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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A038763 Triangular matrix arising in enumeration of catafusenes, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 7, 15, 9, 1, 10, 36, 54, 27, 1, 13, 66, 162, 189, 81, 1, 16, 105, 360, 675, 648, 243, 1, 19, 153, 675, 1755, 2673, 2187, 729, 1, 22, 210, 1134, 3780, 7938, 10206, 7290, 2187, 1, 25, 276, 1764, 7182, 19278, 34020, 37908, 24057, 6561, 1, 28, 351, 2592, 12474, 40824, 91854, 139968, 137781, 78732, 19683
Offset: 0

Views

Author

N. J. A. Sloane, May 03 2000

Keywords

Comments

Triangle T(n,k), 0<=k<=n, read by rows, given by [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 10 2005
Triangle read by rows, n-th row = X^(n-1) * [1, 1, 0, 0, 0, ...] where X = an infinite bidiagonal matrix with (1,1,1,...) in the main diagonal and (3,3,3,...) in the subdiagonal; given row 0 = 1. - Gary W. Adamson, Jul 19 2008
Fusion of polynomial sequences P and Q given by p(n,x)=(x+2)^n and q(n,x)=(2x+1)^n; see A193722 for the definition of fusion of two sequences of polynomials or triangular arrays. - Clark Kimberling, Aug 04 2011

Examples

			Triangle begins:
  1;
  1,  1;
  1,  4,   3;
  1,  7,  15,   9;
  1, 10,  36,  54,   27;
  1, 13,  66, 162,  189,   81;
  1, 16, 105, 360,  675,  648,  243;
  1, 19, 153, 675, 1755, 2673, 2187, 729;
		

Crossrefs

Programs

  • Magma
    A038763:= func< n,k | n eq 0 select 1 else 3^(k-1)*(3*n-2*k)*Binomial(n,k)/n >;
    [A038763(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 27 2023
    
  • Mathematica
    A038763[n_,k_]:= If[n==0, 1, 3^(k-1)*(3*n-2*k)*Binomial[n,k]/n];
    Table[A038763[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 27 2023 *)
  • PARI
    T(n,k) = if ((n<0) || (k<0), return(0)); if ((n==0) && (k==0), return(1)); if (n==1, if (k<=1, return(1))); T(n-1,k) + 3*T(n-1,k-1);
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", "))); \\ Michel Marcus, Jul 25 2023
    
  • SageMath
    def A038763(n,k): return 1 if (n==0) else 3^(k-1)*(3*n-2*k)*binomial(n,k)/n
    flatten([[A038763(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 27 2023

Formula

T(n, 0)=1; T(1, 1)=1; T(n, k)=0 for k>n; T(n, k) = T(n-1, k-1)*3 + T(n-1, k) for n >= 2.
Sum_{k=0..n} T(n,k) = A081294(n). - Philippe Deléham, Sep 22 2006
T(n, k) = A136158(n, n-k). - Philippe Deléham, Dec 17 2007
G.f.: (1-2*x*y)/(1-(3*y+1)*x). - R. J. Mathar, Aug 11 2015
From G. C. Greubel, Dec 27 2023: (Start)
T(n, 0) = A000012(n).
T(n, 1) = A016777(n-1).
T(n, 2) = A062741(n-1).
T(n, 3) = 9*A002411(n-2).
T(n, 4) = 27*A001296(n-3).
T(n, 5) = 81*A051836(n-4).
T(n, n) = A133494(n).
T(n, n-1) = A006234(n+2).
T(n, n-2) = A080420(n-2).
T(n, n-3) = A080421(n-3).
T(n, n-4) = A080422(n-4).
T(n, n-5) = A080423(n-5).
T(2*n, n) = 4*A098399(n-1) + (2/3)*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A006138(n-1) + (2/3)*[n=0].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A110523(n-1) + (4/3)*[n=0]. (End)

Extensions

More terms from Michel Marcus, Jul 25 2023

A081422 Triangle read by rows in which row n consists of the first n+1 n-gonal numbers.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 10, 1, 4, 9, 16, 25, 1, 5, 12, 22, 35, 51, 1, 6, 15, 28, 45, 66, 91, 1, 7, 18, 34, 55, 81, 112, 148, 1, 8, 21, 40, 65, 96, 133, 176, 225, 1, 9, 24, 46, 75, 111, 154, 204, 261, 325, 1, 10, 27, 52, 85, 126, 175, 232, 297, 370, 451
Offset: 0

Views

Author

Paul Barry, Mar 21 2003

Keywords

Examples

			The array starts
  1  1  3 10 ...
  1  2  6 16 ...
  1  3  9 22 ...
  1  4 12 28 ...
The triangle starts
  1;
  1,  1;
  1,  2,  3;
  1,  3,  6, 10;
  1,  4,  9, 16, 25;
  ...
		

