A004006 -id:A004006 - cp's OEIS frontend

A210983 Total number of pairs of states of the first n subshells of the nuclear shell model in which the subshells are ordered by energy level in increasing order.

1, 3, 4, 7, 8, 10, 14, 16, 17, 20, 25, 28, 29, 31, 35, 41, 45, 47, 48, 51, 56, 63, 68, 71, 72, 74, 78, 84, 92, 98, 102, 104, 105, 108, 113, 120, 129, 136, 141, 144, 145, 147, 151, 157, 165, 175, 183, 189, 193, 195, 196, 199, 204, 211, 220, 231

Offset: 1

Omar E. Pol, Jul 14 2012

Additional comments from Omar E. Pol, Sep 02 2012: (Start)
Q: What are energy levels?
A: See the link sections of A212122, A213362, A213372. For example, see this link related to A213372: http://www.flickr.com/photos/mitopencourseware/3772864128/in/set-72157621892931990
Q: What defines the order in A212121?
A: The order of A212121 is defined by A212122.
Note that there are at least five versions of the nuclear shell model in the OEIS:
Goeppert-Mayer (1950): A212012, A004736, A212013, A212014.
Goeppert-Mayer, Jensen (1955): A212122, A212121, A212123, A212124.
Talmi (1993): A213362, A213361, A213363, A213364.
Meyerhof: A213372, A213371, A213373, A213374.
For another version: A162630, A130517, A210983, A210984.
Each version is represented by four sequences: the first sequence is the main entry.
(End)
For additional information see A162630.

			Example 1: written as a triangle in which row i is related to the (i-1)st level of nucleus, the sequence begins:
1;
3,     4;
7,     8,  10;
14,   16,  17,  20;
25,   28,  29,  31,  35;
41,   45,  47,  48,  51,  56;
63,   68,  71,  72,  74,  78,  84;
92,   98, 102, 104, 105, 108, 113, 120;
129, 136, 141, 144, 145, 147, 151, 157, 165;
175, 183, 189, 193, 195, 196, 199, 204, 211, 220;
...
Column 1 gives positive terms of A004006. Right border gives positives terms of A000292.
Example 2: written as an irregular triangle in which row j is related to the j-th shell of nucleus. Note that in this case row 4 has only one term. Triangle begins:
1;
3,     4;
7,     8,  10;
14;
16,   17,  20,  25;
28,   29,  31,  35,  41;
45,   47,  48,  51,  56,  63;
68,   71,  72,  74,  78,  84,  92;
98,  102, 104, 105, 108, 113, 120, 129;
136, 141, 144, 145, 147, 151, 157, 165, 175;
183, 189, 193, 195, 196, 199, 204, 211, 220, 231;
...

Partial sums of A130517 (when that sequence is regarded as a flattened triangle). Other versions are A212013, A212123, A213363, A213373.

Cf. A000292, A004006, A162630, A210984

a(n) = A210984(n)/2.

A011826 f-vectors for simplicial complexes of dimension at most 1 (graphs) on at most n-1 vertices.

Original entry on oeis.org

2, 3, 5, 9, 16, 27, 43, 65, 94, 131, 177, 233, 300, 379, 471, 577, 698, 835, 989, 1161, 1352, 1563, 1795, 2049, 2326, 2627, 2953, 3305, 3684, 4091, 4527, 4993, 5490, 6019, 6581, 7177, 7808, 8475, 9179, 9921, 10702, 11523, 12385, 13289

Offset: 1

Svante Linusson (linusson(AT)math.kth.se)

nonn

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Aviezri S. Fraenkel and Urban Larsson, Playability and arbitrarily large rat games, arXiv:1705.03061 [math.CO], 2017. See p. 27.
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.

Cf. A000124, A004006, A014206, A033547.

a=2;s=3;lst={0,1,s};Do[a+=n;s+=a;AppendTo[lst,s],{n,2,4!,1}];lst+2 (* Vladimir Joseph Stephan Orlovsky, May 24 2009 *)

a(n) = (n^3 - 3n^2 + 8n + 6)/6 fits all listed terms. - John W. Layman, Mar 13 1999

Empirical G.f.: -x*(x^3 - 5*x^2 + 5*x - 2) / (x - 1)^4. - Colin Barker, Sep 19 2012

A057703 a(n) = n(94 + 5n + 25n^2 - 5n^3 + n^4)/120.

