cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A006516 a(n) = 2^(n-1)*(2^n - 1), n >= 0.

Original entry on oeis.org

0, 1, 6, 28, 120, 496, 2016, 8128, 32640, 130816, 523776, 2096128, 8386560, 33550336, 134209536, 536854528, 2147450880, 8589869056, 34359607296, 137438691328, 549755289600, 2199022206976, 8796090925056, 35184367894528, 140737479966720, 562949936644096
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n) is also the number of different lines determined by pair of vertices in an n-dimensional hypercube. The number of these lines modulo being parallel is in A003462. - Ola Veshta (olaveshta(AT)my-deja.com), Feb 15 2001
Let G_n be the elementary Abelian group G_n = (C_2)^n for n >= 1: A006516 is the number of times the number -1 appears in the character table of G_n and A007582 is the number of times the number 1. Together the two sequences cover all the values in the table, i.e., A006516(n) + A007582(n) = 2^(2n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 01 2001
a(n) is the number of n-letter words formed using four distinct letters, one of which appears an odd number of times. - Lekraj Beedassy, Jul 22 2003 [See, e.g., the Balakrishnan reference, problems 2.67 and 2.68, p. 69. - Wolfdieter Lang, Jul 16 2017]
Number of 0's making up the central triangle in a Pascal's triangle mod 2 gasket. - Lekraj Beedassy, May 14 2004
m-th triangular number, where m is the n-th Mersenne number, i.e., a(n)=A000217(A000225(n)). - Lekraj Beedassy, May 25 2004
Number of walks of length 2n+1 between two nodes at distance 3 in the cycle graph C_8. - Herbert Kociemba, Jul 02 2004
The sequence of fractions a(n+1)/(n+1) is the 3rd binomial transform of (1, 0, 1/3, 0, 1/5, 0, 1/7, ...). - Paul Barry, Aug 05 2005
Number of monic irreducible polynomials of degree 2 in GF(2^n)[x]. - Max Alekseyev, Jan 23 2006
(A007582(n))^2 + a(n)^2 = A007582(2n). E.g., A007582(3) = 36, a(3) = 28; A007582(6) = 2080. 36^2 + 28^2 = 2080. - Gary W. Adamson, Jun 17 2006
The sequence 6*a(n), n>=1, gives the number of edges of the Hanoi graph H_4^{n} with 4 pegs and n>=1 discs. - Daniele Parisse, Jul 28 2006
8*a(n) is the total border length of the 4*n masks used when making an order n regular DNA chip, using the bidimensional Gray code suggested by Pevzner in the book "Computational Molecular Biology." - Bruno Petazzoni (bruno(AT)enix.org), Apr 05 2007
If we start with 1 in binary and at each step we prepend 1 and append 0, we construct this sequence: 1 110 11100 1111000 etc.; see A109241(n-1). - Artur Jasinski, Nov 26 2007
Let P(A) be the power set of an n-element set A. Then a(n) = the number of pairs of elements {x,y} of P(A) for which x does not equal y. - Ross La Haye, Jan 02 2008
Wieder calls these "conjoint usual 2-combinations." The set of "conjoint strict k-combinations" is the subset of conjoint usual k-combinations where the empty set and the set itself are excluded from possible selection. These numbers C(2^n - 2,k), which for k = 2 (i.e., {x,y} of the power set of a set) give {1, 0, 1, 15, 91, 435, 1891, 7875, 32131, 129795, 521731, ...}. - Ross La Haye, Jan 15 2008
If n is a member of A000043 then a(n) is also a perfect number (A000396). - Omar E. Pol, Aug 30 2008
a(n) is also the number whose binary representation is A109241(n-1), for n>0. - Omar E. Pol, Aug 31 2008
From Daniel Forgues, Nov 10 2009: (Start)
If we define a spoof-perfect number as:
A spoof-perfect number is a number that would be perfect if some (one or more) of its odd composite factors were wrongly assumed to be prime, i.e., taken as a spoof prime.
And if we define a "strong" spoof-perfect number as:
A "strong" spoof-perfect number is a spoof-perfect number where sigma(n) does not reveal the compositeness of the odd composite factors of n which are wrongly assumed to be prime, i.e., taken as a spoof prime.
The odd composite factors of n which are wrongly assumed to be prime then have to be obtained additively in sigma(n) and not multiplicatively.
Then:
If 2^n-1 is odd composite but taken as a spoof prime then 2^(n-1)*(2^n - 1) is an even spoof perfect number (and moreover "strong" spoof-perfect).
For example:
