A002697 -id:A002697 - cp's OEIS frontend

A258431 Sum over all peaks of Dyck paths of semilength n of the arithmetic mean of the x and y coordinates.

0, 1, 5, 23, 102, 443, 1898, 8054, 33932, 142163, 592962, 2464226, 10209620, 42190558, 173962532, 715908428, 2941192472, 12065310083, 49428043442, 202249741418, 826671597572, 3375609654698, 13771567556012, 56138319705908, 228669994187432, 930803778591278

Offset: 0

Alois P. Heinz, May 29 2015

A Dyck path of semilength n is a (x,y)-lattice path from (0,0) to (2n,0) that does not go below the x-axis and consists of steps U = (1,1) and D = (1,-1). A peak of a Dyck path is any lattice point visited between two consecutive steps UD.

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
Wikipedia, Average, Arithmetic mean
Wikipedia, Lattice path

Cf. A000302, A000346, A000531, A002457, A002697, A002802, A029887.

A258431:= func< n | n eq 0 select 0 else (4^(n-1) + Factorial(2*n-1)/Factorial(n-1)^2)/2 >;
[A258431(n): n in [0..40]]; // G. C. Greubel, Mar 18 2023

a:= proc(n) option remember; `if`(n<3, [0, 1, 5][n+1],
       ((8*n-10)*a(n-1)-(16*n-24)*a(n-2))/(n-1))
    end:
seq(a(n), n=0..30);

a[0]=0; a[1]=1; a[2]=5;
a[n_]:= a[n]= (2*(4*n-5)*a[n-1] - 8*(2*n-3)*a[n-2])/(n-1);
Table[a[n], {n,0,30}] (* Jean-François Alcover, May 31 2018, from Maple *)

def A258431(n): return 0 if (n==0) else (4^(n-1) + factorial(2*n-1)/factorial(n-1)^2)/2
[A258431(n) for n in range(41)] # G. C. Greubel, Mar 18 2023

G.f.: x*(1 + sqrt(1-4*x))/(2*sqrt(1-4*x)^3).
a(n) = (2*(4*n-5)*a(n-1) - 8*(2*n-3)*a(n-2))/(n-1) for n>2, a(0)=0, a(1)=1, a(2)=5.
a(n) = (4^(n-1) + (2*n-1)!/(n-1)!^2)/2 for n>0, a(0) = 0.
a(n) = (A000302(n-1) + A002457(n-1))/2 for n>0, a(0) = 0.
a(n) = (1/2)*binomial(2*n,n)*( 1 + 2*(n-1)/(n+1) + 3*(n-1)*(n-2)/((n+1)*(n+2)) + 4*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + 5*(n-1)*(n-2)*(n-3)*(n-4)/((n+1)*(n+2)*(n+3)*(n+4)) + ...) for n >= 1. - Peter Bala, Feb 17 2022

A038806 Convolution of A008549 with A000302 (powers of 4).

Original entry on oeis.org

0, 1, 10, 69, 406, 2186, 11124, 54445, 259006, 1205790, 5519020, 24918306, 111250140, 492051124, 2159081192, 9409526397, 40766269774, 175707380630, 753876367356, 3221460111958, 13716223138388, 58210889582796

Offset: 0

Wolfdieter Lang

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Hacène Belbachir, Toufik Djellal, Jean-Gabriel Luque, On the self-convolution of generalized Fibonacci numbers, arXiv:1703.00323 [math.CO], 2017.
A. Bernini, F. Disanto, R. Pinzani and S. Rinaldi, Permutations defining convex permutominoes, J. Int. Seq. 10 (2007) # 07.9.7

Cf. A000108, A000346, A008549, A000302, A002697.