Original entry on oeis.org

0, 1, 3, 7, 15, 31, 62, 119, 218, 381, 637, 1023, 1585, 2379, 3472, 4943, 6884, 9401, 12615, 16663, 21699, 27895, 35442, 44551, 55454, 68405, 83681, 101583, 122437, 146595, 174436, 206367, 242824, 284273, 331211, 384167, 443703, 510415, 584934, 667927

Offset: 0

Leonid Broukhis, Oct 24 2000

Previous name was: This sequence is the result of the question: If you have a tall building and 5 plates and you need to find the highest story from which a plate thrown does not break, what is the number of stories you can handle given n tries?

Number of compositions with at most five parts and sum at most n. - Beimar Naranjo, Mar 12 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Boardman, The Egg-Drop Numbers, Mathematics Magazine, 77 (2004), 368-372. [_Parthasarathy Nambi_, Sep 30 2009]
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A004006.

Differences form A055795 + 1 = A000127.

List([0..40], n-> n*(94+5*n+25*n^2-5*n^3+n^4)/120); # G. C. Greubel, Jun 05 2019

[n*(94+5*n+25*n^2-5*n^3+n^4)/120: n in [0..40]]; // G. C. Greubel, Jun 05 2019

seq(sum(binomial(n,k),k=1..5),n=0..38); # Zerinvary Lajos, Dec 13 2007

LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 3, 7, 15, 31}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

vector(40, n, n--; n*(94+5*n+25*n^2-5*n^3+n^4)/120) \\ G. C. Greubel, Jun 05 2019

[n*(94+5*n+25*n^2-5*n^3+n^4)/120 for n in (0..40)] # G. C. Greubel, Jun 05 2019

a(n) = n*(94 + 5*n + 25*n^2 - 5*n^3 + n^4)/120.
a(n) = Sum_{j=1..5} binomial(n, j). - Labos Elemer
G.f.: x*(1 - 3*x + 4*x^2 - 2*x^3 + x^4)/(1-x)^6. - Colin Barker, Apr 15 2012
E.g.f.: x*(120 + 60*x + 20*x^2 + 5*x^3 + x^4)*exp(x)/120. - G. C. Greubel, Jun 05 2019

More terms and formula from James Sellers, Oct 25 2000

Name changed by G. C. Greubel, Jun 06 2019

A185506 Accumulation array, T, of the natural number array A000027, by antidiagonals.

Original entry on oeis.org

1, 3, 4, 7, 11, 10, 14, 23, 26, 20, 25, 42, 51, 50, 35, 41, 70, 88, 94, 85, 56, 63, 109, 140, 156, 155, 133, 84, 92, 161, 210, 240, 250, 237, 196, 120, 129, 228, 301, 350, 375, 374, 343, 276, 165, 175, 312, 416, 490, 535, 550, 532, 476, 375, 220, 231, 415, 558, 664, 735, 771, 770, 728, 639, 495, 286

Offset: 1

Clark Kimberling, Jan 29 2011

Suppose that R={R(n,k) : n>=1, k>=1} is a rectangular array. The accumulation array of R is given by T(n,k) = Sum_{R(i,j): 1<=i<=n, 1<=j<=k}. (See A144112.)

The formula for the integer T(n,k) has denominator 12. The 2nd, 3rd, and 4th accumulation arrays of A000027 have formulas in which the denominators are 144, 2880, and 86400, respectively; see A185507, A185508, and A185509.

			The natural number array A000027 starts with
  1, 2,  4,  7, ...
  3, 5,  8, 12, ...
  6, 9, 13, 18, ...
  ...
T(n,k) is the sum of numbers in the rectangle with corners at (1,1) and (n,k) of A000027, so that a corner of T is as follows:
   1,  3,   7,  14,  25,  41
   4, 11,  23,  42,  70, 109
  10, 26,  51,  88, 140, 210
  20, 50,  94, 156, 240, 350
  35, 85, 155, 250, 375, 535

G. C. Greubel, Table of n, a(n) for the first 50 antidiagonals, flattened

Cf. A000027, A185507, A185508, A185509.