a(8) = 2^(8-1)*(2^8 - 1) = 128*255 = 32640 (where 255 (with factors 3*5*17) is taken as a spoof prime);
sigma(a(8)) = (2^8 - 1)*(255 + 1) = 255*256 = 2*(128*255) = 2*32640 = 2n is spoof-perfect (and also "strong" spoof-perfect since 255 is obtained additively);
a(11) = 2^(11-1)*(2^11 - 1) = 1024*2047 = 2096128 (where 2047 (with factors 23*89) is taken as a spoof prime);
sigma(a(11)) = (2^11 - 1)*(2047 + 1) = 2047*2048 = 2*(1024*2047) = 2*2096128 = 2n is spoof-perfect (and also "strong" spoof-perfect since 2047 is obtained additively).
I did a Google search and didn't find anything about the distinction between "strong" versus "weak" spoof-perfect numbers. Maybe some other terminology is used.
An example of an even "weak" spoof-perfect number would be:
n = 90 = 2*5*9 (where 9 (with factors 3^2) is taken as a spoof prime);
sigma(n) = (1+2)*(1+5)*(1+9) = 3*(2*3)*(2*5) = 2*(2*5*(3^2)) = 2*90 = 2n is spoof-perfect (but is not "strong" spoof-perfect since 9 is obtained multiplicatively as 3^2 and is thus revealed composite).
Euler proved:
If 2^k - 1 is a prime number, then 2^(k-1)*(2^k - 1) is a perfect number and every even perfect number has this form.
The following seems to be true (is there a proof?):
If 2^k - 1 is an odd composite number taken as a spoof prime, then 2^(k-1)*(2^k - 1) is a "strong" spoof-perfect number and every even "strong" spoof-perfect number has this form?
There is only one known odd spoof-perfect number (found by Rene Descartes) but it is a "weak" spoof-perfect number (cf. 'Descartes numbers' and 'Unsolved problems in number theory' links below). (End)
a(n+1) = A173787(2*n+1,n); cf. A020522, A059153. - Reinhard Zumkeller, Feb 28 2010
Also, row sums of triangle A139251. - Omar E. Pol, May 25 2010
Starting with "1" = (1, 1, 2, 4, 8, ...) convolved with A002450: (1, 5, 21, 85, 341, ...); and (1, 3, 7, 15, 31, ...) convolved with A002001: (1, 3, 12, 48, 192, ...). - Gary W. Adamson, Oct 26 2010
a(n) is also the number of toothpicks in the corner toothpick structure of A153006 after 2^n - 1 stages. - Omar E. Pol, Nov 20 2010
The number of n-dimensional odd theta functions of half-integral characteristic. (Gunning, p.22) - Michael Somos, Jan 03 2014
a(n) = A000217((2^n)-1) = 2^(2n-1) - 2^(n-1) is the nearest triangular number below 2^(2n-1); cf. A007582, A233327. - Antti Karttunen, Feb 26 2014
a(n) is the sum of all the remainders when all the odd numbers < 2^n are divided by each of the powers 2,4,8,...,2^n. - J. M. Bergot, May 07 2014
Let b(m,k) = number of ways to form a sequence of m selections, without replacement, from a circular array of m labeled cells, such that the first selection of a cell whose adjacent cells have already been selected (a "first connect") occurs on the k-th selection. b(m,k) is defined for m >=3, and for 3 <= k <= m. Then b(m,k)/2m ignores rotations and reflection. Let m=n+2, then a(n) = b(m,m-1)/2m. Reiterated, a(n) is the (m-1)th column of the triangle b(m,k)/2m, whose initial rows are (1), (1 2), (2 6 4), (6 18 28 8), (24 72 128 120 16), (120 360 672 840 496 32), (720 2160 4128 5760 5312 2016 64); see A249796. Note also that b(m,3)/2m = n!, and b(m,m)/2m = 2^n. Proofs are easy. - Tony Bartoletti, Oct 30 2014
Beginning at a(1) = 1, this sequence is the sum of the first 2^(n-1) numbers of the form 4*k + 1 = A016813(k). For example, a(4) = 120 = 1 + 5 + 9 + 13 + 17 + 21 + 25 + 29. - J. M. Bergot, Dec 07 2014
a(n) is the number of edges in the (2^n - 1)-dimensional simplex. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete plane graph in 2^n points. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of linear elements in a complete parallelotope graph in n dimensions. - Dimitri Boscainos, Oct 05 2015
a(n) is the number of lattices L in Z^n such that the quotient group Z^n / L is C_4. - Álvar Ibeas, Nov 26 2015
a(n) gives the quadratic coefficient of the polynomial ((x + 1)^(2^n) + (x - 1)^(2^n))/2, cf. A201461. - Martin Renner, Jan 14 2017
Let f(x)=x+2*sqrt(x) and g(x)=x-2*sqrt(x). Then f(4^n*x)=b(n)*f(x)+a(n)*g(x) and g(4^n*x)=a(n)*f(x)+b(n)*g(x), where b is A007582. - Luc Rousseau, Dec 06 2018
For n>=1, a(n) is the covering radius of the first order Reed-Muller code RM(1,2n). - Christof Beierle, Dec 22 2021
a(n) =

