 0

Views

Author

Paul D. Hanna, Mar 12 2004

Keywords

Comments

Row sums form A000070, which is the partial sums of the partition numbers (A000041). Rows read backwards converge to the row sums (A000070).
From Alford Arnold, Feb 07 2010: (Start)
The table can also be generated by summing sequences embedded within Table A008284
For example,
1 1 1 1 ... yields 1 2 3 4 ...
1 1 2 2 3 3 ... yields 1 2 4 6 9 12 ...
1 1 2 3 4 5 7 ... yields 1 2 4 7 11 16 ...
(End)
T(n,k) is also count of all 'replacable' cells in the (Ferrers plots of) the partitions on n in exactly k parts. [Wouter Meeussen, Sep 16 2010]
From Wolfdieter Lang, Dec 03 2012: (Start)
The triangle entry T(n,k) is obtained from triangle A072233 by summing the entries of column k up to n (see the partial sum type o.g.f. given by Vladeta Jovovic in the formula section).
Therefore, the o.g.f. for the sequence in column k is x^k/((1-x)* product(1-x^j,j=1..k)).
The triangle with entry a(n,m) = T(n-1,m-1), n >= 1, m = 1, ..., n, is obtained from the partition array A103921 when in row n all entries belonging to part number m are summed (a conjecture). (End)

Examples

			The fourth row (n=3) is {1,3,2,1} and the fourth diagonal is the partial sums of the fourth row: {1,4,6,7,7,7,7,7,...}.
The triangle T(n,k) begins:
n\k 0  1  2  3  4  5  6  7  8  9 10 11 12  ...
0   1
1   1  1
2   1  2  1
3   1  3  2  1
4   1  4  4  2  1
5   1  5  6  4  2  1
6   1  6  9  7  4  2  1
7   1  7 12 11  7  4  2  1
8   1  8 16 16 12  7  4  2  1
9   1  9 20 23 18 12  7  4  2  1
10  1 10 25 31 27 19 12  7  4  2  1
11  1 11 30 41 38 29 19 12  7  4  2  1
12  1 12 36 53 53 42 30 19 12  7  4  2  1
... Reformatted by _Wolfdieter Lang_, Dec 03 2012
T(5,3)=4 because the partitions of 5 in exactly 3 parts are 221 and 311, and they give rise to partitions of 4 in four ways: 221->22 and 211, 311->211 and 31, since both their Ferrers plots have 2 'mobile cells' each. [_Wouter Meeussen_, Sep 16 2010]
T(5,3) = a(6,4) = 4 because the partitions of 6 with 4 parts are 1113 and 1122, with the number of distinct parts 2 and 2, respectively, summing to 4 (see the array A103921). An example for the conjecture given as comment above. - _Wolfdieter Lang_, Dec 03 2012
		

Crossrefs

Antidiagonal sums form the partition numbers (A000041).
Cf. A000070.
Cf. A008284. [Alford Arnold, Feb 07 2010]

Programs

Formula

T(n, k) = sum_{j=0..k} T(n-k, j), with T(n, 0) = 1 for all n>=0. A000070(n) = sum_{k=0..n} T(n, k).
O.g.f.: (1/(1-y))*(1/Product(1-x*y^k, k=1..infinity)). - Vladeta Jovovic, Jan 29 2005

Extensions

Several corrections by Wolfdieter Lang, Dec 03 2012

A288344 Expansion of 1 / ((1-x)^2*(1-x^2)*(1-x^3)*...*(1-x^9)).

Original entry on oeis.org

1, 2, 4, 7, 12, 19, 30, 45, 67, 97, 138, 192, 265, 359, 482, 639, 840, 1092, 1410, 1803, 2291, 2889, 3621, 4508, 5584, 6875, 8424, 10269, 12463, 15055, 18115, 21704, 25910, 30814, 36522, 43137, 50794, 59618, 69774, 81422, 94760, 109984, 127338, 147058, 169438
Offset: 0

Views

Author

Seiichi Manyama, Jun 08 2017

Keywords

Comments

Number of partitions of at most n into at most 9 parts.

Crossrefs

Number of partitions of at most n into at most k parts: A002621 (k=4), A002622 (k=5), A288341 (k=6), A288342 (k=7), A288343 (k=8), this sequence (k=9), A288345 (k=10).
Cf. A288256, A008638 (first differences).

Programs

  • PARI
    x='x+O('x^99); Vec(1/((1-x)*prod(i=1, 9, (1-x^i)))) \\ Altug Alkan, Mar 28 2018

A288345 Expansion of 1 / ((1-x)^2*(1-x^2)*(1-x^3)*...*(1-x^10)).