Crossrefs

Antidiagonals are composed of n-gonal numbers.

Programs

  • GAP
    Flat(List([0..10], n-> List([1..n+1], k-> k*((n-2)*k-(n-4))/2 ))); # G. C. Greubel, Aug 14 2019
  • Magma
    [[k*((n-2)*k-(n-4))/2: k in [1..n+1]]: n in [0..10]]; // G. C. Greubel, Oct 13 2018
    
  • Mathematica
    Table[PolygonalNumber[n,i],{n,0,10},{i,n+1}]//Flatten (* Requires Mathematica version 10.4 or later *) (* Harvey P. Dale, Aug 27 2016 *)
  • PARI
    tabl(nn) = {for (n=0, nn, for (k=1, n+1, print1(k*((n-2)*k-(n-4))/2, ", ");); print(););} \\ Michel Marcus, Jun 22 2015
    
  • Sage
    [[k*((n-2)*k -(n-4))/2 for k in (1..n+1)] for n in (0..10)] # G. C. Greubel, Aug 14 2019
    

Formula

Array of coefficients of x in the expansions of T(k, x) = (1 + k*x -(k-2)*x^2)/(1-x)^4, k > -4.
T(n, k) = k*((n-2)*k -(n-4))/2 (see MathWorld link). - Michel Marcus, Jun 22 2015

A100177 Structured meta-prism numbers, the n-th number from a structured n-gonal prism number sequence.

Original entry on oeis.org

1, 4, 18, 64, 175, 396, 784, 1408, 2349, 3700, 5566, 8064, 11323, 15484, 20700, 27136, 34969, 44388, 55594, 68800, 84231, 102124, 122728, 146304, 173125, 203476, 237654, 275968, 318739, 366300, 418996, 477184, 541233, 611524, 688450, 772416, 863839, 963148, 1070784, 1187200, 1312861, 1448244
Offset: 1

Views

Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Examples

			There are no 1- or 2-gonal prisms, so 1 and (2n) are used as the first and second terms since all the sequences begin as such.
		

Crossrefs

Cf. A002411, A000578, A050509, A006597, A100176, A100177 - structured prisms; A006484 for other meta structured numbers; and A100145 for more on structured numbers.

Programs

  • Magma
    [(1/6)*(3*n^4-9*n^3+12*n^2): n in [1..50] ]; // Vincenzo Librandi, Aug 02 2011
  • Mathematica
    Table[(3n^4-9n^3+12n^2)/6,{n,50}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,4,18,64,175},50] (* Harvey P. Dale, Nov 07 2017 *)
  • PARI
    a(n)=(1/6)*(3*n^4-9*n^3+12*n^2);
    

Formula

a(n) = (1/6)*(3*n^4 - 9*n^3 + 12*n^2).
G.f.: x*(1 - x + 8*x^2 + 4*x^3)/(1-x)^5. - Colin Barker, Jun 08 2012
a(n) = A060354(n) * n = A000124(n-2) * n^2. - Bruce J. Nicholson, Jul 11 2018

A256645 25-gonal pyramidal numbers: a(n) = n*(n+1)*(23*n-20)/6.

Original entry on oeis.org

0, 1, 26, 98, 240, 475, 826, 1316, 1968, 2805, 3850, 5126, 6656, 8463, 10570, 13000, 15776, 18921, 22458, 26410, 30800, 35651, 40986, 46828, 53200, 60125, 67626, 75726, 84448, 93815, 103850, 114576, 126016, 138193, 151130, 164850, 179376, 194731, 210938, 228020
Offset: 0

Views

Author

Luciano Ancora, Apr 07 2015

Keywords

Comments

If b(n,k) = n*(n+1)*((k-2)*n-(k-5))/6 is n-th k-gonal pyramidal number, then b(n,k) = A000292(n) + (k-3)*A000292(n-1) (see Deza in References section, p. 96).
Also, b(n,k) = b(n,k-1) + A000292(n-1) (see Deza in References section, p. 95). Some examples:
for k=4, A000330(n) = A000292(n) + A000292(n-1);
for k=5, A002411(n) = A000330(n) + A000292(n-1);
for k=6, A002412(n) = A002411(n) + A000292(n-1), etc.
This is the case k=25.