Examples

			G.f. = x + 6*x^2 + 28*x^3 + 120*x^4 + 496*x^5 + 2016*x^6 + 8128*x^7 + 32640*x^8 + ...

References

  • V. K. Balakrishnan, Theory and problems of Combinatorics, "Schaum's Outline Series", McGraw-Hill, 1995, p. 69.
  • Martin Gardner, Mathematical Carnival, "Pascal's Triangle", p. 201, Alfred A. Knopf NY, 1975.
  • Richard K. Guy, Unsolved problems in number theory, (p. 72).
  • Ross Honsberger, Mathematical Gems, M.A.A., 1973, p. 113.
  • Clifford A. Pickover, Wonders of Numbers, Chap. 55, Oxford Univ. Press NY 2000.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Equals A006095(n+1) - A006095(n). In other words, A006095 gives the partial sums.
Cf. A000108, A007582, A010036, A016290, A003462, A079598, A134346, A048473, A151821, A010684.
Cf. A000043, A000396. - Omar E. Pol, Aug 30 2008
Cf. A109241, A139251, A153006. - Omar E. Pol, Aug 31 2008, May 25 2010, Nov 20 2010
Cf. A002450, A002001. - Gary W. Adamson, Oct 26 2010
Cf. A049072, A000384, A201461, A005059 (binomial transform, and special 5-letter words), A065442, A211705.
Cf. A171476.

Programs

  • GAP
    List([0..25],n->2^(n-1)*(2^n-1)); # Muniru A Asiru, Dec 06 2018
  • Haskell
    a006516 n = a006516_list !! n
a006516_list = 0 : 1 :
    zipWith (-) (map (* 6) $ tail a006516_list) (map (* 8) a006516_list)
-- Reinhard Zumkeller, Oct 25 2013
  • Magma
    [2^(n-1)*(2^n - 1): n in [0..30]]; // Vincenzo Librandi, Oct 31 2014
  • Maple
    GBC := proc(n,k,q) local i; mul( (q^(n-i)-1)/(q^(k-i)-1),i=0..k-1); end; # define q-ary Gaussian binomial coefficient [ n,k ]_q
[ seq(GBC(n+1,2,2)-GBC(n,2,2), n=0..30) ]; # produces A006516
A006516:=1/(4*z-1)/(2*z-1); # Simon Plouffe in his 1992 dissertation
seq(binomial(2^n, 2), n=0..19); # Zerinvary Lajos, Feb 22 2008
  • Mathematica
    Table[2^(n - 1)(2^n - 1), {n, 0, 30}] (* or *) LinearRecurrence[{6, -8}, {0, 1}, 30] (* Harvey P. Dale, Jul 15 2011 *)
  • Maxima
    A006516(n):=2^(n-1)*(2^n - 1)$ makelist(A006516(n),n,0,30); /* Martin Ettl, Nov 15 2012 */
  • PARI
    a(n)=(1<Charles R Greathouse IV, Jun 10 2011
  • PARI
    vector(100, n, n--; 2^(n-1)*(2^n-1)) \\ Altug Alkan, Oct 06 2015
  • Python
    for n in range(0, 30): print(2**(n-1)*(2**n - 1), end=', ') # Stefano Spezia, Dec 06 2018
  • Sage
    [lucas_number1(n,6,8) for n in range(24)]  # Zerinvary Lajos, Apr 22 2009
  • Sage
    [(4**n - 2**n) / 2 for n in range(24)]  # Zerinvary Lajos, Jun 05 2009