Original entry on oeis.org

1, 2, 4, 7, 12, 19, 30, 45, 67, 97, 139, 194, 269, 366, 494, 658, 870, 1137, 1477, 1900, 2430, 3083, 3890, 4874, 6078, 7533, 9294, 11406, 13940, 16955, 20545, 24787, 29800, 35688, 42600, 50670, 60088, 71024, 83714, 98377, 115305, 134771, 157138, 182746, 212038
Offset: 0

Views

Author

Seiichi Manyama, Jun 08 2017

Keywords

Comments

Number of partitions of at most n into at most 10 parts.

Crossrefs

Number of partitions of at most n into at most k parts: A002621 (k=4), A002622 (k=5), A288341 (k=6), A288342 (k=7), A288343 (k=8), A288344 (k=9), this sequence (k=10).
Cf. A008639 (first differences).

Programs

  • PARI
    x='x+O('x^99); Vec(1/((1-x)*prod(i=1, 10, (1-x^i)))) \\ Altug Alkan, Mar 28 2018

A250207 The number of quartic terms in the multiplicative group modulo n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 5, 1, 3, 3, 1, 1, 4, 3, 9, 1, 3, 5, 11, 1, 5, 3, 9, 3, 7, 1, 15, 2, 5, 4, 3, 3, 9, 9, 3, 1, 10, 3, 21, 5, 3, 11, 23, 1, 21, 5, 4, 3, 13, 9, 5, 3, 9, 7, 29, 1, 15, 15, 9, 4, 3, 5, 33, 4, 11, 3, 35, 3, 18, 9, 5, 9, 15, 3, 39, 1
Offset: 1

Views

Author

R. J. Mathar, Mar 02 2015

Keywords

Comments

In the character table of the multiplicative group modulo n there are phi(n) different characters. [This is made explicit for example by the number of rows in arXiv:1008.2547.] The set of the fourth powers of the characters in all representations has some cardinality, which defines the sequence.

Examples

			For n <= 6, the set of all characters in all representations consists of a subset of +1, -1, +i or -i. Their fourth powers are all +1, a single value, so a(n)=1 then.
For n=7, the set of characters is 1, -1, +-1/2 +- sqrt(3)*i/2, so their fourth powers are 1 or -1/2 +- sqrt(3)*i/2, which are three different values, so a(7)=3.
For n=11, the fourth powers of the characters may be 1, exp(+-2*i*Pi/5) or exp(+-4*i*Pi/5), which are 5 different values.
		

Crossrefs

Programs

  • Maple
    A250207 := proc(n)
        numtheory[phi](n)/A073103(n) ;
    end proc:
  • Mathematica
    a[n_] := EulerPhi[n]/Count[Range[0, n-1]^4 - 1, k_ /; Divisible[k, n]];
    Array[a, 80] (* Jean-François Alcover, Nov 20 2017 *)
    f[p_, e_] := (p - 1)*p^(e - 1)/If[Mod[p, 4] == 1, 4, 2]; f[2, e_] := If[e <= 3, 1, 2^(e - 4)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)
  • PARI
    a(n)=my(f=factor(n)); prod(i=1,#f~, if(f[i,1]==2, 2^max(0,f[i,2]-4), f[i,1]^(f[i,2]-1)*(f[i,1]-1)/if(f[i,1]%4==1,4,2))) \\ Charles R Greathouse IV, Mar 02 2015

Formula

a(n) = A000010(n)/A073103(n).
Multiplicative with a(2^e) = 1 for e<=3; a(2^e) = 2^(e-4) for e>=4; a(p^e) = p^(e-1)*(p-1)/4 for e>=1 and p == 1 (mod 4); a(p^e) = p^(e-1)*(p-1)/2 for e>=1 and p == 3 (mod 4). (Derived from A073103.) - R. J. Mathar, Oct 13 2017

A288342 Expansion of 1 / ((1-x)^2*(1-x^2)*(1-x^3)*...*(1-x^7)).

Original entry on oeis.org

1, 2, 4, 7, 12, 19, 30, 45, 66, 94, 132, 181, 246, 328, 433, 564, 728, 929, 1177, 1477, 1841, 2277, 2799, 3417, 4150, 5010, 6019, 7194, 8561, 10140, 11964, 14057, 16457, 19195, 22315, 25854, 29865, 34391, 39493, 45224, 51654, 58844, 66877, 75823, 85776, 96820
Offset: 0

Views

Author

Seiichi Manyama, Jun 08 2017

Keywords

Comments

Number of partitions of at most n into at most 7 parts.

