A176448 -id:A176448 - cp's OEIS frontend

A028314 Elements in the 5-Pascal triangle A028313 that are not 1.

5, 6, 6, 7, 12, 7, 8, 19, 19, 8, 9, 27, 38, 27, 9, 10, 36, 65, 65, 36, 10, 11, 46, 101, 130, 101, 46, 11, 12, 57, 147, 231, 231, 147, 57, 12, 13, 69, 204, 378, 462, 378, 204, 69, 13, 14, 82, 273, 582, 840, 840, 582, 273, 82, 14, 15, 96, 355, 855, 1422, 1680, 1422, 855, 355, 96, 15

Offset: 0

Mohammad K. Azarian

			Triangle begins as:
   5;
   6,  6;
   7, 12,   7;
   8, 19,  19,   8;
   9, 27,  38,  27,   9;
  10, 36,  65,  65,  36,  10;
  11, 46, 101, 130, 101,  46,  11;
  12, 57, 147, 231, 231, 147,  57,  12;
  13, 69, 204, 378, 462, 378, 204,  69,  13;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000027, A010892, A022112, A028313, A028322,

Cf. A051936, A051937, A121262, A176448.

A028314:= func< n,k | Binomial(n+2,k+1) + 3*Binomial(n,k) >;
[A028314(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 06 2024

A028314[n_, k_]:= Binomial[n+2,k+1] + 3*Binomial[n,k];
Table[A028314[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 06 2024 *)

def A028314(n,k): return binomial(n+2,k+1) + 3*binomial(n,k)
flatten([[A028314(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jan 06 2024

From G. C. Greubel, Jan 06 2024: (Start)
T(n, k) = binomial(n+2, k+1) + 3*binomial(n, k).
T(n, n-k) = T(n, k).
T(n, 0) = T(n, n) = A000027(n+5).
T(n, 1) = T(n, n-1) = A051936(n+4).
T(n, 2) = T(n, n-2) = A051937(n+3).T(2*n, n) = A028322(n+1).
Sum_{k=0..n} T(n, k) = A176448(n).
Sum_{k=0..n} (-1)^k * T(n, k) = 1 + (-1)^n + 3*[n=0].
Sum_{k=0..n} T(n-k, k) = A022112(n+1) - (3-(-1)^n)/2.
Sum_{k=0..n} (-1)^k * T(n-k, k) = 4*A010892(n) - 2*A121262(n+1) - (3 - (-1)^n)/2. (End)
G.f.: (5 - 4*x - 4*x*y + 3*x^2*y)/((1 - x)*(1 - x*y)*(1 - x - x*y)). - Stefano Spezia, Dec 06 2024

More terms from James Sellers

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 *)

A174317 a(0)=1, a(1)=2, a(2)=1; for n>2, a(n) = 7*2^(n-3)-2.

Original entry on oeis.org

1, 2, 1, 5, 12, 26, 54, 110, 222, 446, 894, 1790, 3582, 7166, 14334, 28670, 57342, 114686, 229374, 458750, 917502, 1835006, 3670014, 7340030, 14680062, 29360126, 58720254, 117440510, 234881022, 469762046, 939524094, 1879048190, 3758096382

Offset: 0

Richard Choulet, Mar 15 2010

			a(4) = 14-2 = 12.
a(5) = 7*4-2 = 26.

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

Cf. A174314, A174316

taylor(1+2*z+z^2+5*z^3-((2*z^4)/(1-z))+((14*z^4)/(1-2*z)),z=0,50);

G.f: (1+x-x^2+4*x^3-7*x^4)/(1-x)+(14*x^4)/(1-2*x).

a(n) = A026622(n-1), n>2. a(n) = A176448(n-3), n>2. - R. J. Mathar, Mar 01 2016

Definition edited by Olivier Gérard, Oct 24 2012

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A028314 Elements in the 5-Pascal triangle A028313 that are not 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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A174317 a(0)=1, a(1)=2, a(2)=1; for n>2, a(n) = 7*2^(n-3)-2.

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