A001644 -id:A001644 - cp's OEIS frontend

A112312 Least index k such that the n-th prime divides the k-th tribonacci number.

4, 8, 15, 6, 9, 7, 29, 19, 30, 78, 15, 20, 36, 83, 30, 34, 65, 69, 101, 133, 32, 19, 271, 110, 20, 187, 14, 185, 106, 173, 587, 80, 12, 35, 11, 224, 72, 38, 42, 315, 101, 26, 73, 172, 383, 27, 84, 362, 35, 250, 37, 29, 507, 305, 55, 38, 178, 332, 62, 537, 778, 459, 31, 124

Offset: 1

Jonathan Vos Post, Nov 29 2005

The tribonacci numbers are indexed so that trib(0) = trib(1) = 0, trib(2) = 1, for n>2: trib(n) = trib(n-1) + trib(n-2) + trib(n-3). See A112618 for another version.

Brenner proves that every prime divides some tribonacci number T(n). For the similar 3-step Lucas sequence A001644, there are primes (A106299) that do not divide any term.

			a(1) = 4 because prime(1) = 2 and tribonacci( 4) = 2.
a(2) = 8 because prime(2) = 3 and tribonacci( 8) = 24 = 3 * 2^3.
a(3) = 15 because prime(3) = 5 and tribonacci(15) = 1705 = 5 *(11 * 31).
a(4) = 6 because prime(4) = 7 and tribonacci( 6) = 7.
a(5) = 9 because prime(5) = 11 and tribonacci( 9) = 44 = 11 * 4.
a(6) = 7 because prime(6) = 13 and tribonacci( 7) = 13.
a(7) = 29 because prime(7) = 17 and tribonacci(29) = 8646064 = 17 *(2^4 * 7 * 19 * 239).

J. L. Brenner, Linear Recurrence Relations, Amer. Math. Monthly, Vol. 61 (1954), 171-173.
Eric Weisstein's World of Mathematics, MathWorld: Tribonacci Number

Cf. A000040, A000045, A000073, A000204, A001644, A053028, A106299, A112312.

Cf. also A112618 = this sequence minus 1.

a[0] = a[1] = 0; a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + a[n - 3]; f[n_] := Module[{k = 2, p = Prime[n]}, While[Mod[a[k], p] != 0, k++ ]; k]; Array[f, 64] (* Robert G. Wilson v *)

a(n) = minimum k such that prime(n) | A000073(k) and A000073(k) >= prime(n). a(n) = minimum k such that A000040(n) | A000073(k) and A000073(k) >= A000040(n).

Corrected and extended by Robert G. Wilson v, Dec 01 2005

A191060 Primes that are not squares mod 11.

Original entry on oeis.org

2, 7, 13, 17, 19, 29, 41, 43, 61, 73, 79, 83, 101, 107, 109, 127, 131, 139, 149, 151, 167, 173, 193, 197, 211, 227, 233, 239, 241, 263, 271, 277, 281, 283, 293, 307, 337, 347, 349, 359, 373, 409, 431, 439, 457, 461, 479, 491, 503, 523, 541, 547, 557, 563

Offset: 1

T. D. Noe, May 25 2011

Inert rational primes in the field Q(sqrt(-11)). - N. J. A. Sloane, Dec 25 2017
These are also the primes p for which the polynomial x^3 - x^2 - x - 1 (mod p) has only one integer root. This is important for the Fibonacci and Lucas 3-step recursions, A000073 and A001644. See A106276. - T. D. Noe, Apr 17 2012
It appears that these are the primes p such that the sequence p^(5*n) mod 11 has period length 2, repeating [1, 10]. - Gary Detlefs, May 18 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index to sequences related to decomposition of primes in quadratic fields

[p: p in PrimesUpTo(563) | JacobiSymbol(p, 11) eq -1]; // Vincenzo Librandi, Sep 11 2012

Select[Prime[Range[200]], JacobiSymbol[#, 11] == -1 &]