Cf. A004006 (row 1), A000292 (col 1), A051925 (col 2), A185505 (1st diagonal).

f[n_,k_]:=k*n*(2n^2+3(k+1)*n+2k^2-3k+5)/12;
TableForm[Table[f[n,k],{n,1,10},{k,1,15}]]
Table[f[n-k+1,k],{n,14},{k,n,1,-1}]//Flatten

T(n,k) = k*n*(2*n^2 + 3*(k+1)*n + 2*k^2 - 3*k + 5)/12.

A051576 Order of Burnside group B(3,n) of exponent 3 and rank n.

Original entry on oeis.org

1, 3, 27, 2187, 4782969, 847288609443, 36472996377170786403, 1144561273430837494885949696427, 78551672112789411833022577315290546060373041, 35370553733215749514562618584237555997034634776827523327290883

Offset: 0

N. J. A. Sloane

The Burnside group of exponent e and rank r is B(e,r) := F_r / N where F_r is the free group generated by x_1, ..., x_r and N is the normal subgroup generated by all z^e with z in F_r. The Burnside problem is to determine when B(e,r) is finite. [Warning: Some authors interchange the order of e and r. But the symbol is not symmetric. B(i,j) != B(j,i). - N. J. A. Sloane, Jan 12 2016]
B(1,r), B(2,r), B(3,r), B(4,r) and B(6,r) are all finite: |B(1,r)| = 1, |B(2,r)| = 2^r, |B(3,r)| = A051576, |B(4,r)| = A079682, |B(6,r)| = A079683. |B(5,2)| = 5^34.
Many cases are known where B(e,r) is infinite (see references). Ivanov showed in 1994 that B(e,r) is infinite if r>1, e >= 2^48 and 2^9 divides e if e is even.
It is not known whether B(5,2) is finite or infinite.

Burnside, William. "On an unsettled question in the theory of discontinuous groups." Quart. J. Pure Appl. Math 33.2 (1902): 230-238.
M. Hall, Jr., The Theory of Groups, Macmillan, 1959, Chap. 18.
Havas, G. and Newman, M. F. "Application of Computers to Questions Like Those of Burnside." In Burnside Groups. Proceedings of a Workshop held at the University of Bielefeld, Bielefeld, June-July 1977. New York: Springer-Verlag, pp. 211-230, 1980.
Ivanov, Sergei V. "The free Burnside groups of sufficiently large exponents." International Journal of Algebra and Computation 4.01n02 (1994): 1-308. See Math. Rev. MR 1283947.
W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory, Wiley, 1966, see p. 380.
Novikov, P. S. and Adjan, S. I. "Infinite Periodic Groups I, II, III." Izv. Akad. Nauk SSSR Ser. Mat. 32, 212-244, 251-524, and 709-731, 1968.

Vincenzo Librandi, Table of n, a(n) for n = 0..23
M. Hall, Solution of the Burnside Problem for Exponent Six, Ill. J. Math. 2, 764-786, 1958.
S. V. Ivanov, On the Burnside problem for groups of even exponent, Proc. Internat. Congress of Mathematicians, Vol. II (Berlin, 1998). Doc. Math. 1998, Extra Vol. II, 67-75.
R. C. Lyndon, On Burnside's problem, Transactions of the American Mathematical Society 77, (1954) 202-215.
Todd D. Mateer, A Calculation of an Upper Bound for the Diameter of the Cayley Graph of the Restricted Burnside Group R(2,5)
E. A. O'Brien and M. F. Newman, Application of Computers to Questions Like Those of Burnside, II, Internat. J. Algebra Comput.6, 593-605, 1996.
J. J. O'Connor and E. F. Robertson, History of the Burnside Problem.
D. Rusin, Burnside Problem. [Broken link?]
D. Rusin, Burnside problem [Cached copy]
Eric Weisstein's World of Mathematics, Burnside Problem

Equals 3^A004006(n).

3^Table[n*(n^2 + 5)/6, {n, 0, 10}] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

A051576(n):=3^(n*(n^2+5)/6)$ makelist(A051576(n),n,0,7); /* Martin Ettl, Jan 08 2013 */

a(n) = 3^(n*(n^2+5)/6) for n >= 0.

