A126473 -id:A126473 - cp's OEIS frontend

A054458 Convolution triangle based on A001333(n), n >= 1.

1, 3, 1, 7, 6, 1, 17, 23, 9, 1, 41, 76, 48, 12, 1, 99, 233, 204, 82, 15, 1, 239, 682, 765, 428, 125, 18, 1, 577, 1935, 2649, 1907, 775, 177, 21, 1, 1393, 5368, 8680, 7656, 4010, 1272, 238, 24, 1, 3363, 14641, 27312, 28548, 18358, 7506, 1946, 308, 27, 1

Offset: 0

Wolfdieter Lang, Apr 26 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The G.f. for the row polynomials p(n,x) (increasing powers of x) is LPell(z)/(1-x*z*LPell(z)) with LPell(z) given in 'Formula'.
Column sequences are A001333(n+1), A054459(n), A054460(n) for m=0..2.
Mirror image of triangle in A209696. - Philippe Deléham, Mar 24 2012
Subtriangle of the triangle given by (0, 3, -2/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 25 2012
Riordan array ((1+x)/(1-2*x-x^2), (x+x^2)/(1-2*x-x^2)). - Philippe Deléham, Mar 25 2012

			Fourth row polynomial (n=3): p(3,x)= 17+23*x+9*x^2+x^3.
Triangle begins :
  1
  3, 1
  7, 6, 1
  17, 23, 9, 1
  41, 76, 48, 12, 1
  99, 233, 204, 82, 15, 1
  239, 682, 765, 428, 125, 18, 1. - _Philippe Deléham_, Mar 25 2012
(0, 3, -2/3, -1/3, 0, 0, 0, ...) DELTA (1, 0, 0, 0, ...) begins :
  1
  0, 1
  0, 3, 1
  0, 7, 6, 1
  0, 17, 23, 9, 1
  0, 41, 76, 48, 12, 1
  0, 99, 233, 204, 82, 15, 1
  0, 239, 682, 765, 428, 125, 15, 1. - _Philippe Deléham_, Mar 25 2012

Milan Janjić, Words and Linear Recurrences, J. Int. Seq. 21 (2018), #18.1.4.

Cf. A002203(n+1)/2. Row sums: A055099(n).
Cf. A001333, A054459, A054460.
Cf. A040000, A001333, A055099, A126473, A126501, A126528.

a(n, m) := ((n-m+1)*a(n, m-1) + (2n-m)*a(n-1, m-1) + (n-1)*a(n-2, m-1))/(4*m), n >= m >= 1; a(n, 0)= A001333(n+1); a(n, m) := 0 if n

G.f. for column m: LPell(x)*(x*LPell(x))^m, m >= 0, with LPell(x)= (1+x)/(1-2*x-x^2) = g.f. for A001333(n+1).

G.f.: (1+x)/(1-2*x-y*x-x^2-y*x^2). - Philippe Deléham, Mar 25 2012

T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1), T(0,0) = T(1,1) = 1, T(1,0) = 3 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 25 2012

Sum_{k=0..n} T(n,k)*x^k = A040000(n), A001333(n+1), A055099(n), A126473(n), A126501(n), A126528(n) for x = -1, 0, 1, 2, 3, 4 respectively. - Philippe Deléham, Mar 25 2012

A212835 T(n,k)=Number of 0..k arrays of length n+1 with 0 never adjacent to k.

Original entry on oeis.org

2, 7, 2, 14, 17, 2, 23, 50, 41, 2, 34, 107, 178, 99, 2, 47, 194, 497, 634, 239, 2, 62, 317, 1106, 2309, 2258, 577, 2, 79, 482, 2137, 6306, 10727, 8042, 1393, 2, 98, 695, 3746, 14407, 35954, 49835, 28642, 3363, 2, 119, 962, 6113, 29114, 97127, 204994, 231521

Offset: 1

R. H. Hardin May 28 2012

Table starts
.2.....7......14.......23........34.........47.........62..........79
.2....17......50......107.......194........317........482.........695
.2....41.....178......497......1106.......2137.......3746........6113
.2....99.....634.....2309......6306......14407......29114.......53769
.2...239....2258....10727.....35954......97127.....226274......472943
.2...577....8042....49835....204994.....654797....1758602.....4159927
.2..1393...28642...231521...1168786....4414417...13667858....36590017
.2..3363..102010..1075589...6663906...29760487..106226618...321839625
.2..8119..363314..4996919..37994674..200635007..825593474..2830847119
.2.19601.1293962.23214443.216628994.1352612477.6416514026.24899654327