A212634 Triangle read by rows: T(n,k) is the number of dominating subsets with cardinality k of the cycle C_n (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 0, 6, 4, 1, 0, 5, 10, 5, 1, 0, 3, 14, 15, 6, 1, 0, 0, 14, 28, 21, 7, 1, 0, 0, 8, 38, 48, 28, 8, 1, 0, 0, 3, 36, 81, 75, 36, 9, 1, 0, 0, 0, 25, 102, 150, 110, 45, 10, 1, 0, 0, 0, 11, 99, 231, 253, 154, 55, 11, 1

Offset: 1

Emeric Deutsch, Jun 14 2012

The entries in row n are the coefficients of the domination polynomial of the cycle C_n (see the Alikhani and Peng reference in Opuscula Math).
Sum of entries in row n = A001644(n) (number of dominating subsets; tribonacci numbers with initial values 1,3,7).
From Petros Hadjicostas, Jan 26 2019: (Start)
Let L(n, r, K) be the number of r-combinations of the n consecutive integers 1, 2, ..., n displaced on a circle, with no K integers consecutive (assuming that n and 1 are consecutive integers).
If on the circle where the n integers 1, 2, ..., n are displaced we put 1 for any integer that is included in the r-combination and 0 otherwise, we get a marked cyclic sequence of r ones and n-r zeros with no K consecutive ones.
Let L_{n,K}(x) = Sum_{r=0..floor(n-(n/K))} L(n, r, K)*x^(n-r) for n, K >= 1. Charalambides (1991) proved that L_{n,K}(x) = x*(n + Sum_{j=1..n-1} L_{n-j, K}(x)) for K >= 1 and 1 <= n <= K, and L_{n,K}(x) = x*Sum_{j=1..K} L_{n-j, K}(x) for K >= 1 and n >= K + 1. See Theorem 2.1 (p. 291) in his paper.
He also proved that L_{n,K}(x) = -1 + Sum_{j=0..floor(n/(K + 1))} (-1)^j*(n/(n-j*K))*binomial(n - j*K, j)*x^j*(1+x)^(n-j*(K+1)). See Theorem 2.3 (p. 293) in his paper.
He also produced other formulas for L(n, r, K) and L_{n,K}(x) including the explicit formulas of Moser and Abramson (1969) (regarding L(n, r, K)).
For the current sequence, we have T(n, k) = L(n, r = n-k, K = 3) for 1 <= n <= k. In other words, T(n, n-k) is the number of k-combinations of the n consecutive integers 1, 2, ..., n displaced on a circle, with no K = 3 consecutive integers (assuming n and 1 are consecutive).
(End)

			Row 4 is [0,6,4,1] because the cycle A-B-C-D-A has dominating subsets AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD, and ABCD.
Triangle starts:
1;
2, 1;
3, 3,  1;
0, 6,  4,  1;
0, 5, 10,  5, 1;
0, 3, 14, 15, 6, 1;
...
From _Petros Hadjicostas_, Jan 27 2019: (Start)
Let n=6 and 1 <= k <= 6. Then T(n, k) is the number of (n-k)-combinations of the integers 1, 2, 3, 4, 5, 6 displaced on a circle with no K=3 consecutive integers (assuming 6 and 1 are consecutive). Equivalently, T(n, k) is the number of marked cyclic sequences consisting of n-k ones and k zeros with no K=3 consecutive ones.
For each k below we give the corresponding (n-k)-combinations and the equivalent marked sequences of 0's and 1's.
k=1, n-k = 5: none; T(n=6, k=1) = 0.
k=2, n-k = 4: 1245 <-> 110110, 2356 <-> 011011, 1346 <-> 101101; T(n=6, k=2) = 3.
k=3, n-k = 3: 124 <-> 110100, 125 <-> 110010, 134 <-> 101100, 135 <-> 101010, 136 <-> 101001, 145 <-> 100110, 146 <-> 100101, 235 <-> 011010, 236 <-> 011001, 245 <-> 010110, 246 <-> 010101, 256 <-> 010011, 346 <-> 001101, 356 <-> 001011; T(n=6, k=3) = 14.
k=4, n-k=2: all 2-combinations of the integers 1,2,3,4,5,6 and all marked cyclic sequences with exactly 2 ones and 4 zeros; hence, T(n=6, k=4) = binomial(6, 2) = 15.
k=5, n-k=1: all 1-combinations of the integers 1,2,3,4,5,6 and all marked cyclic sequences with exactly 1 one and 5 zeros; hence, T(n=6, k=5) = binomial(6, 1) = 6.
k=6, n-k=0: empty combination <-> 000000; T(n=6, k=6) = 1.
(End)