Entry revised by N. J. A. Sloane, Jan 12 2016 and Jan 15 2016

A079682 Order of Burnside group B(4,n) of exponent 4 and rank n.

Original entry on oeis.org

1, 4, 4096, 590295810358705651712

Offset: 0

N. J. A. Sloane, Jan 31 2003

nonn

The Burnside group of exponent e and rank r is B(e,r) := F_r / N where F_r is the free group generated by x_1, ..., x_r and N is the normal subgroup generated by all z^e with z in F_r. The Burnside problem is to determine when B(e,r) is finite. [Warning: Some authors interchange the order of e and r. But the symbol is not symmetric. B(i,j) != B(j,i). - N. J. A. Sloane, Jan 12 2016]
B(1,r), B(2,r), B(3,r), B(4,r) and B(6,r) are all finite: |B(1,r)| = 1, |B(2,r)| = 2^r, |B(3,r)| = A051576, |B(4,r)| = A079682, |B(6,r)| = A079683. |B(5,2)| = 5^34.
Many cases are known where B(e,r) is infinite (see references). Ivanov showed in 1994 that B(e,r) is infinite if r>1, e >= 2^48 and 2^9 divides e if e is even.
It is not known whether B(5,2) is finite or infinite.
See A051576 for additional references.

Bayes, A. J.; Kautsky, J.; and Wamsley, J. W. "Computation in Nilpotent Groups (Application)." In Proceedings of the Second International Conference on the Theory of Groups. Held at the Australian National University, Canberra, August 13-24, 1973(Ed. M. F. Newman). New York: Springer-Verlag, pp. 82-89, 1974.
Burnside, William. "On an unsettled question in the theory of discontinuous groups." Quart. J. Pure Appl. Math 33.2 (1902): 230-238.
M. Hall, Jr., The Theory of Groups, Macmillan, 1959, Chap. 18.
Havas, G. and Newman, M. F. "Application of Computers to Questions Like Those of Burnside." In Burnside Groups. Proceedings of a Workshop held at the University of Bielefeld, Bielefeld, June-July 1977. New York: Springer-Verlag, pp. 211-230, 1980.
W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory, Wiley, 1966, see p. 380.
Tobin, J. J. On Groups with Exponent 4. Thesis. Manchester, England: University of Manchester, 1954.

N. J. A. Sloane, Table of n, a(n) for n = 0..5
S. V. Ivanov, On the Burnside problem for groups of even exponent, Proc. Internat. Congress of Mathematicians, Vol. II (Berlin, 1998). Doc. Math. 1998, Extra Vol. II, 67-75.
E. A. O'Brien and M. F. Newman, Application of Computers to Questions Like Those of Burnside, II, Internat. J. Algebra Comput.6, 593-605, 1996.
J. J. O'Connor and E. F. Robertson, History of the Burnside Problem
Eric Weisstein's World of Mathematics, Burnside Problem

Cf. A051576, A004006, A116398, A079682.

The first few terms are 2 to the powers 0, 2, 12, 69, 422, 2728, that is, 2^A116398(n).

Entry revised by N. J. A. Sloane, Jan 12 2016 and Jan 15 2016

A123736 Triangle T(n,k) = Sum_{j=0..k/2} binomial(n-j-1,k-2*j), read by rows.

Original entry on oeis.org

1, 0, 1, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 3, 4, 3, 2, 1, 1, 0, 1, 4, 7, 7, 5, 3, 2, 1, 1, 0, 1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1, 0, 1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1, 0, 1, 7, 22, 41, 51, 46, 33, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1, 8, 29, 63, 92, 97, 79, 54, 34, 21, 13, 8, 5, 3, 2, 1, 1, 0, 1