[(n+3)*4^n -(n+2)*Binomial(2*n+3, n+1)/2: n in [0..25]]; // Vincenzo Librandi, Jun 09 2011

CoefficientList[Series[x ((1 - Sqrt[1 - 4 x])/(2 x)/(1 - 4 x))^2, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 29 2014 *)

a(n) = (n+3)*4^n -(n+2)*binomial(2*n+3, n+1)/2.
G.f.: x*(c(x)/(1-4*x))^2, where c(x) = g.f. for Catalan numbers A000108.
a(n+1), n >= 0 is convolution of A000346 with itself; a(n+1), n >= 0 is convolution of Catalan numbers A000108 C(n+1), n >= 0 with A002697; a(-1)=0.
Asymptotics: a(n) ~ 4^n*(n+1-4*sqrt(n/Pi)). - Fung Lam, Mar 28 2014
Recurrence: (n-1)*(n+1)*a(n) = 2*(n+1)*(4*n-3)*a(n-1) - 8*n*(2*n+1)*a(n-2). - Vaclav Kotesovec, Mar 28 2014

A042941 Convolution of Catalan numbers A000108 with A038845.

Original entry on oeis.org

1, 13, 110, 765, 4746, 27314, 149052, 781725, 3975730, 19730150, 95973956, 459145778, 2165937060, 10095323460, 46566906872, 212857023069, 965208806082, 4345780250270, 19442667426420, 86489687956518

Offset: 0

Wolfdieter Lang

Also convolution of A018218(n+1), n >= 0, with A000302 (powers of 4); also convolution of A000346 with A002697.

Fung Lam, Table of n, a(n) for n = 0..1000

CoefficientList[Series[(1-Sqrt[1-4*x])/(2*x*(1-4*x)^3), {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 16 2014 *)

a(n) = binomial(n+3, 2)*(4^(n+1) - A000984(n+3)/A000984(2)) / 2.
G.f.: c(x)/(1-4*x)^3, where c(x) is the g.f. for Catalan numbers.
Recurrence: (n+1)*a(n) = 128*(1-2*n)*a(n-4) + 32*(8*n-1)*a(n-3) - 24*(4*n+1)*a(n-2) + 2*(8*n+5)*a(n-1). - Fung Lam, Apr 13 2014
a(n) ~ 2^(2*n)*(n^2 - 8*n^(3/2)/(3*sqrt(Pi))). - Fung Lam, Apr 13 2014
Recurrence: n*(n+1)*a(n) = 2*n*(4*n+9)*a(n-1) - 8*(n+2)*(2*n+3)*a(n-2). - Vaclav Kotesovec, Apr 16 2014

A336952 E.g.f.: 1 / (1 - x * exp(4*x)).

Original entry on oeis.org

1, 1, 10, 102, 1336, 22200, 443664, 10334128, 275060608, 8236914048, 274069953280, 10031110907136, 400520747437056, 17324601073921024, 807023462798608384, 40278407730378332160, 2144307919689898491904, 121291661335680615284736, 7264376142168665821741056

Offset: 0

Ilya Gutkovskiy, Aug 08 2020

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..375

Column k=4 of A351790.

Cf. A002697, A006153, A235328, A326324, A328183, A336950, A336951.

nmax = 18; CoefficientList[Series[1/(1 - x Exp[4 x]), {x, 0, nmax}], x] Range[0, nmax]!
Join[{1}, Table[n! Sum[(4 (n - k))^k/k!, {k, 0, n}], {n, 1, 18}]]
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] k 4^(k - 1) a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 18}]

seq(n)={ Vec(serlaplace(1 / (1 - x*exp(4*x + O(x^n))))) } \\ Andrew Howroyd, Aug 08 2020

a(n) = n! * Sum_{k=0..n} (4 * (n-k))^k / k!.
a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * k * 4^(k-1) * a(n-k).
a(n) ~ n! * (4/LambertW(4))^n / (1 + LambertW(4)). - Vaclav Kotesovec, Aug 09 2021

A133224 Let P(A) be the power set of an n-element set A and let B be the Cartesian product of P(A) with itself. Remove (y,x) from B when (x,y) is in B and x <> y and let R35 denote the reduced set B. Then a(n) = the sum of the sizes of the union of x and y for every (x,y) in R35.

