A051633 -id:A051633 - cp's OEIS frontend

A118654 Square array T(n,k) read by antidiagonals: T(n,k) = 2^n*Fibonacci(k) - Fibonacci(k-2).

1, 1, 0, 1, 1, 1, 1, 3, 2, 1, 1, 7, 4, 3, 2, 1, 15, 8, 7, 5, 3, 1, 31, 16, 15, 11, 8, 5, 1, 63, 32, 31, 23, 18, 13, 8, 1, 127, 64, 63, 47, 38, 29, 21, 13, 1, 255, 128, 127, 95, 78, 61, 47, 34, 21, 1, 511, 256, 255, 191, 158, 125, 99, 76, 55, 34

Offset: 0

Ross La Haye, May 17 2006

Inverse binomial transform (by columns) of A090888.

			T(2,3) = 7 because 2^2(Fibonacci(3)) - Fibonacci(3-2) = 4*2 - 1 = 7.
{1};
{1,  0};
{1,  1,  1};
{1,  3,  2,  1};
{1,  7,  4,  3,  2};
{1, 15,  8,  7,  5,  3};
{1, 31, 16, 15, 11,  8,  5};
{1, 63, 32, 31, 23, 18, 13,  8};

Rows: T(0,k) = A000045(k-1), for k > 0; T(1,k) = A000045(k+1); T(2,k) = A000032(k+1); T(3,k) = A022097(k); T(4,k) = A022105(k); T(5,k) = A022401(k).

Columns: T(n,1) = A000225(n); T(n,2) = A000079(n); T(n,3) = A000225(n+1); T(n,4) = A055010(n+1); T(n,5) = A051633(n); a(T,6) = A036563(n+3).

T(n,k) = 2^n*Fibonacci(k) - Fibonacci(k-2).
T(n,k) = (2^n-2)*Fibonacci(k) + Fibonacci(k+1).
T(n,0) = 1; T(n,1) = 2^n - 1; T(n,k) = T(n,k-1) + T(n,k-2), for k > 1.
T(0,k) = Fibonacci(k-1); T(1,k) = Fibonacci(k+1); T(n,k) = 3T(n-1,k) - 2T(n-2,k), for n > 1.
T(n,k) = 2T(n-1,k) + Fibonacci(k-2), for n > 0.
T(n,k) = A109754(2^n-2, k+1) = A101220(2^n-2, 0, k+1), for n > 0.
O.g.f. (by rows) = (1+(-2+2^n)x)/(1-x-x^2).
Sum_{k=0..n} T(n-k,k) = A119587(n+1). - Ross La Haye, May 31 2006

A131051 Row sums of triangle A133805.

Original entry on oeis.org

1, 3, 8, 18, 38, 78, 158, 318, 638, 1278, 2558, 5118, 10238, 20478, 40958, 81918, 163838, 327678, 655358, 1310718, 2621438, 5242878, 10485758, 20971518, 41943038, 83886078, 167772158, 335544318, 671088638, 1342177278, 2684354558

Offset: 1

Gary W. Adamson, Sep 23 2007

Last digit of a(n) is 8 for n > 2. - Jon Perry, Nov 19 2012

			a(4) = 18 = sum of row 4 terms of triangle A133805: (7 + 6 + 4 + 1).
a(4) = 18 = (1, 3, 3, 1), dot (1, 2, 3, 2) = (1 + 6 + 9 + 2).

Essentially a duplicate of A051633.
Cf. A133805.
Cf. A083329, A053209.

a:=[1]; for n in [2..31] do Append(~a,2*n-2+&+[a[i]:i in [1..n-1]]); end for; a; // Marius A. Burtea, Oct 15 2019

R:=PowerSeriesRing(Integers(), 31); Coefficients(R!( (1+x^2)/((1-x)*(1-2*x)))); // Marius A. Burtea, Oct 15 2019

Binomial transform of [1, 2, 3, 2, 3, 2, 3, ...].
O.g.f.: (1+x^2)/((1-x)(1-2*x)). a(n)=A051633(n-2). - R. J. Mathar, Jun 13 2008
a(n) = 5*2^(n-2)-2, n>1. - Gary Detlefs, Jun 22 2010
a(n) = 2(n-1) + Sum_{i=1..n-1} a(i). - Jon Perry, Nov 19 2012

More terms from R. J. Mathar, Jun 13 2008

A123208 Start with 1, then alternately add 2 or double.