S. Alikhani and Y. H. Peng, Introduction to domination polynomial of a graph, arXiv:0905.2251 [math.CO], 2009.
S. Alikhani and Y. H. Peng, Dominating sets and domination polynomials of paths, International J. Math. and Math. Sci., Vol. 2009, Article ID542040.
S. Alikhani and Y. H. Peng, Dominating sets and domination polynomials of certain graphs, II, Opuscula Math., 30, No. 1, 2010, 37-51.
J. L. Arocha, B. Llano, The number of dominating k-sets of paths, cycles and wheels, arXiv preprint arXiv:1601.01268 [math.CO], 2016.
C. A. Charalambides, Lucas numbers and polynomials of order k and the length of the longest circular success run, The Fibonacci Quarterly, 29 (1991), 290-297.
W. O. J. Moser and M. Abramson, Enumeration of combinations with restricted differences and cospan, J. Combin. Theory, 7 (1969), 162-170.
Eric Weisstein's World of Mathematics, Cycle Graph
Eric Weisstein's World of Mathematics, Domination Polynomial

Cf. A001644, A125127, A284966, A322057.

Cf. A212635 (corresponding sequence for wheel graphs). - Eric W. Weisstein, Apr 06 2017

p := proc (n) if n = 1 then x elif n = 2 then x^2+2*x elif n = 3 then x^3+3*x^2+3*x else sort(expand(x*(p(n-1)+p(n-2)+p(n-3)))) end if end proc: for n to 15 do seq(coeff(p(n), x, k), k = 1 .. n) end do; # yields sequence in triangular form

CoefficientList[LinearRecurrence[{x, x, x}, {1, 2 + x, 3 + 3 x + x^2}, 10], x] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)

If p(n)=p(n,x) denotes the generating polynomial of row n (called the domination polynomial of the cycle C_n), then p(1)=x, p(2) = 2x + x^2 , p(3) = 3x + 3x^2 + x^3 and p(n) = x*[p(n-1) + p(n-2) + p(n-3)] for n >= 4 (see Theorem 4.5 in the Alikhani & Peng reference in Opuscula Math.).
From Petros Hadjicostas, Jan 26 2019: (Start)
T(n, k) = binomial(n, k) for 1 <= k <= n and n = 1, 2.
T(n, k) = Sum_{j=0..floor(n/3)-1} (-1)^j*binomial(k, j)*(n/(n - 3*j))*binomial(n - 3*j, k) for n - floor(2*n/3) <= k <= n and n >= 3. (This is a special case of a corrected version of formula (2.1) in Charalambides (1991) and equation (14) in Moser and Abramson (1969).)
T(n, k) = 0 for 1 <= k < n - floor(2*n/3) and n >= 4. (Thus, the number of initial zeros in row n is n - floor(2*n/3) - 1.)
G.f. for row n: p(n, x) = -1 + Sum_{j=0..floor(n/4)} (-1)^j*(n/(n-3*j))*binomial(n - 3*j, j)*x^j*(1+x)^(n-4*j). It satisfies the recurrence given above (found in Alikhani and Peng (2010) and Charalambides (1991)).
(End)

A127208 Union of all n-step Lucas sequences, that is, all sequences s(1-n) = s(2-n) = ... = s(-1) = -1, s(0) = n and for k > 0, s(k) = s(k-1) + ... + s(k-n).