Original entry on oeis.org

0, 2, 14, 78, 400, 1960, 9312, 43232, 197120, 885888, 3934720, 17307136, 75509760, 327182336, 1409343488, 6039920640, 25770065920, 109522223104, 463857647616, 1958507577344, 8246342451200

Offset: 0

Ross La Haye, Dec 30 2007, Jan 03 2008

A082134 is the analogous sequence if "union" is replaced by "intersection" and A002697 is the analogous sequence if "union" is replaced by "symmetric difference". Here, X union Y = Y union X are considered as the same Cartesian product [Relation (37): U_Q(n) in document of Ross La Haye in reference], if we want to consider that X Union Y and Y Union X are two distinct Cartesian products, see A212698. [Bernard Schott, Jan 11 2013]

			a(2) = 14 because for P(A) = {{},{1},{2},{1,2}} |{} union {1}| = 1, |{} union {2}| = 1, |{} union {1,2}| = 2, |{1} union {2}| = 2, |{1} union {1,2}| = 2 and |{2} union {1,2}| = 2, |{} union {}| = 0, |{1} union {1}| = 1, |{2} union {2}| = 1, |{1,2} union {1,2}| = 2, which sums to 14.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6.
Index entries for linear recurrences with constant coefficients, signature (12,-52,96,-64).

Cf. A027471, A002697, A082134, A212698.

[n*(2^(n-2) + 3*2^(2*n-3)): n in [0..30]]; // Vincenzo Librandi, Jun 10 2011

LinearRecurrence[{12,-52,96,-64},{0,2,14,78},30] (* Harvey P. Dale, Jan 24 2019 *)

a(n) = n*(2^(n-2) + 3*2^(2*n-3)).
G.f.: 2*x*(7*x^2-5*x+1) / ((2*x-1)^2*(4*x-1)^2). [Colin Barker, Dec 10 2012]
E.g.f.: exp(2*x)*(1 + 3*exp(2*x))*x. - Stefano Spezia, Aug 04 2022

A300332 Integers of the form Sum_{j in 0:p-1} x^j*y^(p-j-1) where x and y are positive integers with max(x, y) >= 2 and p is some prime.

Original entry on oeis.org

3, 4, 7, 12, 13, 19, 21, 27, 28, 31, 37, 39, 43, 48, 49, 52, 57, 61, 63, 67, 73, 75, 76, 79, 80, 84, 91, 93, 97, 103, 108, 109, 111, 112, 117, 121, 124, 127, 129, 133, 139, 147, 148, 151, 156, 157, 163, 169, 171, 172, 175, 181, 183, 189, 192, 193, 196, 199

Offset: 1

Peter Luschny, Mar 03 2018

nonn

Equivalently these are the integers represented by a cyclotomic binary form Phi_p(x,y) where p is prime and x and y are positive integers with max(x,y) >= 2. A cyclotomic binary form (over Z) is a homogeneous polynomial in two variables of the form f(x, y) = y^phi(k)*Phi(k, x/y) where Phi(k, z) is a cyclotomic polynomial of index k and phi is Euler's totient function.

An efficient and safe calculation of this sequence requires a precise knowledge of the range of possible solutions of the associated Diophantine equations. The bounds used in the Julia program below were specified by Fouvry, Levesque and Waldschmidt.

			Let p denote an odd prime. Subsequences are numbers of the form
2^p - 1,         (A001348) (x = 1, y = 2) (Mersenne numbers),
p*2^(p - 1),     (A299795) (x = 2, y = 2),
(3^p - 1)/2,     (A003462) (x = 1, y = 3),
3^p - 2^p,       (A135171) (x = 2, y = 3),
p*3^(p - 1),     (A027471) (x = 3, y = 3),
(4^p - 1)/3,     (A002450) (x = 1, y = 4),
2^(p-1)*(2^p-1), (A006516) (x = 2, y = 4),
4^p - 3^p,       (A005061) (x = 3, y = 4),
p*4^(p - 1),     (A002697) (x = 4, y = 4),
(p^p-1)/(p-1),   (A023037),
p^p,             (A000312, A051674).
.
The generalized cuban primes A007645 are a subsequence, as are the quintan primes A002649, the septan primes and so on.
All primes in this sequence less than 1031 are generalized cuban primes. 1031 is an element because 1031 = f(5,2) where f(x,y) = x^4 + y*x^3 + y^2*x^2 + y^3*x + y^4, however 1031 is not a cuban prime because 1030 is not divisible by 6.