Offset: 1

Roger L. Bagula, Nov 14 2006

Row sums give: A000225

			The triangle starts in row n=1 with columns 0 <= k < 2*n:
  1, 0;
  1, 1,  1,  0;
  1, 2,  2,  1,  1,  0;
  1, 3,  4,  3,  2,  1,  1,  0;
  1, 4,  7,  7,  5,  3,  2,  1,  1,  0;
  1, 5, 11, 14, 12,  8,  5,  3,  2,  1,  1, 0;
  1, 6, 16, 25, 26, 20, 13,  8,  5,  3,  2, 1, 1, 0;
  1, 7, 22, 41, 51, 46, 33, 21, 13,  8,  5, 3, 2, 1, 1, 0;
  1, 8, 29, 63, 92, 97, 79, 54, 34, 21, 13, 8, 5, 3, 2, 1, 1, 0;

G. C. Greubel, Rows n = 1..100 of triangle, flattened
Eric Weisstein's World of Mathematics, Steenrod Algebra

Cf. A136431 (antidiagonals), A027926 (row-reversed), A004006 (column m=3)

Flat(List([1..10], n-> List([0..2*n-1], k-> Sum([0..Int(k/2)], j-> Binomial(n-j-1, k-2*j) )))); # G. C. Greubel, Sep 05 2019

[&+[Binomial(n-j-1, k-2*j): j in [0..Floor(k/2)]]: k in [0..2*n-1], n in [1..10]]; // G. C. Greubel, Sep 05 2019

seq(seq(sum(binomial(n-j-1, k-2*j), j=0..floor(k/2)), k=0..2*n-1), n=1..10); # G. C. Greubel, Sep 05 2019

Table[Sum[Binomial[n-j-1, k-2*j], {j,0,Floor[k/2]}], {n, 10}, {k, 0, 2*n-1}]//Flatten (* modified by G. C. Greubel, Sep 05 2019 *)

T(n,k) = sum(j=0, k\2, binomial(n-j-1, k-2*j));
for(n=1,10, for(k=0,2*n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Sep 05 2019

[[sum(binomial(n-j-1, k-2*j) for j in (0..floor(k/2))) for k in (0..2*n-1)] for n in (1..10)] # G. C. Greubel, Sep 05 2019

A175287 Partial sums of ceiling(n^2/4).

Original entry on oeis.org

0, 1, 2, 5, 9, 16, 25, 38, 54, 75, 100, 131, 167, 210, 259, 316, 380, 453, 534, 625, 725, 836, 957, 1090, 1234, 1391, 1560, 1743, 1939, 2150, 2375, 2616, 2872, 3145, 3434, 3741, 4065, 4408, 4769, 5150, 5550, 5971, 6412, 6875, 7359, 7866, 8395, 8948, 9524, 10125, 10750

Offset: 0

Mircea Merca, Dec 03 2010

a(n) is the number of 1243-avoiding odd Grassmannian permutations of size n+1. Avoiding any of the patterns 2134, 2341, or 4123, gives the same sequence. - Juan B. Gil, Mar 09 2023

Conjecture: a(n) is the number of perimeter-magic (hollow) triangles of order 3 with magic sum n+2. Order 3 means each of the 3 edges has 3 elements >=1; the triangle has 6 elements. The elements do not need to be distinct, and triangles obtained by rotations are counted only once. The triangle (read ccw) for magic sum 3 has elements 1 1 1 1 1 1. The 2 triangles with magic sum 4 are 1 1 2 1 1 2 and 1 2 1 2 1 2. - R. J. Mathar, Mar 08 2025

			a(4) = ceil(0/4)+ceil(1/4)+ceil(4/4)+ceil(9/4)+ceil(16/4) = 0+1+1+3+4=9.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
L. Colmenarejo, Combinatorics on several families of Kronecker coefficients related to plane partitions, arXiv:1604.00803 [math.CO], 2016. See Table 1 p. 5.
Juan B. Gil and Jessica A. Tomasko, Pattern-avoiding even and odd Grassmannian permutations, arXiv:2207.12617 [math.CO], 2022.
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Partial sums of A004652.

Cf. A361272.