			Some solutions for n=5 k=4
..1....4....1....1....1....3....2....1....1....4....3....1....0....2....2....3
..1....3....0....4....2....4....1....1....2....2....3....4....1....3....3....1
..1....3....3....1....3....3....2....2....2....4....3....3....1....1....0....3
..1....3....0....4....2....3....3....3....4....2....2....0....4....4....2....1
..1....4....0....3....2....2....3....3....4....4....0....1....4....3....4....2
..1....2....0....4....3....1....0....2....4....2....3....2....1....4....3....4

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 2 is A001333(n+2)
Column 3 is A055099(n+1)
Column 4 is A126473(n+1)
Column 5 is A126501(n+1)
Column 6 is A126528(n+1)
Row 1 is A008865(n+1)

Empirical for column k: a(n) = k*a(n-1) +(k-1)*a(n-2)
Empirical for rows:
n=1: a(k) = k^2 + 2*k - 1
n=2: a(k) = k^3 + 3*k^2 - k - 1
n=3: a(k) = k^4 + 4*k^3 - 4*k + 1
n=4: a(k) = k^5 + 5*k^4 + 2*k^3 - 8*k^2 + k + 1
n=5: a(k) = k^6 + 6*k^5 + 5*k^4 - 12*k^3 - 3*k^2 + 6*k - 1
n=6: a(k) = k^7 + 7*k^6 + 9*k^5 - 15*k^4 - 13*k^3 + 15*k^2 - k - 1
n=7: a(k) = k^8 + 8*k^7 + 14*k^6 - 16*k^5 - 30*k^4 + 24*k^3 + 8*k^2 - 8*k + 1

A108851 a(n) = 4a(n-1) + 3a(n-2), a(0) = 1, a(1) = 2.

Original entry on oeis.org

1, 2, 11, 50, 233, 1082, 5027, 23354, 108497, 504050, 2341691, 10878914, 50540729, 234799658, 1090820819, 5067682250, 23543191457, 109375812578, 508132824683, 2360658736466, 10967033419913, 50950109889050

Offset: 0

Philippe Deléham, Jul 11 2005

Binomial transform of A083098, second binomial transform of (1, 0, 7, 0, 49, 0, 243, 0, ...).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,3).

Cf. A080042. - Zerinvary Lajos, May 14 2009
Cf. A015530, A083098.
Appears in A179596, A179597 and A126473. - Johannes W. Meijer, Aug 01 2010

[Floor(((2 + Sqrt(7))^n + (2 - Sqrt(7))^n) / 2): n in [0..30]]; // Vincenzo Librandi, Jul 18 2011

LinearRecurrence[{4,3},{1,2},30] (* Harvey P. Dale, Jan 02 2022 *)

a(n)=round(((2+sqrt(7))^n+(2-sqrt(7))^n)/2) \\ Charles R Greathouse IV, Dec 06 2011

[lucas_number2(n,4,-3)/2 for n in range(0, 22)] # Zerinvary Lajos, May 14 2009

a(n) = ((2 + sqrt(7))^n + (2 - sqrt(7))^n) / 2.
G.f.: (1 - 2*x) / (1 - 4*x - 3*x^2).
E.g.f.: exp(2*x)*cosh(sqrt(7)*x).
a(n+1)/a(n) converges to 2 + sqrt(7) = 4.645751311064...
Limit_{k->oo} a(n+k)/a(k) = A108851(n) + A015530(n)*sqrt(7); also lim_{n->oo} A108851(n)/A015530(n) = sqrt(7). - Johannes W. Meijer, Aug 01 2010
a(n) = Sum_{k=0..n} A201730(n,k)*6^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(7*k-4)/(x*(7*k+3) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 27 2013
a(n) = (2 + sqrt(7))^n - A015530(n)*sqrt(7). - Robert FERREOL, Aug 04 2025

A287839 Number of words of length n over the alphabet {0,1,...,10} such that no two consecutive terms have distance 9.