Formula

G.f.: x/((1 - 2*x)*(1 - 4*x)).
E.g.f. for a(n+1), n>=0: 2*exp(4*x) - exp(2*x).
a(n) = 2^(n-1)*Stirling2(n+1,2), n>=0, with Stirling2(n,m)=A008277(n,m).
Second column of triangle A075497.
a(n) = Stirling2(2^n,2^n-1) = binomial(2^n,2). - Ross La Haye, Jan 12 2008
a(n+1) = 4*a(n) + 2^n. - Philippe Deléham, Feb 20 2004
Convolution of 4^n and 2^n. - Ross La Haye, Oct 29 2004
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} 4^(n-j)*binomial(j,k). - Paul Barry, Aug 05 2005
a(n+2) = 6*a(n+1) - 8*a(n), a(1) = 1, a(2) = 6. - Daniele Parisse, Jul 28 2006 [Typo corrected by Yosu Yurramendi, Aug 06 2008]
Row sums of triangle A134346. Also, binomial transform of A048473: (1, 5, 17, 53, 161, ...); double bt of A151821: (1, 4, 8, 16, 32, 64, ...) and triple bt of A010684: (1, 3, 1, 3, 1, 3, ...). - Gary W. Adamson, Oct 21 2007
a(n) = 3*Stirling2(n+1,4) + Stirling2(n+2,3). - Ross La Haye, Jun 01 2008
a(n) = (4^n - 2^n)/2.
a(n) = A153006(2^n-1). - Omar E. Pol, Nov 20 2010
Sum_{n>=1} 1/a(n) = 2 * (A065442 - 1) = A211705 - 2. - Amiram Eldar, Dec 24 2020
a(n) = binomial(2*n+2, n+1) - Catalan(n+2). - N. J. A. Sloane, Apr 01 2021
a(n) = A171476(n-1), for n >= 1, and a(0) = 0. - Wolfdieter Lang, Jul 27 2022

A110200 Triangle, read by rows, where T(n,k) equals the sum of squares of numbers < 2^n having exactly k ones in their binary expansion.

Original entry on oeis.org

1, 5, 9, 21, 70, 49, 85, 395, 535, 225, 341, 1984, 3906, 3224, 961, 1365, 9429, 24066, 29274, 17241, 3969, 5461, 43434, 135255, 215900, 188595, 86106, 16129, 21845, 196095, 717825, 1412275, 1628175, 1106445, 411995, 65025, 87381, 872788
Offset: 1

Views

Author

Paul D. Hanna, Jul 16 2005

Keywords

Comments

Compare to triangle A110205 (sum of cubes).

Examples

			Row 4 is formed by sums of squares of numbers < 2^4:
T(4,1) = 1^2 + 2^2 + 4^2 + 8^2 = 85;
T(4,2) = 3^2 + 5^2 + 6^2 + 9^2 + 10^2 + 12^2 = 395;
T(4,3) = 7^2 + 11^2 + 13^2 + 14^2 = 535;
T(4,4) = 15^2 = 225.
Triangle begins:
1;
5, 9;
21, 70, 49;
85, 395, 535, 225;
341, 1984, 3906, 3224, 961;
1365, 9429, 24066, 29274, 17241, 3969;
5461, 43434, 135255, 215900, 188595, 86106, 16129;
21845, 196095, 717825, 1412275, 1628175, 1106445, 411995, 65025;
87381, 872788, 3662848, 8541876, 12197570, 10974236, 6095208, 1915228, 261121; ...
Row g.f.s are:
row 1: (1 + 1*x)/(1+x);
row 2: (5 + 9*x);
row 3: (21 + 49*x)*(1+x);
row 4: (85 + 225*x)*(1+x)^2.
G.f. for row n is:
((4^n-1)/3 + (2^n-1)^2*x)*(1+x)^(n-2).

Links

Crossrefs

Cf. A110201 (central terms), A002450 (column 1), A110202 (column 2), A110203 (column 3), A110204 (column 4), A016290 (row sums), A110205.

Programs

  • PARI
    T(n,k)=(4^n-1)/3*binomial(n-2,k-1)+(2^n-1)^2*binomial(n-2,k-2)
for(n=1, 15, for(k=1, n, print1(T(n, k), ", ")); print(""))
  • PARI
    /* Using G.f. of A(x,y): */
T(n,k)=my(X=x+x*O(x^n),Y=y+y*O(y^k));if(n
  • PARI
    /* Sum of Squares of numbers<2^n with k 1-bits: */
T(n,k)=my(B=vector(n+1));if(n

Formula

T(n,k) = (4^n-1)/3 * C(n-2, k-1) + (2^n-1)^2 * C(n-2, k-2).
G.f.: A(x,y) = x*y*(1-2*x*(1-y)) / ((1-x*(1+y))*(1-2*x*(1+y))*(1-4*x*(1+y))).
G.f. for row n: ((4^n-1)/3 + (2^n-1)^2*x)*(1+x)^(n-2).