Peter Luschny, Table of n, a(n) for n = 1..10000
Étienne Fouvry, Claude Levesque, Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.

Indices of the nonzero values of A300333.

Cf. A001348, A299795, A003462, A135171, A027471, A002450, A006516, A005061, A002697, A000312, A051674, A023037, A007645.

using Primes
function isA300332(n)
    logn = log(n)^1.161
    K = Int(floor(5.383*logn))
    M = Int(floor(2*(n/3)^(1/2)))
    k = 2
    while k <= K
        if k == 7
            K = Int(floor(4.864*logn))
            M = Int(ceil(2*(n/11)^(1/4)))
        end
        for y in 2:M, x in 1:y
            r = x == y ? k*y^(k - 1) : div(x^k - y^k, x - y)
            n == r && return true
        end
        k = nextprime(k+1)
    end
    return false
end
A300332list(upto) = [n for n in 1:upto if isA300332(n)]
println(A300332list(200))

A104002 Triangle T(n,k) read by rows: number of permutations in S_n avoiding all k-length patterns that start with 1 except one fixed pattern and containing it exactly once.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 12, 6, 1, 5, 32, 27, 8, 1, 6, 80, 108, 48, 10, 1, 7, 192, 405, 256, 75, 12, 1, 8, 448, 1458, 1280, 500, 108, 14, 1, 9, 1024, 5103, 6144, 3125, 864, 147, 16, 1, 10, 2304, 17496, 28672, 18750, 6480, 1372, 192, 18, 1, 11, 5120, 59049, 131072

Offset: 2

Ralf Stephan, Feb 26 2005

T(n+k,k+1) = total number of occurrences of any given letter in all possible n-length words on a k-letter alphabet. For example, with the 2 letter alphabet {0,1} there are 4 possible 2-length words: {00,01,10,11}. The letter 0 occurs 4 times altogether, as does the letter 1. T(4,3) = 4. - Ross La Haye, Jan 03 2007

Table T(n,k) = k*n^(k-1) n,k > 0 read by antidiagonals. - Boris Putievskiy, Dec 17 2012

			Triangle begins:
  1;
  2,   1;
  3,   4,    1;
  4,  12,    6,    1;
  5,  32,   27,    8,   1;
  6,  80,  108,   48,  10,   1;
  7, 192,  405,  256,  75,  12,  1;
  8, 448, 1458, 1280, 500, 108, 14, 1;

Michael De Vlieger, Table of n, a(n) for n = 2..11176 (rows 2 <= n <= 150).
T. Mansour, Permutations containing and avoiding certain patterns, arXiv:math/9911243 [math.CO], 1999-2000.
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.

Cf. Left-edge columns include A001787, A027471, A002697, A053464, A053469, A027473, A053539, A053540, A053541, A081127, A081128.

Table[(n - k + 1) (k - 1)^(n - k), {n, 2, 12}, {k, 2, n}] // Flatten (* Michael De Vlieger, Aug 22 2018 *)

T(n, k) = (n-k+1) * (k-1)^(n-k), k<=n.

As a linear array, the sequence is a(n) = A004736(n)*A002260(n)^(A004736(n)-1) or a(n) = ((t*t+3*t+4)/2-n)*(n-(t*(t+1)/2))^((t*t+3*t+4)/2-n-1), where t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 17 2012

A119733 Offsets of the terms of the nodes of the reverse Collatz function.

