A131051 -id:A131051 - cp's OEIS frontend

A131047 (1/2) * (A007318 - A007318^(-1)).

1, 0, 2, 1, 0, 3, 0, 4, 0, 4, 1, 0, 10, 0, 5, 0, 6, 0, 20, 0, 6, 1, 0, 21, 0, 35, 0, 7, 0, 8, 0, 56, 0, 56, 0, 8, 1, 0, 36, 0, 126, 0, 84, 0, 9

Offset: 1

Gary W. Adamson, Jun 12 2007

Row sums = (1, 2, 4, 8, ...). A131047 * (1,2,3, ...) = A087447 starting (1, 4, 10, 24, 56, ...). A generalized set of analogous triangles: (1/(Q+1)) * (P^Q - 1/P), Q an integer, generates triangles with row sums = powers of (Q+1). Cf. A131048, A131049, A131050, A131051 for triangles having Q = 2,3,4 and 5, respectively.

A007318, Pascal's triangle, = this triangle + A119467, since one triangle = the zeros or masks of the other. - Gary W. Adamson, Jun 12 2007

			First few rows of the triangle:
  1;
  0, 2;
  1, 0,  3;
  0, 4,  0,  4;
  1, 0, 10,  0,  5;
  0, 6,  0, 20,  0, 6;
  1, 0, 21,  0, 35, 0, 7;
  ...

Cf. A131048, A131049, A131050, A131051.

Cf. A119467.

Let A007318 (Pascal's triangle) = P, then A131047 = (1/2) * (P - 1/P); deleting the right border of zeros.

A051633 a(n) = 5*2^n - 2.

Original entry on oeis.org

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, 5368709118, 10737418238, 21474836478

Offset: 0

Asher Auel

			a(5) = 5*2^4 - 2 = 80 - 2 = 78.

Stefano Spezia, Table of n, a(n) for n = 0..3300
Index entries for linear recurrences with constant coefficients, signature (3,-2).

Cf. A000079, A020714, A118654.

Cf. A123208, A048487, A131051, A153894.

LinearRecurrence[{3, -2},{3, 8},30] (* Ray Chandler, Jul 18 2020 *)

a(n) = A118654(n, 5).
a(n) = A000079(n)*5 - 2 = A020714(n) - 2. - Omar E. Pol, Dec 23 2008
a(n) = 2*(a(n-1)+1) with a(0)=3. - Vincenzo Librandi, Aug 06 2010
a(n) = A123208(2*n+1) = A048487(n)+2 = A131051(n+2) = A153894(n)-1. - Philippe Deléham, Apr 15 2013
G.f.: ( 3-x ) / ( (2*x-1)*(x-1) ). - R. J. Mathar, Mar 23 2023
E.g.f.: exp(x)*(5*exp(x) - 2). - Stefano Spezia, Oct 03 2023

A053209 Row sums of A051598.

Original entry on oeis.org

1, 5, 14, 32, 68, 140, 284, 572, 1148, 2300, 4604, 9212, 18428, 36860, 73724, 147452, 294908, 589820, 1179644, 2359292, 4718588, 9437180, 18874364, 37748732, 75497468, 150994940, 301989884, 603979772, 1207959548, 2415919100

Offset: 0

Asher Auel, Dec 14 1999

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

Cf. A051598, A053208.
Cf. A083329, A131051.
Cf. A048491.

I:=[5,14]; [1] cat [n le 2 select I[n] else 3*Self(n-1) - 2*Self(n-2): n in [1..30]]; // G. C. Greubel, Sep 03 2018

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+x)^2)/((1-x)*(1-2*x))); // Marius A. Burtea, Oct 15 2019