References

  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93 (23rd row of the table).

Crossrefs

Partial sums of A255184.
Cf. similar sequences listed in A237616.

Programs

  • Magma
    k:=25; [n*(n+1)*((k-2)*n-(k-5))/6: n in [0..40]]; // Vincenzo Librandi, Apr 08 2015
  • Mathematica
    Table[n (n + 1) (23 n - 20)/6, {n, 0, 40}]
    LinearRecurrence[{4, -6, 4, -1}, {0, 1, 26, 98}, 40] (* Vincenzo Librandi, Apr 08 2015 *)
  • PARI
    concat(0, Vec(x*(1 + 22*x)/(1 - x)^4 + O(x^100))) \\ Colin Barker, Apr 07 2015
    

Formula

G.f.: x*(1 + 22*x)/(1 - x)^4.
a(n) = A000292(n) + 22*A000292(n-1) = A256716(n) + A000292(n-1), see comments.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Colin Barker, Apr 07 2015
E.g.f.: exp(x)*x*(6 + 72*x + 23*x^2)/6. - Elmo R. Oliveira, Aug 04 2025

A337414 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the edges of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 70, 1, 5, 40, 1407, 8200, 1, 6, 75, 12480, 9080559, 12804908, 1, 7, 126, 69050, 1503323520, 4906480368591, 304899216832, 1, 8, 196, 281946, 81461669375, 48226825456539776, 187380251418565888983, 103685962258536432, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is an octahedron with 12 edges. The number of edges is 2n*(n-1) for n>1.
Also the number of achiral colorings of the regular (n-2)-dimensional orthotopes (hypercubes) in a regular n-dimensional orthotope.

Examples

			Table begins with T(1,1):
1  2    3     4     5      6      7       8       9       10 ...
1  6   18    40    75    126    196     288     405      550 ...
1 70 1407 12480 69050 281946 931490 2632512 6598935 15041950 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337411 (oriented), A337412 (unoriented), A337413 (chiral).
Rows 1-4 are A000027, A002411, A331351, A331357.
Cf. A327086 (simplex edges), A337410 (orthotope edges), A325007 (orthoplex vertices).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}];DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3,n}]],0,(per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]],1,j2], 2j2], {j2,n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[])]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0,cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[m]=b;
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,8}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = 2*A337412(n,k) - A337411(n,k) = A337411(n,k) - 2*A337413(n,k) = A337412(n,k) - A337413(n,k).

A121997 Count up to n, n times.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7
Offset: 1

Views

Author

Keywords

Comments

The n-th block consists of n subblocks, each of which counts from 1 to n.
This a fractal sequence: removing the first instance of each value leaves the original sequence.
The first comment implies that this gives the column index of the n-th element of a sequence whose terms are coefficients, read by rows, of a sequence of matrices of size 1 X 1, 2 X 2, 3 X 3, etc.; cf. example. The row index is given by A238013(n), and the size of the matrix by A074279(n). - M. F. Hasler, Feb 16 2014

Examples

			Sequence begins:
  1;
  1,2;
  1,2;
  1,2,3;
  1,2,3;
  1,2,3;
  ...
The blocks of n subblocks of n terms (n=1,2,3,...) can be cast into a square matrices of order n; then the terms are equal to the index of the column they fall into.
		

Crossrefs

Cf. A081489 (locations of new values), A075349 (locations of 1's).
Cf. A000290 (row lengths), A002411 (row sums), A036740 (row products).
Cf. A002024 and references there, esp. in PROG section.
Cf. A238013.