Original entry on oeis.org

1, 3, 4, 7, 11, 15, 18, 21, 26, 29, 31, 39, 47, 51, 57, 63, 71, 76, 99, 113, 120, 123, 127, 131, 191, 199, 223, 239, 241, 247, 255, 322, 367, 439, 443, 475, 493, 502, 511, 521, 708, 815, 843, 863, 943, 983, 1003, 1013, 1023, 1364, 1365, 1499, 1695, 1871, 1959

Offset: 1

T. D. Noe, Jan 09 2007

nonn

Noe and Post conjectured that the only positive terms that are common to any two distinct n-step Lucas sequences are the Mersenne numbers (A001348) that begin each sequence and 7 and 11 (in 2- and 3-step) and 5071 (in 3- and 4-step). The intersection of this sequence with the union of all the n-step Fibonacci sequences (A124168) appears to consist of 4, 21, 29, the Mersenne numbers 2^n-1 for all n and the infinite set of Eulerian numbers in A127232.

T. D. Noe, Table of n, a(n) for n=1..1000
Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4

Cf. A227885.

LucasSequence[n_,kMax_] := Module[{a,s,lst={}}, a=Join[Table[ -1,{n-1}],{n}]; While[s=Plus@@a; a=RotateLeft[a]; a[[n]]=s; s<=kMax, AppendTo[lst,s]]; lst]; nn=10; t={}; Do[t=Union[t,LucasSequence[n,2^(nn+1)]], {n,2,nn}]; t

Union(A000032, A001644, A073817, A074048, A074584, A104621, A105754, A105755,...)

A145027 a(n) = a(n-1) + a(n-2) + a(n-3) with a(1) = 2, a(2) = 3, a(3) = 4.

Original entry on oeis.org

2, 3, 4, 9, 16, 29, 54, 99, 182, 335, 616, 1133, 2084, 3833, 7050, 12967, 23850, 43867, 80684, 148401, 272952, 502037, 923390, 1698379, 3123806, 5745575, 10567760, 19437141, 35750476, 65755377, 120942994, 222448847, 409147218

Offset: 1

Vladimir Joseph Stephan Orlovsky, Sep 30 2008

nonn

If the conjectured recurrence in A000382 is correct, then a(n) = A000382(n+2) - A000382(n+1), n>=4. - R. J. Mathar, Jan 30 2011

G. C. Greubel, Table of n, a(n) for n = 1..1000
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Index entries for linear recurrences with constant coefficients, signature (1,1,1).

Cf. A000073, A000213, A001590, A003265, A056816, A081172, A001644.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x*(1+x)*(2-x)/(1-x-x^2-x^3) )); // G. C. Greubel, Apr 22 2019

LinearRecurrence[{1,1,1},{2,3,4},33] (* Ray Chandler, Dec 08 2013 *)

my(x='x+O('x^30)); Vec(x*(1+x)*(2-x)/(1-x-x^2-x^3)) \\ G. C. Greubel, Apr 22 2019

a=(x*(1+x)*(2-x)/(1-x-x^2-x^3)).series(x, 40).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Apr 22 2019

From R. J. Mathar, Jan 30 2011: (Start)
a(n) = -A000073(n-1) + A000073(n) + 2*A000073(n+1).
G.f. x*(1+x)*(2-x)/(1-x-x^2-x^3). (End)

A247505 Generalized Lucas numbers: square array A(n,k) read by antidiagonals, A(n,k)=(-1)^(k+1)k[x^k](-log((1+sum_{j=1..n}(-1)^(j+1)*x^j)^(-1))), (n>=0, k>=0).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 3, 1, 0, 0, 1, 3, 4, 1, 0, 0, 1, 3, 7, 7, 1, 0, 0, 1, 3, 7, 11, 11, 1, 0, 0, 1, 3, 7, 15, 21, 18, 1, 0, 0, 1, 3, 7, 15, 26, 39, 29, 1, 0, 0, 1, 3, 7, 15, 31, 51, 71, 47, 1, 0, 0, 1, 3, 7, 15, 31, 57, 99, 131, 76, 1, 0