Crossrefs

Number of partitions of at most n into at most k parts: A002621 (k=4), A002622 (k=5), A288341 (k=6), this sequence (k=7), A288343 (k=8), A288344 (k=9), A288345 (k=10).
Cf. A288254.

Programs

  • PARI
    x='x+O('x^99); Vec(1/((1-x)*prod(i=1, 7, (1-x^i)))) \\ Altug Alkan, Mar 28 2018

A288343 Expansion of 1 / ((1-x)^2*(1-x^2)*(1-x^3)*...*(1-x^8)).

Original entry on oeis.org

1, 2, 4, 7, 12, 19, 30, 45, 67, 96, 136, 188, 258, 347, 463, 609, 795, 1025, 1313, 1665, 2099, 2624, 3262, 4026, 4945, 6035, 7332, 8859, 10660, 12764, 15226, 18083, 21402, 25230, 29647, 34713, 40525, 47155, 54719, 63307, 73056, 84074, 96524, 110536, 126301
Offset: 0

Views

Author

Seiichi Manyama, Jun 08 2017

Keywords

Comments

Number of partitions of at most n into at most 8 parts.

Crossrefs

Number of partitions of at most n into at most k parts: A002621 (k=4), A002622 (k=5), A288341 (k=6), A288342 (k=7), this sequence (k=8), A288344 (k=9), A288345 (k=10).
Cf. A288255.

Programs

  • Mathematica
    CoefficientList[Series[1/((1-x)Times@@(1-x^Range[8])),{x,0,50}],x] (* Harvey P. Dale, Dec 06 2017 *)
  • PARI
    x='x+O('x^99); Vec(1/((1-x)*prod(i=1, 8, (1-x^i)))) \\ Altug Alkan, Mar 28 2018

A293483 The number of 6th powers in the multiplicative group modulo n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 5, 1, 2, 1, 2, 2, 8, 1, 3, 2, 1, 5, 11, 1, 10, 2, 3, 1, 14, 2, 5, 4, 5, 8, 2, 1, 6, 3, 2, 2, 20, 1, 7, 5, 2, 11, 23, 2, 7, 10, 8, 2, 26, 3, 10, 1, 3, 14, 29, 2, 10, 5, 1, 8, 4, 5, 11, 8, 11, 2, 35, 1, 12, 6, 10, 3, 5, 2, 13, 4, 9, 20
Offset: 1

Views

Author

R. J. Mathar, Oct 10 2017

Keywords

Comments

The size of the set of numbers j^6 mod n, gcd(j,n)=1, 1 <= j <= n.
A000010(n) / a(n) is another multiplicative integer sequence.

Crossrefs

The number of k-th powers in the multiplicative group modulo n: A046073 (k=2), A087692 (k=3), A250207 (k=4), A293482 (k=5), this sequence (k=6), A293484 (k=7), A293485 (k=8).

Programs

  • Maple
    A293483 := proc(n)
        local r,j;
        r := {} ;
        for j from 1 to n do
            if igcd(j,n)= 1 then
                r := r union { modp(j &^ 6,n) } ;
            end if;
        end do:
        nops(r) ;
    end proc:
    seq(A293483(n),n=1..120) ;
  • Mathematica
    a[n_] := EulerPhi[n]/Count[Range[0, n - 1]^6 - 1, k_ /; Divisible[k, n]];
    Array[a, 100] (* Jean-François Alcover, May 24 2023 *)
    f[p_, e_] := (p - 1)*p^(e - 1)/If[Mod[p, 6] == 1, 6, 2]; f[2, e_] := If[e <= 3, 1, 2^(e - 3)]; f[3, e_] := If[e <= 2, 1, 3^(e - 2)]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Aug 10 2023 *)

Formula

Conjecture: a(2^e) = 1 for e <= 3; a(2^e) = 2^(e-3) for e >= 3; a(3^e) = 1 for e <= 2; a(3^e) = 3^(e-2) for e >= 2; a(p^e) = (p-1)*p^(e-1)/2 for p == 5 (mod 6); a(p^e) = (p-1)*p^(e-1)/6 for p == 1 (mod 6). - R. J. Mathar, Oct 13 2017
a(n) = A000010(n)/A319100(n). This implies that the conjecture above is true. - Jianing Song, Nov 10 2019
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