A340312 Triangle read by rows: T(n,k) is the number of subsets of {0..2^n-1} with k elements such that the bitwise-xor of all the subset members gives zero, 0 <= k <= 2^n.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 7, 14, 7, 0, 1, 1, 1, 1, 0, 35, 140, 273, 448, 715, 870, 715, 448, 273, 140, 35, 0, 1, 1, 1, 1, 0, 155, 1240, 6293, 27776, 105183, 330460, 876525, 2011776, 4032015, 7063784, 10855425, 14721280, 17678835, 18796230, 17678835, 14721280, 10855425, 7063784, 4032015, 2011776, 876525, 330460, 105183, 27776, 6293, 1240, 155, 0, 1, 1
Offset: 0

Views

Author

Jianing Song, Jan 04 2021

Keywords

Comments

Sum_{k=0..2^n} T(n, k) gives the total number of subsets with bitwise-xor of all the subset members zero. There are in total 2^(2^n - n) such subsets of {0, 1, ..., 2^n-1}, see A300361 and the Mathematics Stack Exchange link below.
Equivalently, T(n, k) is the number of subsets of the vector space (F_2)^n such that the sum of elements in the subset is the zero vector.
T(n, k) is symmetric, that is, T(n, k) = T(n, 2^n-k) for k = 0..2^n, since if the bitwise-xor of the members in S is zero, then the complement of S in {0, 1, ..., 2^n-1} also has this property.

Examples

			Triangle begins:
[0]  1, 1;
[1]  1, 1, 0;
[2]  1, 1, 0, 1, 1;
[3]  1, 1, 0, 7, 14, 7, 0, 1, 1;
[4]  1, 1, 0, 35, 140, 273, 448, 715, 870, 715, 448, 273, 140, 35, 0, 1, 1;
[5]  1, 1, 0, 155, 1240, 6293, 27776, 105183, 330460, 876525, 2011776, 4032015, 7063784, 10855425, 14721280, 17678835, 18796230, 17678835, 14721280, 10855425, 7063784, 4032015, 2011776, 876525, 330460, 105183, 27776, 6293, 1240, 155, 0, 1, 1;
T(n,0) = 1 since the bitwise-xor of all the elements in the empty set is the identity of bitwise-xor (0), hence the empty set meets the requirement.
T(n,1) = 1 since the only such subset is {0}.
T(n,2) = 0 since no distinct a, b have a ^ b = 0.
T(n,3) = A006095(n): if distinct a, b, c have a ^ b ^ c = 0, then c = a ^ b, and a, b must both be nonzero since a = 0 implies b = c. On the other hand, if a, b are nonequal and are both nonzero, then c = a ^ b has c != a and c != b since c = a implies b = 0. So the total number of triples (a, b, c) is (2^n-1)*(2^n-2), and the total number of subsets {a, b, c} is (2^n-1)*(2^n-2)/3! = A006095(n).
T(n,4) = A016290(n-2): if distinct a, b, c, d have a ^ b ^ c ^ d = 0, then d = a ^ b ^ c. On the other hand, if a, b, c are distinct, then d = a ^ b ^ c has d != a, d != b, d != c since d = a implies b = c. So the total number of quadruples (a, b, c, d) is 2^n*(2^n-1)*(2^n-2), and the total number of subsets {a, b, c, d} is 2^n*(2^n-1)*(2^n-2)/4! = A016290(n-2).

Links

Crossrefs

Cf. A000120 (hamming weight of n), A300361 (row sums).
Cf. A340263 (irreducible (?) factor if T(n,k) is seen as representing polynomials).
Cf. A340259(n) = T(n, 2^(n-1)), the central term of row n.
Cf. A340030 (case that only nonzero elements allowed).
Cf. A054669, A058878.
Cf. A006095 (k=3 column), A016290 (k=4 column); cf. also A010080-A010084 and A281123. - Jon E. Schoenfield, Jan 06 2021