Original entry on oeis.org

1, 11, 117, 1247, 13289, 141619, 1509213, 16083463, 171399121, 1826575451, 19465548357, 207441511727, 2210673955769, 23558830139779, 251063019088173, 2675542001860183, 28512861152219041, 303857405535211691, 3238164083417650197, 34508642672922983807

Offset: 0

David Nacin, Jun 07 2017

In general, the number of sequences on {0,1,...,10} such that no two consecutive terms have distance 6+k for k in {0,1,2,3,4} has generating function (-1 - x)/(-1 + 10*x + (2*k+1)*x^2).

Colin Barker, Table of n, a(n) for n = 0..900
Index entries for linear recurrences with constant coefficients, signature (10,7).

Cf. A040000, A003945, A083318, A078057, A003946, A126358, A003946, A055099, A003947, A015448, A126473. A287804-A287819. A287825-A287839.

a:=proc(n) option remember; if n=0 then 1 elif n=1 then 11 elif n=2 then 117 else 10*a(n-1)+7*a(n-2); fi; end: seq(a(n), n=0..30); # Wesley Ivan Hurt, Nov 25 2017

LinearRecurrence[{10, 7}, {1, 11, 117}, 20]

Vec((1 + x) / (1 - 10*x - 7*x^2) + O(x^30)) \\ Colin Barker, Nov 25 2017

def a(n):
 if n in [0,1,2]:
  return [1, 11, 117][n]
 return 10*a(n-1) + 7*a(n-2)

For n>2, a(n) = 10*a(n-1) + 7*a(n-2), a(0)=1, a(1)=11, a(2)=117.
G.f.: (-1 - x)/(-1 + 10 x + 7 x^2).
a(n) = (((5-4*sqrt(2))^n*(-3+2*sqrt(2)) + (3+2*sqrt(2))*(5+4*sqrt(2))^n)) / (4*sqrt(2)). - Colin Barker, Nov 25 2017

A254600 Numbers of words on alphabet {0,1,...,10} with no subwords ii, for i from {0,1}.

Original entry on oeis.org

1, 11, 119, 1289, 13961, 151211, 1637759, 17738489, 192124721, 2080893611, 22538058599, 244108628489, 2643928812281, 28636265779211, 310158017102639, 3359306563039289, 36384487784316641, 394078636910520011, 4268246759164049879, 46229175323835178889

Offset: 0

Milan Janjic, Feb 02 2015

a(n) equals the number of sequences over the alphabet {0,1,...,9,10} such that no two consecutive terms have distance 10. - David Nacin, Jun 02 2017

Colin Barker, Table of n, a(n) for n = 0..950
Index entries for linear recurrences with constant coefficients, signature (10,9).

Cf. A015591, A055099, A126473, A126501, A126528.

[n le 1 select 11^n else 10*Self(n)+9*Self(n-1): n in [0..20]]; // Bruno Berselli, Feb 03 2015

RecurrenceTable[{a[0]==1, a[1]==11, a[n]== 10a[n-1] +9a[n-2]}, a[n], {n, 0, 25}]
Table[(-3 I)^(n-1)*(ChebyshevU[n-1, 5*I/3] - 3*I*ChebyshevU[n, 5*I/3]), {n,0,25}] (* G. C. Greubel, Feb 13 2021 *)

Vec((x+1) / (1-10*x-9*x^2) + O(x^30)) \\ Colin Barker, Jan 21 2017

[(-3*i)^(n-1)*( chebyshev_U(n-1, 5*i/3) -3*i*chebyshev_U(n, 5*i/3) ) for n in (0..30)] # G. C. Greubel, Feb 13 2021

G.f.: (1+x)/(1-10*x-9*x^2).
a(n) = 10*a(n-1) + 9*a(n-2) with n>1, a(0) = 1, a(1) = 11.
a(n) = ((5-sqrt(34))^n*(-6+sqrt(34)) + (5+sqrt(34))^n*(6+sqrt(34))) / (2*sqrt(34)). - Colin Barker, Jan 21 2017
a(n) = (-3*i)^(n-1) * (ChebyshevU(n-1, 5*i/3) - 3*i*ChebyshevU(n, 5*i/3)). - G. C. Greubel, Feb 13 2021

A287831 Number of sequences over the alphabet {0,1,...,9} such that no two consecutive terms have distance 8.