Original entry on oeis.org

1, 3, 6, 8, 16, 18, 36, 38, 76, 78, 156, 158, 316, 318, 636, 638, 1276, 1278, 2556, 2558, 5116, 5118, 10236, 10238, 20476, 20478, 40956, 40958, 81916, 81918, 163836, 163838, 327676, 327678, 655356, 655358, 1310716, 1310718, 2621436, 2621438, 5242876, 5242878

Offset: 0

Philippe Deléham, Oct 04 2006

			1, 1+2=3, 3*2=6, 6+2=8, 8*2=16, ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-2).

Cf. A048487, A051633.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x+3*x^2-x^3)/((1-x^2)*(1-2*x^2)))); // Vincenzo Librandi, Jun 25 2013

a:=proc(n) if n mod 2 = 0 then 5*2^(n/2)-4 else 5*2^((n-1)/2)-2 fi end: seq(a(n),n=0..45); # Emeric Deutsch, Oct 10 2006

nxt[{a_,b_}]:={b+2,2(b+2)}; Rest[Flatten[NestList[nxt,{1,1},20]]] (* or *) LinearRecurrence[{0,3,0,-2},{1,3,6,8},40] (* Harvey P. Dale, Oct 10 2012 *)
CoefficientList[Series[(1 + 3 x + 3 x^2 - x^3) / ((1 - x) (1 + x) (1 - 2 x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Jun 25 2013 *)

a(2n) = 5*2^n - 4; a(2n+1) = 5*2^n - 2 (n >= 0). - Emeric Deutsch, Oct 10 2006
From Colin Barker, Sep 10 2012: (Start)
a(n) = 3*a(n-2) - 2*a(n-4).
G.f.: (1+3*x+3*x^2-x^3)/((1-x)*(1+x)*(1-2*x^2)). (End)
a(2n) = A048487(n); a(2n+1) = A051633(n). - Philippe Deléham, Apr 15 2013
E.g.f.: 5*cosh(sqrt(2)*x) - 4*cosh(x) + 5*sinh(sqrt(2)*x)/sqrt(2) - 2*sinh(x). - Stefano Spezia, Oct 03 2023

More terms from Emeric Deutsch, Oct 10 2006

A028263 Elements in 3-Pascal triangle A028262 (by row) that are not 1.

Original entry on oeis.org

3, 4, 4, 5, 8, 5, 6, 13, 13, 6, 7, 19, 26, 19, 7, 8, 26, 45, 45, 26, 8, 9, 34, 71, 90, 71, 34, 9, 10, 43, 105, 161, 161, 105, 43, 10, 11, 53, 148, 266, 322, 266, 148, 53, 11, 12, 64, 201, 414, 588, 588, 414, 201, 64, 12, 13, 76, 265, 615, 1002, 1176, 1002, 615, 265, 76, 13

Offset: 0

Mohammad K. Azarian

Rows of triangle formed using Pascal's rule except begin and end n-th row with n+3.

			The triangle T(n,k) begins
n\k  0  1   2   3    4    5    6   7   8  9 10 ...
0:   3
1:   4  4
2:   5  8   5
3:   6 13  13   6
4:   7 19  26  19    7
5:   8 26  45  45   26    8
6:   9 34  71  90   71   34    9
7:  10 43 105 161  161  105   43  10
8:  11 53 148 266  322  266  148  53  11
9:  12 64 201 414  588  588  414 201  64 12
10: 13 76 265 615 1002 1176 1002 615 265 76 13
... Reformatted. - _Wolfdieter Lang_, Jun 28 2015

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Row sums give A051633(n).

a028263 n k = a028263_tabl !! n !! k
a028263_row n = a028263_tabl !! n
a028263_tabl = zipWith (zipWith (+)) a007318_tabl a014410_tabl
-- Reinhard Zumkeller, Mar 12 2012

T(n,k) = A007318(n,k) + A014410(n+2,k+1). [Reinhard Zumkeller, Mar 12 2012]

More terms from James Sellers

A134062 Row sums of triangle A134061.