Offset: 0

Peter Luschny, Nov 02 2014

			n\k[0][1][2][3] [4] [5] [6]  [7]  [8]  [9]  [10]  [11]  [12]
[0] 0, 0, 0, 0,  0,  0,  0,   0,   0,   0,    0,    0,    0
[1] 0, 1, 1, 1,  1,  1,  1,   1,   1,   1,    1,    1,    1
[2] 0, 1, 3, 4,  7, 11, 18,  29,  47,  76,  123,  199,  322 [A000032]
[3] 0, 1, 3, 7, 11, 21, 39,  71, 131, 241,  443,  815, 1499 [A001644]
[4] 0, 1, 3, 7, 15, 26, 51,  99, 191, 367,  708, 1365, 2631 [A073817]
[5] 0, 1, 3, 7, 15, 31, 57, 113, 223, 439,  863, 1695, 3333 [A074048]
[6] 0, 1, 3, 7, 15, 31, 63, 120, 239, 475,  943, 1871, 3711 [A074584]
[7] 0, 1, 3, 7, 15, 31, 63, 127, 247, 493,  983, 1959, 3903 [A104621]
[8] 0, 1, 3, 7, 15, 31, 63, 127, 255, 502, 1003, 2003, 3999 [A105754]
[.] .  .  .  .   .   .   .    .    .    .     .     .     .
oo] 0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095 [A000225]
'
As a triangular array, starts:
0,
0, 0,
0, 1, 0,
0, 1, 1, 0,
0, 1, 3, 1, 0,
0, 1, 3, 4, 1, 0,
0, 1, 3, 7, 7, 1,  0,
0, 1, 3, 7, 11, 11, 1, 0,
0, 1, 3, 7, 15, 21, 18, 1, 0,
0, 1, 3, 7, 15, 26, 39, 29, 1, 0,

Cf. A247506, A000225, A000032, A001644, A073817, A074048, A074584, A104621, A105754.

Cf. A125127.

A := proc(n, k) f := -log((1+add((-1)^(j+1)*x^j, j=1..n))^(-1));
(-1)^(k+1)*k*coeff(series(f,x,k+2),x,k) end:
seq(print(seq(A(n,k), k=0..12)), n=0..8);

A[n_, k_] := Module[{f, x}, f = -Log[(1+Sum[(-1)^(j+1) x^j, {j, 1, n}] )^(-1)]; (-1)^(k+1) k SeriesCoefficient[f, {x, 0, k}]];
Table[A[n-k, k], {n, 0, 12}, {k, 0, n}] (* Jean-François Alcover, Jun 28 2019, from Maple *)

A323949 Number of set partitions of {1, ..., n} with no block containing three distinct cyclically successive vertices.

Original entry on oeis.org

1, 1, 2, 4, 10, 36, 145, 631, 3015, 15563, 86144, 508311, 3180930, 21018999, 146111543, 1065040886, 8117566366, 64531949885, 533880211566, 4587373155544, 40865048111424, 376788283806743, 3590485953393739, 35312436594162173, 357995171351223109, 3736806713651177702

Offset: 0

Gus Wiseman, Feb 10 2019

nonn

Cyclically successive means 1 is a successor of n.

			The a(1) = 1 through a(4) = 10 set partitions:
  {{1}}  {{1,2}}    {{1},{2,3}}    {{1,2},{3,4}}
         {{1},{2}}  {{1,2},{3}}    {{1,3},{2,4}}
                    {{1,3},{2}}    {{1,4},{2,3}}
                    {{1},{2},{3}}  {{1},{2},{3,4}}
                                   {{1},{2,3},{4}}
                                   {{1,2},{3},{4}}
                                   {{1},{2,4},{3}}
                                   {{1,3},{2},{4}}
                                   {{1,4},{2},{3}}
                                   {{1},{2},{3},{4}}

Alois P. Heinz, Table of n, a(n) for n = 0..250

Cf. A000085, A000110, A000126, A000296, A000325, A001610, A001644, A078012, A169985.