[Floor((n+1)*(2*n^2+n+9)/24): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

a:= n-> round((2*n^(3)+3*n^(2)+10*n)/24): seq(a(n), n=0..20);

Table[Sum[Ceiling[i^2/4], {i, 0, n}], {n, 0, 49}] (* or *) Table[(2n(2n^2 + 3n + 10) -9(-1)^n + 9)/48, {n, 0, 49}] (* Alonso del Arte, Dec 03 2010 *)
CoefficientList[Series[(x^3 - x^2 + x)/(x^5 - 3 x^4 + 2 x^3 + 2 x^2 - 3 x + 1), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)
Accumulate[Ceiling[Range[0,50]^2/4]] (* or *) LinearRecurrence[{3,-2,-2,3,-1},{0,1,2,5,9},60] (* Harvey P. Dale, Nov 19 2014 *)

x='x+O('x^99); concat(0, Vec((x^3-x^2+x)/ (x^5-3*x^4+2*x^3+2*x^2-3*x+1))) \\ Altug Alkan, Apr 05 2016

a(n) = round((2*n+1)*(2*n^2+2*n+9)/48).
a(n) = floor((n+1)*(2*n^2+n+9)/24).
a(n) = ceiling((2*n^3+3*n^2+10*n)/24).
a(n) = round((2*n^3+3*n^2+10*n)/24).
a(n) = a(n-4)+n^2-3*n+5 , n>3.
G.f.: x*(1-x+x^2) / ( (1+x)*(x-1)^4 ).
a(n) = (2*n*(2*n^2+3*n+10)-9*(-1)^n+9)/48. - Bruno Berselli, Dec 03 2010
a(n)+a(n+1) = A004006(n+1). - R. J. Mathar, Mar 08 2025

A341091 Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .

Original entry on oeis.org

1, 0, 2, 1, -1, 3, 0, 3, -3, 4, 1, -2, 7, -6, 5, 0, 4, -8, 14, -10, 6, 1, -3, 13, -21, 25, -15, 7, 0, 5, -15, 35, -45, 41, -21, 8, 1, -4, 21, -49, 81, -85, 63, -28, 9, 0, 6, -24, 71, -129, 167, -147, 92, -36, 10, 1, -5, 31, -94, 201, -295, 315, -238, 129, -45, 11

Offset: 0

Thomas Scheuerle, Feb 13 2022

If we want to calculate the sum of finite differences for a sequence b(n):
  b(0)*T(0, n) + ... + b(n)*T(n, n) = b(0) + b(1) + ... + b(n) + (b(1) - b(0)) + ... + (b(n) - b(n-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... This sum includes the sequence b(n) itself. This defines an invertible linear sequence transformation with a deep connection to Bernoulli numbers and other interesting sequences of rational numbers.
From Thomas Scheuerle, Apr 29 2024: (Start)
These are the coefficients of the polynomials defined by the recurrence: P(k, x) = P(k - 1, x) + (x^2 - x)*P(k - 2, x) + 1, with P(-1, x) = 0 and P(0, x) = 1. This can also be expressed as P(k, x) = Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^2 - x)^(m - 1) = Sum_{n=0..k} T(n, k)*x^(k-n). If we would evaluate P(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1+(t-1)*x)*(1 - t*x)).
We may replace (x^2 - x) by (x^(-2) - x^(-1)) to get the coefficients in reverse order: x^k*Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^(-2) - x^(-1))^(m - 1) = Sum_{n=0..k} T(n, k)*x^n = F(k, x). If we would evaluate F(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1 - (t-1)*x)*(1 - t*x)). (End)

			Triangle begins with T(n, k):
   n=   0,  1,   2,   3,   4,   5,   6,   7,   8
  k=0   1
  k=1   0,  2
  k=2   1, -1,   3
  k=3   0,  3,  -3,   4
  k=4   1, -2,   7,  -6,   5
  k=5   0,  4,  -8,  14, -10,   6
  k=6   1, -3,  13, -21,  25, -15,   7
  k=7   0,  5, -15,  35, -45,  41, -21,   8
  k=8   1, -4,  21, -49,  81, -85,  63, -28,   9
  ...

Cf. A000027, A000032, A000045, A000110, A000217.
Cf. A001045, A004006, A032346, A051744, A052875.
Cf. A084416, A130595, A145766.
Cf. A027642, A164555 (Numerators and denominators of Bernoulli numbers).
Cf. A001008, A002805 (Numerators and denominators of harmonic numbers).
Cf. A344037, A372245.
Sequences below will be obtained by evaluation of the associated polynomials:
Cf. A000295, A000392, A000975, A016208.
Cf. A016218, A016228, A016241, A016249.
Cf. A016256, A016261, A249997.