Original entry on oeis.org

1, 8, 18, 38, 78, 158, 318, 638, 1278, 2558, 5118, 10238, 20478, 40958, 81918, 163838, 327678, 655358, 1310718, 2621438, 5242878, 10485758, 20971518, 41943038, 83886078, 167772158, 335544318, 671088638, 1342177278, 2684354558, 5368709118, 10737418238

Offset: 0

Gary W. Adamson, Oct 05 2007

a(n) = bottom term of the matrix-vector product M^n*V, where M = the 3 X 3 matrix [1,0,0; 0,1,0; 1,1,2] and V = [1,1,3].
Binomial transform of (1,7,3,7,3,7,3,...).
Essentially the same as A131051 and A051633. - R. J. Mathar, Mar 28 2012

			a(2) = 18 = sum of row 2 terms, triangle A134061: (3 + 10 + 5).
a(3) = 38 = (1, 3, 3, 1) dot (1, 7, 3, 7) = (1 + 21 + 9 + 7).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A134061.

a=8; lst={1, a}; k=10; Do[a+=k; AppendTo[lst, a]; k+=k, {n, 0, 5!}]; lst (* Vladimir Joseph Stephan Orlovsky, Dec 17 2008 *)
Flatten[{1, Table[5*2^n-2, {n, 1, 40}]}] (* Vaclav Kotesovec, Jan 26 2015 *)
LinearRecurrence[{3,-2},{1,8,18},40] (* Harvey P. Dale, Dec 17 2024 *)

Vec(-(4*x^2-5*x-1)/((x-1)*(2*x-1)) + O(x^100)) \\ Colin Barker, Nov 17 2015

For n > 0, a(n) = 5*2^n - 2. - Vaclav Kotesovec, Jan 26 2015
From Colin Barker, Nov 17 2015: (Start)
a(n) = 3*a(n-1) - 2*a(n-2) for n>2.
G.f.: -(4*x^2-5*x-1) / ((x-1)*(2*x-1)). (End)

More terms from Jon E. Schoenfield, Jan 25 2015

A159290 A generalized Jacobsthal sequence.

Original entry on oeis.org

3, 5, 13, 25, 53, 105, 213, 425, 853, 1705, 3413, 6825, 13653, 27305, 54613, 109225, 218453, 436905, 873813, 1747625, 3495253, 6990505, 13981013, 27962025, 55924053, 111848105, 223696213, 447392425, 894784853, 1789569705, 3579139413

Offset: 0

Creighton Dement, Apr 08 2009

Sequence generated by the floretion: X*Y with X = 0.5('i + 'j + 'k + 'ee') and Y = 0.5(i' + j' + k' + 'ij' + 'ik' + 'ji' + 'jk' + 'ki' + 'kj' + 'ee')

G. C. Greubel, Table of n, a(n) for n = 0..1000
Creighton Dement, Online Floretion Multiplier [broken link]
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

A083943, A068156

[-1 + (2*(-1)^n + 5*2^(n+1))/3: n in [0..50]]; // G. C. Greubel, Jun 27 2018

LinearRecurrence[{2, 1, -2}, {3, 5, 13}, 50] (* or *) Table[-1 + (2*(-1)^n + 5*2^(n+1))/3, {n,0,30}] (* G. C. Greubel, Jun 27 2018 *)

x='x+O('x^50); Vec((3-x)/(-x^2+1-2*x+2*x^3)) \\ G. C. Greubel, Jun 27 2018

a(n) = -1 + (2*(-1)^n + 5*2^(n+1))/3.
G.f.: (3-x)/((1-x)*(1+x)*(1-2*x)).
a(n) = 3*A000975(n+1) - A000975(n). - R. J. Mathar, Sep 11 2019
a(n)+a(n+1) = A051633(n+1). - R. J. Mathar, Mar 23 2023