Cf. A306351, A306357, A323950, A323951, A323953, A323954, A323955, A323956.

spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@spsu[Select[foo,Complement[#,Complement[set,s]]=={}&],Complement[set,s]]]/@Cases[foo,{i,_}];
Table[Length[spsu[Select[Subsets[Range[n]],Select[Partition[Range[n],3,1,1],Function[ed,UnsameQ@@ed&&Complement[ed,#]=={}]]=={}&],Range[n]]],{n,8}]

a(12)-a(25) from Alois P. Heinz, Feb 10 2019

A323955 Regular triangle read by rows where T(n, k) is the number of set partitions of {1, ..., n} with no block containing k cyclically successive vertices, n >= 1, 2 <= k <= n + 1.

Original entry on oeis.org

1, 1, 2, 1, 4, 5, 4, 10, 14, 15, 11, 36, 46, 51, 52, 41, 145, 184, 196, 202, 203, 162, 631, 806, 855, 869, 876, 877, 715, 3015, 3847, 4059, 4115, 4131, 4139, 4140, 3425, 15563, 19805, 20813, 21056, 21119, 21137, 21146, 21147, 17722, 86144, 109339, 114469

Offset: 1

Gus Wiseman, Feb 10 2019

Cyclically successive means 1 is a successor of n.

			Triangle begins:
    1
    1    2
    1    4    5
    4   10   14   15
   11   36   46   51   52
   41  145  184  196  202  203
  162  631  806  855  869  876  877
  715 3015 3847 4059 4115 4131 4139 4140
Row 4 counts the following partitions:
  {{13}{24}}      {{12}{34}}      {{1}{234}}      {{1234}}
  {{1}{24}{3}}    {{13}{24}}      {{12}{34}}      {{1}{234}}
  {{13}{2}{4}}    {{14}{23}}      {{123}{4}}      {{12}{34}}
  {{1}{2}{3}{4}}  {{1}{2}{34}}    {{124}{3}}      {{123}{4}}
                  {{1}{23}{4}}    {{13}{24}}      {{124}{3}}
                  {{12}{3}{4}}    {{134}{2}}      {{13}{24}}
                  {{1}{24}{3}}    {{14}{23}}      {{134}{2}}
                  {{13}{2}{4}}    {{1}{2}{34}}    {{14}{23}}
                  {{14}{2}{3}}    {{1}{23}{4}}    {{1}{2}{34}}
                  {{1}{2}{3}{4}}  {{12}{3}{4}}    {{1}{23}{4}}
                                  {{1}{24}{3}}    {{12}{3}{4}}
                                  {{13}{2}{4}}    {{1}{24}{3}}
                                  {{14}{2}{3}}    {{13}{2}{4}}
                                  {{1}{2}{3}{4}}  {{14}{2}{3}}
                                                  {{1}{2}{3}{4}}

First column (k = 2) is A000296. Second column (k = 3) is A323949. Rightmost terms are A000110. Second to rightmost terms are A058692.

Cf. A000126, A000325, A001610, A001644, A169985, A306351, A306357, A323950, A323954.

spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@spsu[Select[foo,Complement[#,Complement[set,s]]=={}&],Complement[set,s]]]/@Cases[foo,{i,_}];
Table[Length[spsu[Select[Subsets[Range[n]],Select[Partition[Range[n],k,1,1],Function[ed,UnsameQ@@ed&&Complement[ed,#]=={}]]=={}&],Range[n]]],{n,7},{k,2,n+1}]

A323956 Triangle read by rows: T(n, k) = 1 + n * (n - k) for 1 <= k <= n.