Original entry on oeis.org

1, 10, 96, 924, 8892, 85572, 823500, 7924932, 76265388, 733938084, 7063035084, 67970944260, 654116708844, 6294876045156, 60578584659468, 582976518206148, 5610260171812140, 53990200655546148, 519573366930788172, 5000101506310370436, 48118353758378062956

Offset: 0

David Nacin, Jun 02 2017

In general, the number of sequences over the alphabet {0,1,...,9} such that no two consecutive terms have distance 5+k for k in {0,1,2,3,4} is given by a(n) = 9*a(n-1) + 2*k*a(n-2), a(0)=1, a(1)=10.

Index entries for linear recurrences with constant coefficients, signature (9,6).

Cf. A040000, A003945, A083318, A078057, A003946, A126358, A003946, A055099, A003947, A015448, A126473. A287804-A287819. A287825-A287831.

Mathematica
```
LinearRecurrence[{9, 6}, {1, 10}, 30]
```

def a(n):
 if n in [0, 1]:
  return [1, 10][n]
 return 9*a(n-1)+6*a(n-2)

a(n) = 9*a(n-1) + 6*a(n-2), a(0)=1, a(1)=10.
G.f.: (-1 - x)/(-1 + 9*x + 6*x^2).
a(n) = ((1 - 11/sqrt(105))/2)*((9 - sqrt(105))/2)^n + ((1 + 11/sqrt(105))/2)*((9 + sqrt(105))/2)^n.

A254657 Numbers of words on alphabet {0,1,...,8} with no subwords ii, where i is from {0,1,2}.

Original entry on oeis.org

1, 9, 78, 678, 5892, 51204, 444984, 3867096, 33606672, 292055952, 2538087648, 22057036896, 191684821056, 1665820789824, 14476675244928, 125808326698368, 1093326665056512, 9501463280642304, 82571666235477504, 717582109567673856, 6236086873954255872

Offset: 0

Milan Janjic, Feb 04 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,6).

Cf. A055099, A126473, A126501, A126528, A190560, A254598, A254602.

[n le 1 select 9^n else 8*Self(n)+6*Self(n-1): n in [0..20]];

RecurrenceTable[{a[0] == 1, a[1] == 9, a[n] == 8 a[n - 1] + 6 a[n - 2]}, a[n], {n, 0, 20}]

Vec((1+x)/(1-8*x-6*x^2) + O(x^30)) \\ Colin Barker, Nov 16 2016

G.f.: (1 + x)/(1 - 8*x - 6*x^2).
a(n) = 8*a(n-1) + 6*a(n-2) with n>1, a(0) = 1, a(1) = 9.
a(n) = (((4-sqrt(22))^n*(-5+sqrt(22)) + (4+sqrt(22))^n*(5+sqrt(22))))/(2*sqrt(22)). - Colin Barker, Nov 16 2016

A254601 Numbers of n-length words on alphabet {0,1,...,6} with no subwords ii, where i is from {0,1,2}.

Original entry on oeis.org

1, 7, 46, 304, 2008, 13264, 87616, 578752, 3822976, 25252864, 166809088, 1101865984, 7278432256, 48078057472, 317582073856, 2097804673024, 13857156333568, 91534156693504, 604633565495296, 3993938019745792, 26382162380455936, 174268726361718784

Offset: 0

Milan Janjic, Feb 02 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,4).

Cf. A055099, A126473, A126501, A126528, A135032, A190976 (shifted bin. trans).

[n le 1 select 7^n else 6*Self(n)+4*Self(n-1): n in [0..25]]; // Bruno Berselli, Feb 03 2015

RecurrenceTable[{a[0] == 1, a[1] == 7, a[n] == 6 a[n - 1] + 4 a[n - 2]}, a[n], {n, 0, 25}]
LinearRecurrence[{6,4},{1,7},30] (* Harvey P. Dale, Oct 10 2017 *)

Vec((1 + x)/(1 - 6*x - 4*x^2) + O(x^30)) \\ Colin Barker, Jan 22 2017

G.f.: (1 + x)/(1 - 6*x - 4*x^2).
a(n) =  6*a(n-1) + 4*a(n-2) with n>1, a(0) = 1, a(1) = 7.
a(n) = ((3-r)^n*(-4+r) + (3+r)^n*(4+r)) / (2*r), where r=sqrt(13). - Colin Barker, Jan 22 2017
a(n) = A135032(n-1)+A135032(n). - R. J. Mathar, Apr 07 2022

A254658 Numbers of words on alphabet {0,1,...,7} with no subwords ii, where i is from {0,1,2,3}.