A360692 a(0) = 0. Thereafter a(n+1) = a(a(n)) if a(n) has not occurred previously, otherwise a(n+1) = n - 1 - a(n-1).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 3, 1, 4, 0, 5, 2, 6, 0, 7, 3, 8, 1, 9, 4, 10, 0, 11, 5, 12, 2, 13, 6, 14, 0, 15, 7, 16, 3, 17, 8, 18, 1, 19, 9, 20, 4, 21, 10, 22, 0, 23, 11, 24, 5, 25, 12, 26, 2, 27, 13, 28, 6, 29, 14, 30, 0, 31, 15, 32, 7, 33, 16, 34, 3, 35, 17, 36, 8

Offset: 0

David James Sycamore, Feb 16 2023

nonn

An inductive argument shows that a(n) <= n for all n, with equality iff n = 0. It follows that a(n) is well defined, and the sequence is infinite.
Apart from a(1) = 0 every repeat term is followed by a novel term, and vice versa.
Every nonnegative integer appears infinitely many times.
The proper subsequence given by a(2*k) for k >= 2 is the sequence itself, which is therefore fractal.
Starting from a(1) = 0 the sequence is the nonnegative integers interleaved with itself.

			a(0) = 0 is a novel term so a(1) = a(a(0)) = 0. Since a(1) is a repeat term a(2) = 0 - a(0) = 0 - 0 = 0. a(1,2) = 0,0 is the only case of consecutive repeat terms.
Since a(2) = 0 is a repeat term, a(3) = 1 - a(1) = 1 - 0 = 1, a novel term so a(4) = a(a(1)) = 0, and so on.
a(16) = 3, a repeat term (last seen at a(7)), so a(17) = 15 - a(15) = 15 - 7 = 8.

Winston de Greef, Table of n, a(n) for n = 0..10000
Rémy Sigrist, C program

Cf. A000225, A027383, A051633, A101279, A176488, A176449.

C
```
// See Links section.
```

nn = 10^4; c[] = False; a[0] = i = j = 0; Reap[Do[If[c[j], k = n - 2 - i, k = a[j]]; Set[{a[n], c[j], i, j}, {k, True, j, k}]; Sow[k], {n, nn}]][[-1, -1]] (* _Michael De Vlieger, Feb 18 2023 *)

a(2*n + 1) = n for all n >= 0.
A027383(n) = 0. (n >= 0) gives the positions of all zeros after a(0) = 0.
a((2*k + 3)*2^n - 2) = k (n >= 0) gives the positions of all k > 0.
The number of nonnegative terms occurring between consecutive zeros is 0,0,1,1,3,3,7,7,15,15,... (A000225(n), repeat).
a(n) = A101279(n+2) - 1. - Rémy Sigrist, Feb 18 2023

More terms from Rémy Sigrist, Feb 18 2023

cp's OEIS Frontend

A118654 Square array T(n,k) read by antidiagonals: T(n,k) = 2^n*Fibonacci(k) - Fibonacci(k-2).

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A131051 Row sums of triangle A133805.

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Magma

Magma

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A123208 Start with 1, then alternately add 2 or double.

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Magma

Maple

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A028263 Elements in 3-Pascal triangle A028262 (by row) that are not 1.

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Haskell

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A134062 Row sums of triangle A134061.

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Mathematica

PARI

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A159290 A generalized Jacobsthal sequence.

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Magma

Mathematica

PARI

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A360692 a(0) = 0. Thereafter a(n+1) = a(a(n)) if a(n) has not occurred previously, otherwise a(n+1) = n - 1 - a(n-1).

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