A341091(n, k) = sum(m=n, k,(-1)^(m+n)*binomial(m+1, n))

A341091(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*hypergeom([1,k+3],k+3-n,-1)+(-1/2)^n*(2^(n+1)-1)) \\ Thomas Scheuerle, Apr 29 2024

b(0)*T(0, m) + b(1)*T(1, m) + ... + b(m)*T(m, m)
  = Sum_{j=0..m} Sum_{n=0..m-j} Sum_{k=0..n} (-1)^k*binomial(n, k)*b(j+n-k)
  = Sum_{n=0..m} b(n)*Sum_{j=n..m}(-1)^(j+n)*binomial(j+1, n).
T(n, k) = Sum_{m=n..k}(-1)^(m+n)*binomial(m+1, n).
T(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*Hypergeometric2F1(1, k+3, k+3-n, -1)+(-1/2)^n*(2^(n+1) - 1)), where Hypergeometric2F1 is the Gaussian hypergeometric function 2F1 as defined in Mathematica. - Thomas Scheuerle, Apr 29 2024
T(k, k) = A000027(k+1) The positive integers.
|T(k-1, k)| = A000217(k) The triangular numbers.
T(k-2, k) = A004006(k).
|T(k-3, k)| = A051744(k).
T(0, k*2) = 1.
T(0, k*2 + 1) = 0.
T(1, k*2 + 1) = k + 2.
T(1, k*2 + 2) = -(k + 1).
T(n, k) with constant n and variable k, a linear recurrence relation with characteristic polynomial (x-1)*(x+1)^(n+1).
Sum_{n=0..k} T(n, k)*B_n = 1. B_n is the n-th Bernoulli number with B_1 = 1/2. B_n = A164555(n)/A027642(n).
Sum_{n=0..k} T(n, k)*(1 - B_n) = k.
Sum_{n=0..k} T(n, k)*(2*n - 3+3*B_n) = k^2.
Sum_{n=0..k} T(n, k)*A032346(n) = A032346(k+1).
From Thomas Scheuerle, Apr 29 2024: (Start)
Sum_{n=0..k} T(n, k)*A000110(n+1) = A000110(k+2) - 1.
Sum_{n=0..k} T(n, k)*(1/(1+n)) = H(1+floor(k/2)), where H(k) is the harmonic number A001008(k)/A002805(k). (End)
Sum_{n=0..k} T(n, k)*c(n) = c(k). C(k) = {-1, 0, 1/2, 1/2, 1/8, -7/20, ...} this sequence of rational numbers can be defined recursively: c(0) = -1, c(m) = (-c(m-1) + Sum_{k=0..m-1} A130595(m+1, k)*c(k))/m.
  c(m) is an eigensequence of this transformation, all eigensequences are c(m) multiplied by any factor.
Sum_{n=0..k} T(n, k)*A000045(n) = 2*(A000045(2*floor((k+1)/2) - 1) - 1). A000045 are the Fibonacci numbers.
Sum_{n=0..k} T(n, k)*A000032(n) = A000032(2*floor(k/2)+2) - 2. A000032 are the Lucas numbers.
Sum_{n=0..k} T(n, k)*A001045(n) = A145766(floor((k+1)/2)). A001045 is the Jacobsthal sequence.
This sequence acting as an operator onto a monomial n^w:
Sum_{n=0..k} T(n, k)*n^w = (1/(w+1))*k^(w+1) + Sum_{v=1..w} ((v+B_v)*(w)_v/v!)*k^(w+1-v) - A052875(w) + O_k(w) (w)_v is the falling factorial. If k > w-1 then O_k(w) = 0. If k <= w-1 then O_k(w) is A084416(w, 2+k), the sequence with the exponential generating function: (e^x-1)^(2+k)/(2-e^x).
From Thomas Scheuerle, Apr 29 2024: (Start)
This sequence acting by its inverse operator onto a monomial k^w:
Sum_{n=0..k} T(n, k)*( Sum_{m=0..k} ((-1)^(1+m+k)*binomial(k, m)*(2^(k-m) - 1)*n^m + A344037(m)*B_n) ) = k^w - A372245(w, k+3), note that A372245(w, k+3) = 0 if k+3 > w. B_n is the n-th Bernoulli number with B_1 = 1/2.
How this sequence will act as an operator onto a Dirichlet series may be developed by the formulas below:
Sum_{n=0..k} T(n, k)*2^n = A000295(k+2).
Sum_{n=0..k} T(n, k)*3^n = A000392(k+3).
Sum_{n=0..k} T(n, k)*4^n = A016208(k).
Sum_{n=0..k} T(n, k)*5^n = A016218(k).
Sum_{n=0..k} T(n, k)*6^n = A016228(k).
Sum_{n=0..k} T(n, k)*7^n = A016241(k).
Sum_{n=0..k} T(n, k)*8^n = A016249(k).
Sum_{n=0..k} T(n, k)*9^n = A016256(k).
Sum_{n=0..k} T(n, k)*10^n = A016261(k).
Sum_{n=0..k} T(n, k)*m^n = m^2*m^k/(m-1) - (m-1)^2*(m-1)^k/(m-2) + 1/((m-1)*(m-2)), for m > 2.
Sum_{n=0..k} T(n, k)*( m*B_n + (m-1)*Sum_{t=1..m} t^n )*(1/m^2) = m^k, for m > 0. B_n is the n-th Bernoulli number with B_1 = 1/2.
Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j) = (-1)^(k+1)*LerchPhi(-1, 1, k+3) - 1 + log(2).
Sum_{n=0..k} T(k - n, k)*2^n = A000975(k+1)
Sum_{n=0..k} T(k - n, k)*3^n = A091002(k+2)
Sum_{n=0..k} T(k - n, k)*4^n = A249997(k). (End)