Join[{1}, LinearRecurrence[{3, -2}, {5, 14}, 50]] (* G. C. Greubel, Sep 03 2018 *)

m=30; v=concat([5,14], vector(m-2)); for(n=3, m, v[n] = 3*v[n-1] -2*v[n-2]); concat([1], v) \\ G. C. Greubel, Sep 03 2018

a(0) = 1, a(1) = 5, a(n+1) = 2*a(n) + 4, for n >= 1.
a(n) = 9*2^(n-1) - 4, n >= 1.
a(n) =  4*n + Sum[i = 0, n - 1] a(i). - Jon Perry, Nov 20 2012
a(n) = A048491(n)/2, n>0. - Philippe Deléham, Apr 15 2013
G.f.: (1+x)^2/((1-x)*(1-2*x)). - Philippe Deléham, Apr 15 2013
a(n) = 3*a(n-1) - 2*a(n-2) with a(0)=1, a(1)=5, a(2)=14. - Philippe Deléham, Apr 15 2013
E.g.f.: (1 - 8*exp(x) + 9*exp(2*x))/2. - Stefano Spezia, Sep 28 2022

A131048 (1/3) * (A007318^2 - A007318^(-1)).

Original entry on oeis.org

1, 1, 2, 3, 3, 3, 5, 12, 6, 4, 11, 25, 30, 10, 5, 21, 66, 75, 60, 15, 6, 43, 147, 231, 175, 105, 21, 7, 85, 344, 588, 616, 350, 168, 28, 8, 171, 765, 1548, 1764, 1386, 630, 252, 36, 9

Offset: 1

Gary W. Adamson, Jun 12 2007

Left border = A001045: (1, 1, 3, 5, 11, 21, 43, 85, ...).
Row sums = (1, 3, 9, 27, ...).
Analogous triangles for other powers of P are: A131047, A131049, A131050 and A131051.

			First few rows of the triangle:
   1;
   1,   2;
   3,   3,   3;
   5,  12,   6,   4;
  11,  25,  30,  10,   5;
  21,  66,  75,  60,  15,  6;
  43, 147, 231, 175, 105, 21, 7;
  ...

Cf. A131047, A131049, A131050, A131051, A001045, A007318.

Cf. A001045, A094440, A132148.

Let A007318 (Pascal's triangle) = P. then A131048 = (1/3) * (P^2 - 1/P). Delete right border of zeros.
From Peter Bala, Oct 24 2007: (Start)
O.g.f.: 1/(1 - (2*x + 1)*t + (x^2 + x - 2)*t^2) = 1 + (1 + 2*x)*t + (3 + 3*x + 3*x^2)*t^2 + ....
T(n,n-k) = (1/3)*C(n,k)*(2^k - (-1)^k) = C(n,k)*A001045(k).
The row polynomials R(n,x) := Sum_{k = 0..n} T(n,n-k)*x^(n-k) = (1/3)*((x + 2)^n - (x - 1)^n) and have the divisibility property R(n,x) divides R(m,x) in the polynomial ring Z[x] if n divides m.
The polynomials R(n,-x), n >= 2, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re x = 1/2 in the complex plane. Compare with A094440. (End)

A131050 (1/5) * (A007318^4 - A007318^(-1)).

Original entry on oeis.org

1, 3, 2, 13, 9, 3, 51, 52, 18, 4, 205, 255, 130, 30, 5, 819, 1230, 765, 260, 45, 6, 3277, 5733, 4305, 1785, 455, 63, 7, 13107, 26216, 22932, 11480, 3570, 728, 84, 8

Offset: 1

Gary W. Adamson, Jun 12 2007

Row sums = powers of 5: (1, 5, 25, 125, ...).

Left border = A015521: (1, 3, 13, 51, 205, 819, ...).

			First few rows of the triangle:
    1;
    3,   2;
   13,   9,   3;
   51,  52,  18,  4;
  205, 255, 130, 30, 5;
  ...

Cf. A015521, A131047, A131048, A131049, A131051.

Let P = Pascal's triangle, A007318. Then A131050 = (1/5) * (P^4 - 1/P); deleting the right border of zeros.

A131049 (1/4) * (A007318^3 - A007318^(-1)).

Original entry on oeis.org

1, 2, 2, 7, 6, 3, 20, 28, 12, 4, 61, 100, 70, 20, 5, 182, 366, 300, 140, 30, 6, 547, 1274, 1281, 700, 245, 42, 7, 1640, 4376, 5096, 3416, 1400, 392, 56, 8, 4921, 14760, 19692, 15288, 7686, 2520, 588, 72, 9

Offset: 1

Gary W. Adamson, Jun 12 2007

Row sums = powers of 4: (1, 4, 16, 64, ...).