Original entry on oeis.org

0, 1, 2, 5, 4, 7, 10, 19, 8, 11, 14, 23, 20, 29, 38, 65, 16, 19, 22, 31, 28, 37, 46, 73, 40, 49, 58, 85, 76, 103, 130, 211, 32, 35, 38, 47, 44, 53, 62, 89, 56, 65, 74, 101, 92, 119, 146, 227, 80, 89, 98, 125, 116, 143, 170, 251, 152, 179, 206, 287, 260, 341, 422, 665, 64, 67

Offset: 0

William Entriken, Jun 14 2006

nonn

Create a binary tree starting with x. To follow 0 from the root, apply f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting with x, the string 010 {also known as f(g(f(x)))}, you would get (8x-2)/3. These expressions represent the reverse Collatz function and will provide numbers whose Collatz path may include x. These expressions will all be of the form (2^a*x-b)/3^c. This sequence concerns b. What makes b interesting is that if you draw the tree, each level of the tree will have the same sequence of values for b. The root of the tree x, can be written as (2^0*x-0)/3^0, which has the first value for b. Each subsequent level contains twice as many values of b.
This sequence is 0 followed by a permutation of A213539, and therefore consists of 0 plus the elements of A116640 multiplied by 2^k, where k >= 0. E.g., 1, 5, 7, 19 becomes 0, 2^0*1, 2^1*1, 2^0*5, 2^2*1, 2^0*7, 2^1*5, 2^0*19, ... - Joe Slater, Dec 19 2016
When this sequence is arranged as an irregular triangle the sum of each row a(2^k)...a(2^(k+1)-1) equals A081039(2^k). The cumulative sum from a(0) to a(2^k-1) equals A002697(k). - Joe Slater, Apr 12 2018

			a(1) = 1 = 2 * 0 + 3^0 since 0 written in binary contains no 1's.

Alois P. Heinz, Table of n, a(n) for n = 0..16383
Víctor Martín Chabrera, An algebraic fractal approach to Collatz Conjecture, Bachelor tesis, Universitat Politècnica de Catalunya (Barcelona, 2019), see lemma 6.1 with a(n) = r(A005836(n)).

Cf. A002697, A081039, A116623, A116640, A116641, A213539, A226383.

a:= proc(n) `if`(n=0, 0, `if`(irem(n, 2, 'r')=0, 0,
      3^add(i, i=convert(r, base, 2)))+2*a(r))
    end:
seq(a(n), n=0..127);  # Alois P. Heinz, Aug 13 2017

a[0] := 0; a[n_?OddQ] := 2a[(n - 1)/2] + 3^Plus@@IntegerDigits[(n - 1)/2, 2]; a[n_?EvenQ] := 2a[n/2]; Table[a[n], {n, 0, 65}] (* Alonso del Arte, Apr 21 2011 *)

a(n) = my(ret=0); if(n, for(i=0,logint(n,2), if(bittest(n,i), ret=3*ret+1<Kevin Ryde, Oct 22 2021

# call with n to get 2^n values
$depth=shift; sub funct { my ($i, $b, $c) = @_; if ($i < $depth) { funct($i+1, $b*2, $c); funct($i+1, 2*$b+$c, $c*3); } else { print "$b, "; } } funct(0, 0, 1); print " ";

from sympy.core.cache import cacheit
@cacheit
def a(n): return 0 if n==0 else 2*a((n - 1)//2) + 3**bin((n - 1)//2).count('1') if n%2 else 2*a(n//2)
print([a(n) for n in range(131)]) # Indranil Ghosh, Aug 13 2017

a(0) = 0, a(2*n + 1) = 2*a(n) + 3^wt(n) = 2*a(n) + A048883(n), a(2*n) = 2*a(n), where wt(n) = A000120(n) = the number 1's in the binary representation of n.
a(k) = [z^k] 1 + (1/(1-z)) * Sum_{s=0..n-1} 2^s*z^(2^s)*(1 - z^(2^s)) * Product_{r=s+1..n-1} (1 + 3*z^(2^r)), for 0 < k <= 2^n-1. - Wolfgang Hintze, Jul 28 2017
a(n) = Sum_{i=0..k} 2^e[i] * 3^i where binary expansion n = 2^e[0] + 2^e[1] + ... + 2^e[k] with descending e[0] > e[1] > ... > e[k] (A272011). [Martín Chabrera lemma 6.1, adapting index i] - Kevin Ryde, Oct 22 2021

A128235 Triangle read by rows: T(n,k) is the number of sequences of length n on the alphabet {0,1,2,3}, containing k subsequences 00 (0<=k<=n-1).