A079683 Order of Burnside group B(6,n) of exponent 6 and rank n.

Original entry on oeis.org

1, 6, 227442304239437611008

Offset: 0

N. J. A. Sloane, Jan 31 2003

The Burnside group of exponent e and rank r is B(e,r) := F_r / N where F_r is the free group generated by x_1, ..., x_r and N is the normal subgroup generated by all z^e with z in F_r. The Burnside problem is to determine when B(e,r) is finite. [Warning: Some authors interchange the order of e and r. But the symbol is not symmetric. B(i,j) != B(j,i). - N. J. A. Sloane, Jan 12 2016]
B(1,r), B(2,r), B(3,r), B(4,r) and B(6,r) are all finite: |B(1,r)| = 1, |B(2,r)| = 2^r, |B(3,r)| = A051576, |B(4,r)| = A079682, |B(6,r)| = A079683. |B(5,2)| = 5^34.
Many cases are known where B(e,r) is infinite (see references). Ivanov showed in 1994 that B(e,r) is infinite if r>1, e >= 2^48 and 2^9 divides e if e is even.
It is not known whether B(5,2) is finite or infinite.
The next term, a(3), is 2^4375*3^833. - N. J. A. Sloane, Jan 12 2016
See A051576 for additional references.

M. Hall, Jr., The Theory of Groups, Macmillan, 1959, Chap. 18.
W. Magnus, A. Karrass and D. Solitar, Combinatorial Group Theory, Wiley, 1966, see p. 380.

S. V. Ivanov, On the Burnside problem for groups of even exponent, Proc. Internat. Congress of Mathematicians, Vol. II (Berlin, 1998). Doc. Math. 1998, Extra Vol. II, 67-75.
J. J. O'Connor and E. F. Robertson, History of the Burnside Problem

Cf. A051576, A004006, A079682, A116398.

B6n:=proc(n) local a,b,c;
b:=1+(n-1)*2^n;
c:=n+binomial(n,2)+binomial(n,3);
a:=1+(n-1)*3^c;
2^a*3^(b+binomial(b,2)+binomial(b,3));
end; # N. J. A. Sloane, Jan 12 2016

The formula for a(n) was found by Marshall Hall, Jr.: a(n) = 2^i 3^(j + (j choose 2) + (j choose 3)) where i = 1 + (n-1)3^(n + (n choose 2) + (n choose 3)) and j = 1 + (n-1)2^n. (See also the Maple code.)

Entry revised by N. J. A. Sloane, Jan 12 2016 and Jan 15 2016

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