Left border = A015518: (1, 2, 7, 20, 61, 182, ...).

			First few rows of the triangle:
    1;
    2,    2;
    7,    6,    3;
   20,   28,   12,   4;
   61,  100,   70,  20,   5;
  182,  366,  300, 140,  30,  6;
  547, 1274, 1281, 700, 245, 42, 7;
  ...

Cf. A015518, A007318, A131047, A131048, A131050, A131051.

(1/4) * (P^3 - 1/P), where P = Pascal's triangle, A007318. Delete right border of zeros.

A277215 a(n) is the smallest even number not congruent to 1 modulo 3 that starts a (2n+1)-element alternating sequence of x/2 and (3x+1) iterations ending in the maximum of its Collatz trajectory.

Original entry on oeis.org

0, 26, 6, 14, 30, 1214, 1662, 254, 510, 1022, 2046, 28670, 40958, 180222, 32766, 65534, 131070, 1835006, 5767166, 1048574, 2097150, 4194302, 8388606, 16777214, 33554430, 469762046, 671088638, 268435454, 536870910, 7516192766, 2147483646, 4294967294, 8589934590, 17179869182, 34359738366, 755914244094

Offset: 0

Hartmut F. W. Hoft, Nov 03 2016

nonn

a(n) starts a maximal alternating Collatz sequence v_0, ..., v_2n of 2n+1 elements and must have the form v_0 = 2*(q*2^n - 1) where q is the smallest odd number not a multiple of 3 such that v_(2n) = 2*(q*3^n - 1) is the maximum of its Collatz trajectory.
The intermediate elements of the sequence for 1 <= j <= n are v_(2j-1) = q * 2^(n-j+1) * 3^(j-1) - 1, which is odd, and v_(2j) = 2 * (q * 2^(n-j) * 3^j - 1), which is congruent to 2 modulo 4 except for j=n.
A277875(n) is the odd multiplier q in the expression for a(n).
Subsequences of a(n) are related to subsequences of the following sequences depending on the value of q:
a(n) = 2*A000225(n) = A000918(n+1) when A277875(n) = 1;
a(n) = 2*A153894(n) = A131051(n+3) when A277875(n) = 5;
a(n) = 2*A086224(n) = A086224(n+1)-1 = A176448(n+1) when A277875(n) = 7;
a(n) = A086225(n+1)-1 when A277875(n) = 11;
a(n) = A198274(n+1)-1 when A277875(n) = 13;
a(n) = A198276(n+1)-1 when A277875(n) = 19;
For small q > 1, the positions of 2*(q*2^n - 1) among the first 200 numbers in the sequence are:
q = 5: 12, 26, 36, 46, 58, 62, 174;
q = 7: 1, 11, 17, 25, 29, 45, 49, 53, 57, 61, 65, 77, 93, 103, 109, 113, 117, 125, 139, 141, 145, 157, 165, 173, 187, 189, 193;
q = 11: 13, 18, 35, 59, 69, 75, 83, 114, 133, 179;
q = 13: 6, 118;
q = 19: 5;
and among the first 400 numbers are:
q = 17: 222, 229, 230, 268;
(see A277875).
Conjecture: For every n there is an odd number q such that the alternating sequence ends in v_(2n), the maximum of the Collatz trajectory of a(n)=v_0.

			a(0) = 0 = 2*(1*2^0 - 1) since it is the start and end of the first alternating sequence of 1 element and the maximum of its trajectory.
a(1) = 26 = 2*(7*2^1 - 1) since sequence 26, 13, 40 has 3 elements and ends in the maximum of its trajectory and since 2, 10 and 18 do not satisfy the conditions for a(1).
a(5) = 1214 = 2*(19*2^5 - 1) starts the alternating sequence of 11 elements - 1214, 607, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232 - that ends in the trajectory maximum 9232 while the 11-element alternating sequences starting at 2*(q*2^5 - 1) with odd q<19 either do not end at the trajectory maximum or are congruent to 1 modulo 3 and therefore do not have maximal length.