Original entry on oeis.org

1, 4, 15, 1, 57, 6, 1, 216, 33, 6, 1, 819, 162, 36, 6, 1, 3105, 756, 189, 39, 6, 1, 11772, 3402, 945, 216, 42, 6, 1, 44631, 14931, 4536, 1143, 243, 45, 6, 1, 169209, 64314, 21168, 5778, 1350, 270, 48, 6, 1, 641520, 273051, 96633, 28323, 7128, 1566, 297, 51

Offset: 0

Emeric Deutsch, Feb 27 2007

Row n has n terms (n>=1). T(n,0) = A125145(n). Sum(k*T(n,k), k=0..n-1) = (n-1)*4^(n-2) = A002697(n-1).

			T(4,2) = 6 because we have 0001, 0002, 0003, 1000, 2000 and 3000.
Triangle starts:
1;
4;
15,    1;
57,    6,  1;
216,  33,  6, 1;
819, 162, 36, 6, 1;

Alois P. Heinz, Rows n = 0..150, flattened

Cf. A125145, A002697.

G:=(1+z-t*z)/(1-3*z-3*z^2-t*z+3*t*z^2): Gser:=simplify(series(G,z=0,14)): for n from 0 to 11 do P[n]:=sort(coeff(Gser,z,n)) od: 1; for n from 1 to 11 do seq(coeff(P[n],t,j),j=0..n-1) od; # yields sequence in triangular form

G.f.: (1+z-tz)/(1-3z-3z^2-tz+3tz^2).

A174720 Triangle T(n, k, q) = (1-q^n)*( binomial(n, k) - 1 ) + 1, with q = 4, read by rows.

Original entry on oeis.org

1, 1, 1, 1, -14, 1, 1, -125, -125, 1, 1, -764, -1274, -764, 1, 1, -4091, -9206, -9206, -4091, 1, 1, -20474, -57329, -77804, -57329, -20474, 1, 1, -98297, -327659, -557021, -557021, -327659, -98297, 1, 1, -458744, -1769444, -3604424, -4521914, -3604424, -1769444, -458744, 1

Offset: 0

Roger L. Bagula, Mar 28 2010

The row sums of this class of sequences, for varying q, is given by Sum_{k=0..n} T(n, k, q) = q^n * (n+1) + 2^n * (1 - q^n). - G. C. Greubel, Feb 09 2021

			Triangle begins as:
  1;
  1,       1;
  1,     -14,        1;
  1,    -125,     -125,        1;
  1,    -764,    -1274,     -764,        1;
  1,   -4091,    -9206,    -9206,    -4091,        1;
  1,  -20474,   -57329,   -77804,   -57329,   -20474,        1;
  1,  -98297,  -327659,  -557021,  -557021,  -327659,   -98297,       1;
  1, -458744, -1769444, -3604424, -4521914, -3604424, -1769444, -458744, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A000012 (q=1), A174718 (q=2), A174719 (q=3), this sequence (q=4).

Cf. A002697, A248217.

T:= func< n,k,q | 1 + (1-q^n)*(Binomial(n,k) -1) >;
[T(n,k,4): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 09 2021

T[n_, k_, q_]:= 1 +(1-q^n)*(Binomial[n, k] -1);
Table[T[n,k,4], {n,0,12}, {k,0,n}]//Flatten

def T(n,k,q): return 1 + (1-q^n)*(binomial(n,k) - 1)
flatten([[T(n,k,4) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 09 2021

T(n, k, q) = (1-q^n)*( binomial(n, k) - 1 ) + 1, with q=4.

Sum_{k=0..n} T(n, k, 4) = 4^n*(n+1) + 2^n*(1 - 4^n) = A002697(n+1) - A248217(n). - G. C. Greubel, Feb 09 2021

Edited by G. C. Greubel, Feb 09 2021

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