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 13, 9, 5, 1, 21, 16, 11, 6, 1, 31, 25, 19, 13, 7, 1, 43, 36, 29, 22, 15, 8, 1, 57, 49, 41, 33, 25, 17, 9, 1, 73, 64, 55, 46, 37, 28, 19, 10, 1, 91, 81, 71, 61, 51, 41, 31, 21, 11, 1, 111, 100, 89, 78, 67, 56, 45, 34, 23, 12, 1

Offset: 1

Gus Wiseman, Feb 10 2019

			Triangle begins:
  n\k:   1   2   3   4   5   6   7   8   9  10  11  12
  ====================================================
    1:   1
    2:   3   1
    3:   7   4   1
    4:  13   9   5   1
    5:  21  16  11   6   1
    6:  31  25  19  13   7   1
    7:  43  36  29  22  15   8   1
    8:  57  49  41  33  25  17   9   1
    9:  73  64  55  46  37  28  19  10   1
   10:  91  81  71  61  51  41  31  21  11   1
   11: 111 100  89  78  67  56  45  34  23  12   1
   12: 133 121 109  97  85  73  61  49  37  25  13   1
  etc.

G. C. Greubel, Rows n = 1..100 of triangle, flattened

First column is A002061. Second column is A000290. Third column is A028387.

Cf. A000126, A001610, A001644, A006000, A169985, A306351, A323952, A323955.

[[1+n*(n-k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, Apr 22 2019

Mathematica
```
Table[1+n*(n-k),{n,12},{k,n}]//Flatten
```

{T(n,k) = 1+n*(n-k)}; \\ G. C. Greubel, Apr 22 2019

[[1+n*(n-k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Apr 22 2019

From Werner Schulte, Feb 12 2019: (Start)
G.f.: Sum_{n>0,k=1..n} T(n,k)*x^k*t^n = x*t*((1-t+2*t^2)*(1-x*t) + (1-t)*t)/((1-t)^3*(1-x*t)^2).
Row sums: Sum_{k=1..n} T(n,k) = A006000(n-1) for n > 0.
Recurrence: T(n,k) = T(n,k-1) - n for 1 < k <= n with initial values T(n,1) = n^2-n+1 for n > 0.
Recurrence: T(n,k) = T(n-1,k) + 2*n-k-1 for 1 <= k < n with initial values T(n,n) = 1 for n > 0.
(End)

A001643 A Fielder sequence.

Original entry on oeis.org

1, 3, 4, 11, 21, 42, 71, 131, 238, 443, 815, 1502, 2757, 5071, 9324, 17155, 31553, 58038, 106743, 196331, 361106, 664183, 1221623, 2246918, 4132721, 7601259, 13980892, 25714875, 47297029, 86992802, 160004703, 294294531, 541292030, 995591267, 1831177831

Offset: 1

N. J. A. Sloane

nonn

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..1000
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Index entries for linear recurrences with constant coefficients, signature (1, 1, 0, 1, 1, 1).

Cf. A001644.

I:=[1,3,4,11,21,42]; [n le 6 select I[n] else Self(n-1) + Self(n-2) + Self(n-4) + Self(n-5) + Self(n-6): n in [1..30]]; // G. C. Greubel, Jan 09 2018

A001643:=-(1+2*z+4*z**3+5*z**4+6*z**5)/(z+1)/(z**3+z**2+z-1)/(z**2-z+1); [Conjectured by Simon Plouffe in his 1992 dissertation.]

LinearRecurrence[{1, 1, 0, 1, 1, 1}, {1, 3, 4, 11, 21, 42}, 50] (* T. D. Noe, Aug 09 2012 *)

a(n)=if(n<0,0,polcoeff(x*(1+2*x+4*x^3+5*x^4+6*x^5)/(1-x-x^2-x^4-x^5-x^6)+x*O(x^n),n))

G.f.: x*(1+2*x+4*x^3+5*x^4+6*x^5)/(1-x-x^2-x^4-x^5-x^6).
From Greg Dresden, Jul 07 2021: (Start)
a(3*n+1) = A001644(3*n+1).
a(3*n+2) = A001644(3*n+2).
a(3*n+3) = A001644(3*n+3) - 3*(-1)^n. (End)

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