Cf. A000225, A000918, A086224, A086225, A131051, A153894, A176448, A198274, A198276, A277875.

collatz[n_] := If[OddQ[n], 3n+1, n/2]
altdata[low_, high_] := Module[{n, q, notDone, v, a, m, list={}}, For[n=low, n<=high, n++, q=-1; notDone=True; While[notDone, q+=2; v=2*(q*2^n-1); If[Mod[v, 3]!=1, a=NestWhile[collatz, v, Mod[#,4]!=0&]; m=Max[NestWhileList[collatz, a, #!=1&]]; notDone=(a!=m)]]; AppendTo[list, {n, q, v, a}]]; list]/;(low>1)
a277215[n_]:=Map[#[[3]]&, altdata[2,n]]
Join[{0,26}, a277215[35]] (* sequence data *)

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

A268896 Start at a(0)=1. a(n) = a(n-1)+2 if n == 1,2 (mod 3) and a(n)=a(n-1)+a(n-3) if n == 0 (mod 3).

Original entry on oeis.org

1, 3, 5, 6, 8, 10, 16, 18, 20, 36, 38, 40, 76, 78, 80, 156, 158, 160, 316, 318, 320, 636, 638, 640, 1276, 1278, 1280, 2556, 2558, 2560, 5116, 5118, 5120, 10236, 10238, 10240, 20476, 20478, 20480, 40956, 40958

Offset: 0

Ravesh Sukhram, Feb 27 2016

See Mathematica section for an explicit formula for the n-th term. - Benedict W. J. Irwin, May 30 2016

Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-2).

Simplify[Table[1/6 (10 (2^n)^(1/3) + 4 (-3 + 5 2^(n/3)) Cos[(2 n Pi)/3] + 5 2^((4 + n)/3)Sin[(n Pi)/3] (Sqrt[3] (-1 + 2^(1/3)) Cos[(n Pi)/3] + (1 + 2^(1/3)) Sin[(n Pi)/3]) - 4 (3 + Sqrt[3] Sin[(2 n Pi)/3])), {n, 0, 20}]] (* Benedict W. J. Irwin, May 30 2016 *)

G.f.: ( 1+3*x+5*x^2+3*x^3-x^4-5*x^5 ) / ( (x-1)*(2*x^3-1)*(1+x+x^2) ). - R. J. Mathar, Apr 16 2016

a(3n) = A048487(n). a(3n+1) = A131051(n+1). a(3n+2)=A020714(n). - R. J. Mathar, Apr 16 2016

A359312 a(1) = 1; for n >= 1, a(2n) = A000005(a(n)), a(2n + 1) = A000005(a(n)) + 1.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 1, 2, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2

Offset: 1

Ctibor O. Zizka, Dec 25 2022

nonn

			a(1) = 1;
a(2) = A000005(a(1)) = 1;
a(3) = A000005(a(1)) + 1 = 2;
a(4) = A000005(a(2)) = 1;
a(5) = A000005(a(2)) + 1 = 2;
and so on.

Cf. A000005, A131051.

a[1] = 1; a[n_] := a[n] = If[EvenQ[n], DivisorSigma[0, a[n/2]], DivisorSigma[0, a[(n - 1)/2]] + 1]; Array[a, 100] (* Amiram Eldar, Dec 25 2022 *)

Sum_{i = 2^k..2^(k + 1) - 1} a(i) = 5*2^(k - 1) - 2, for k >= 1.

a(2^k) = 1.

cp's OEIS Frontend

A131047 (1/2) * (A007318 - A007318^(-1)).

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A051633 a(n) = 5*2^n - 2.

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A053209 Row sums of A051598.

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A131048 (1/3) * (A007318^2 - A007318^(-1)).

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A131050 (1/5) * (A007318^4 - A007318^(-1)).

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A131049 (1/4) * (A007318^3 - A007318^(-1)).

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A277215 a(n) is the smallest even number not congruent to 1 modulo 3 that starts a (2n+1)-element alternating sequence of x/2 and (3x+1) iterations ending in the maximum of its Collatz trajectory.

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A134062 Row sums of triangle A134061.

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A268896 Start at a(0)=1. a(n) = a(n-1)+2 if n == 1,2 (mod 3) and a(n)=a(n-1)+a(n-3) if n == 0 (mod 3).

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A359312 a(1) = 1; for n >= 1, a(2n) = A000005(a(n)), a(2n + 1) = A000005(a(n)) + 1.