Programs

  • PARI
    A121997(N=9)=concat(vector(N,i,concat(vector(i,j,vector(i,k,k))))) \\ Note: this creates a vector; use A121997()[n] to get the n-th term. - M. F. Hasler, Feb 16 2014
    
  • Python
    from sympy import integer_nthroot
    def A121997(n): return 1+(n-(k:=(m:=integer_nthroot(3*n,3)[0])+(6*n>m*(m+1)*((m<<1)+1)))*(k-1)*((k<<1)-1)//6-1)%k # Chai Wah Wu, Nov 04 2024

A255211 a(n) = n*(n+1)*(7*n+2)/6.

Original entry on oeis.org

0, 3, 16, 46, 100, 185, 308, 476, 696, 975, 1320, 1738, 2236, 2821, 3500, 4280, 5168, 6171, 7296, 8550, 9940, 11473, 13156, 14996, 17000, 19175, 21528, 24066, 26796, 29725, 32860, 36208, 39776, 43571, 47600, 51870, 56388, 61161, 66196, 71500, 77080, 82943
Offset: 0

Views

Author

Luce ETIENNE, Feb 17 2015

Keywords

Comments

a(n) is the number of triangles of all sizes in a polyiamond of trapezoid shape with 3 sides of length n and the base of length 2*n. The number of triangular cells in the trapezoid is 3*n^2. This is half of a regular hexagon with side lengths n.
The number of triangles oriented with their bases aligned with the base of the trapezoid is n*(n+1)*(2*n+1)/3 and the number oriented in the opposite direction is n^2*(n+1)/2. a(n) is the sum of these two.

Examples

			From the second comment: a(1)= 2+1, a(2)= 10+6, a(3)= 28+18, a(4)= 60+40.
		

Crossrefs

Partial sums of A022264.

Programs

  • Magma
    [n*(n+1)*(7*n+2)/6 : n in [0..50]]; // Wesley Ivan Hurt, Apr 11 2021
  • Mathematica
    Table[n (n + 1) (7 n + 2)/6, {n, 0, 50}] (* Bruno Berselli, Feb 17 2015 *)
  • PARI
    concat(0, Vec(x*(4*x+3)/(x-1)^4 + O(x^100))) \\ Colin Barker, Feb 17 2015
    
  • PARI
    vector(50, n, n--; n*(n+1)*(7*n+2)/6) \\ Bruno Berselli, Feb 17 2015
    

Formula

G.f.: x*(3 + 4*x) / (1 - x)^4. - Colin Barker, Feb 17 2015
a(n) = Sum_{j=0..n-1} (n-j)*(3*n-2*j) = Sum_{j=1..n} j*(n+2*j) for n>0.
a(n) = A000292(2*n) - A000292(n). - Bruno Berselli, Sep 22 2016
Sum_{n>=1} 1/a(n) = 21*HarmonicNumber(2/7)/5 - 6/5 = 0.44513027538601361333... . - Vaclav Kotesovec, Sep 22 2016
E.g.f.: exp(x)*x*(18 + 30*x + 7*x^2)/6. - Stefano Spezia, Mar 02 2025

Extensions

Edited and extended by Bruno Berselli, Dec 01 2016

A294842 Expansion of Product_{k>=1} (1 + x^k)^(k^2*(k+1)/2).

Original entry on oeis.org

1, 1, 6, 24, 73, 238, 722, 2175, 6343, 18177, 50982, 140671, 382227, 1023623, 2706184, 7067324, 18250671, 46635309, 117997008, 295794098, 735030985, 1811435607, 4429226677, 10749552338, 25903858181, 62000039513, 147435739522, 348431110651, 818549931526, 1912010876019, 4441687009798
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 09 2017

Keywords

Comments

Weigh transform of the pentagonal pyramidal numbers (A002411).

Crossrefs

Programs

  • Mathematica
    nmax = 30; CoefficientList[Series[Product[(1 + x^k)^(k^2 (k + 1)/2), {k, 1, nmax}], {x, 0, nmax}], x]
    a[n_] := a[n] = If[n == 0, 1, Sum[Sum[(-1)^(k/d + 1) d^3 (d + 1)/2, {d, Divisors[k]}] a[n - k], {k, 1, n}]/n]; Table[a[n], {n, 0, 30}]