Original entry on oeis.org

1, 8, 60, 452, 3404, 25636, 193068, 1454020, 10950412, 82468964, 621084396, 4677466628, 35226603980, 265296094372, 1997979076524, 15047037913156, 113321181698188, 853436423539940, 6427339691572332, 48405123535166084, 364545223512451916, 2745437058727827748

Offset: 0

Milan Janjic, Feb 04 2015

a(n) equals the number of octonary sequences of length n such that no two consecutive terms differ by 6. - David Nacin, May 31 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,4).

Cf. A015561, A055099, A126473, A126501, A126528, A254598, A254602.

[n le 1 select 8^n else 7*Self(n)+4*Self(n-1): n in [0..20]];

RecurrenceTable[{a[0] == 1, a[1] == 8, a[n] == 7 a[n - 1] + 4 a[n - 2]}, a[n], {n, 0, 20}]
LinearRecurrence[{7,4},{1,8},30] (* Harvey P. Dale, Jan 21 2023 *)

Vec((1 + x) / (1 - 7*x -4*x^2) + O(x^30)) \\ Colin Barker, Jan 21 2017

G.f.: (1 + x)/(1 - 7*x -4*x^2).
a(n) = 7*a(n-1) + 4*a(n-2) with n>1, a(0) = 1, a(1) = 8.
a(n) = (2^(-1-n)*((7-sqrt(65))^n*(-9+sqrt(65)) + (7+sqrt(65))^n*(9+sqrt(65)))) / sqrt(65). - Colin Barker, Jan 21 2017

A254660 Numbers of words on alphabet {0,1,...,6} with no subwords ii, where i is from {0,1,...,4}.

Original entry on oeis.org

1, 7, 44, 278, 1756, 11092, 70064, 442568, 2795536, 17658352, 111541184, 704563808, 4450465216, 28111918912, 177572443904, 1121658501248, 7085095895296, 44753892374272, 282693546036224, 1785669060965888, 11279401457867776, 71247746869138432

Offset: 0

Milan Janjic, Feb 04 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,2).

Cf. A135030, A126473, A126501, A126528, A254598, A254602.

RecurrenceTable[{a[0] == 1, a[1] == 7, a[n] == 6 a[n - 1] + 2 a[n - 2]}, a[n], {n, 0, 20}]
LinearRecurrence[{6,2},{1,7},30] (* Harvey P. Dale, Sep 11 2024 *)

Vec((1 + x) / (1 - 6*x -2*x^2) + O(x^30)) \\ Colin Barker, Jan 21 2017

G.f.: (1 + x)/(1 - 6*x -2*x^2).
a(n) = 6*a(n-1) + 2*a(n-2) with n>1, a(0) = 1, a(1) = 7.
a(n) = ((3-sqrt(11))^n*(-4+sqrt(11)) + (3+sqrt(11))^n*(4+sqrt(11))) / (2*sqrt(11)). - Colin Barker, Jan 21 2017

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cp's OEIS Frontend

A054458 Convolution triangle based on A001333(n), n >= 1.

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A212835 T(n,k)=Number of 0..k arrays of length n+1 with 0 never adjacent to k.

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A108851 a(n) = 4*a(n-1) + 3*a(n-2), a(0) = 1, a(1) = 2.

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A287839 Number of words of length n over the alphabet {0,1,...,10} such that no two consecutive terms have distance 9.

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A254600 Numbers of words on alphabet {0,1,...,10} with no subwords ii, for i from {0,1}.

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A287831 Number of sequences over the alphabet {0,1,...,9} such that no two consecutive terms have distance 8.

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A254657 Numbers of words on alphabet {0,1,...,8} with no subwords ii, where i is from {0,1,2}.

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A108851 a(n) = 4a(n-1) + 3a(n-2), a(0) = 1, a(1) = 2.