Formula

G.f.: Product_{k>=1} (1 + x^k)^A002411(k).
a(n) ~ exp(-2401 * Pi^16 / (2^12 * 3^11 * 5^8 * Zeta(5)^3) + (343 * Pi^12 / (2^(38/5) * 3^(37/5) * 5^(36/5) * Zeta(5)^(11/5))) * n^(1/5) - (49*Pi^8 / (2^(31/5) * 3^(24/5) * 5^(22/5) * Zeta(5)^(7/5))) * n^(2/5) + (7*Pi^4 / (2^(14/5) * 3^(16/5) * 5^(8/5) * Zeta(5)^(3/5))) * n^(3/5) + (5 * 3^(2/5) * (5*Zeta(5))^(1/5) / 2^(12/5)) * n^(4/5)) * 3^(1/5) * Zeta(5)^(1/10) / (2^(167/240) * 5^(2/5) * sqrt(Pi) * n^(3/5)). - Vaclav Kotesovec, Nov 10 2017

A337410 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 70, 1, 5, 40, 1407, 93024, 1, 6, 75, 12480, 294157089, 47823602694208, 1, 7, 126, 69050, 91983927296, 67514530382043163023924, 443077371786837979607993095063601152, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of achiral colorings of the regular (n-2)-dimensional simplexes in a regular n-dimensional orthoplex.

Examples

			Table begins with T(1,1):
1  2    3     4     5      6      7       8       9       10 ...
1  6   18    40    75    126    196     288     405      550 ...
1 70 1407 12480 69050 281946 931490 2632512 6598935 15041950 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337407 (oriented), A337408 (unoriented), A337409 (chiral).
Rows 1-4 are A000027, A002411, A331351, A331361.
Cf. A327086 (simplex edges), A337414 (orthoplex edges), A325015 (orthotope vertices).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, n - m]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}];DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3,n}]],0,(per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]],1,j2], 2j2], {j2,n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[])]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0,cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,7}, {n,m,d+m-1}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = 2*A337408(n,k) - A337407(n,k) = A337407(n,k) - 2*A337409(n,k) = A337408(n,k) - A337409(n,k).

A363636 Indices of numbers of the form k^2+1, k >= 0, that can be written as a product of smaller numbers of that same form.

Original entry on oeis.org

0, 3, 7, 13, 17, 18, 21, 31, 38, 43, 47, 57, 68, 73, 91, 99, 111, 117, 123, 132, 133, 157, 183, 211, 241, 242, 253, 255, 268, 273, 293, 302, 307, 313, 322, 327, 343, 381, 413, 421, 438, 443, 463, 487, 507, 515, 553, 557, 577, 593, 601, 651, 693, 697, 703, 707
Offset: 1

Views

Author

Pontus von Brömssen, Jun 19 2023

Keywords

Comments

For the corresponding sequence for numbers of the form k^3+1 instead of k^2+1, the only terms known to me are 0 and 26, with 26^3+1 = (2^3+1)^2*(6^3+1).

Examples

			0 is a term because 0^2+1 = 1 equals the empty product.
3 is a term because 3^2+1 = 10 = 2*5 = (1^2+1)*(2^2+1).
38 is a term because 38^2+1 = 1445 = 5*17*17 = (2^2+1)*(4^2+1)^2. (This is the first term that requires more than two factors.)
		

Crossrefs

Sequences that list those terms (or their indices or some other key) of a given sequence that are products of smaller terms of the same sequence (in other words, the nonprimitive terms of the multiplicative closure of the sequence):
this sequence (A002522),

Programs

  • Mathematica
    g[lst_, p_] :=
      Module[{t, i, j},
       Union[Flatten[Table[t = lst[[i]]; t[[j]] = p*t[[j]];
          Sort[t], {i, Length[lst]}, {j, Length[lst[[i]]]}], 1],
        Table[Sort[Append[lst[[i]], p]], {i, Length[lst]}]]];
    multPartition[n_] :=
      Module[{i, j, p, e, lst = {{}}}, {p, e} =
        Transpose[FactorInteger[n]];
       Do[lst = g[lst, p[[i]]], {i, Length[p]}, {j, e[[i]]}]; lst];
    output = Join[{0}, Flatten[Position[Table[
         test = Sqrt[multPartition[n^2 + 1][[2 ;; All]] - 1];
         Count[AllTrue[#, IntegerQ] & /@ test, True] > 0
         , {n, 707}], True]]]
    (* David Trimas, Jul 23 2023 *)
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