Programs

  • C
    /* Generating program for T(4,k), see link. */
  • Maple
    A340312_row := proc(n) local a, b, c; c := 2^(n-1);
if n = 0 then return [1, 1] fi;
b := n -> add(binomial(2^n, 2*k)*x^(2*k), k = 0..2^n);
a := n -> x*mul(b(k), k = 0..n);
(x + 1)^c*(b(n-1) - (c-1)*a(n-2));
[seq(coeff(expand(%), x, j), j = 0..2*c)] end:
for n from 0 to 6 do A340312_row(n) od; # Peter Luschny, Jan 06 2021
  • Mathematica
    T[n_, k_] := Binomial[2^n, k]/2^n + If[EvenQ[k], (-1)^(k/2)*(1-1/2^n)* Binomial[2^(n-1), k/2], 0];
Table[T[n, k], {n, 0, 5}, {k, 0, 2^n}] // Flatten (* Jean-François Alcover, Jan 14 2021, after Andrew Howroyd *)
  • PARI
    T(n, k)={binomial(2^n, k)/2^n + if(k%2==0, (-1)^(k/2)*(1-1/2^n)*binomial(2^(n-1), k/2))} \\ Andrew Howroyd, Jan 09 2021
  • SageMath
    def A340312():
    a, b, c = 1, 1, 1
    yield [1, 1]
    yield [1, 1, 0]
    while True:
        c *= 2
        a *= b
        b = sum(binomial(c, 2 * k) * x^(2 * k) for k in range(c + 1))
        p = (x + 1)^c * (b - (c - 1) * x * a)
        yield expand(p).list()
A340312_row = A340312()
for _ in range(6):
    print(next(A340312_row)) # Peter Luschny, Jan 07 2021

Formula

T(n, k) = [x^k] p(n; x) where p(n; x) = (x + 1)^c*(b(n-1) - (c-1)*a(n-2)), b(n) = Sum_{k=0..2^n} binomial(2^n, 2*k)*x^(2*k), a(n) = x*Product_{k=0..n} b(k) and c = 2^(n-1), for n >= 1. - Peter Luschny, Jan 06 2021
T(n+1, k) = [x^k] (x+1)^(2^n)*p_n(x) where p_n(x) are the polynomials defined in A340263. - Peter Luschny, Jan 06 2021
From Andrew Howroyd, Jan 09 2021: (Start)
First take any subset of k-1 elements and append the bitwise-xor of the elements. The final element will either be a duplicate or not and consideration of the two cases leads to a formula linking T(n,k) and T(n,k-2) with binomial(2^n,k-1).
T(n, k) = (1/k)*(binomial(2^n,k-1) - (2^n-(k-2))*T(n,k-2)) for k >= 2.
T(n, k) = binomial(2^n, k)/2^n for odd k.
T(n, k) = binomial(2^n, k)/2^n + (-1)^(k/2)*(1-1/2^n)*binomial(2^(n-1), k/2) for even k.
T(n, k) = [x^k] ((1+x)^(2^n) + (2^n-1)*(1-x^2)^(2^(n-1)))/2^n.
T(n, k) = A340030(n,k-1) + A340030(n,k).
(End)

Extensions

More terms from Andrew Howroyd and Jon E. Schoenfield.

A028258 Expansion of 1/((1-2*x)*(1-4*x)(1-8*x)(1-16*x)).

Original entry on oeis.org

1, 30, 620, 11160, 188976, 3108960, 50434240, 812507520, 13044728576, 209073047040, 3348029967360, 53591377582080, 857645259698176, 13723790036459520, 219592368170516480, 3513571713573027840, 56217898008516427776, 899492372901406310400
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A006516, A016290, A006097.

Programs

  • Mathematica
    CoefficientList[Series[1/((1-2x)(1-4x)(1-8x)(1-16x)),{x,0,50}],x] (* or *) LinearRecurrence[{30,-280,960,-1024},{1,30,620,11160},50] (* or *) Table[(2^(n-1)(2^n-1)(2^(n+1)-1)(2^(n+2)-1))/21,{n,20}] (* Harvey P. Dale, Jun 18 2011 *)
  • PARI
    a(n)=(2^n*(2^(n+1)-1)*(2^((n+1)+1)-1)*(2^(n+3)-1))/21 \\ Charles R Greathouse IV, Feb 10 2017

Formula

Difference of Gaussian binomial coefficients [ n+1, 4 ]-[ n, 4 ] (n >= 3).
a(n) = 30*a(n-1)-280*a(n-2)+960*a(n-3)-1024*a(n-4), with a(0)=1, a(1)=30, a(2)=620, a(3)=11160. - Harvey P. Dale, Jun 18 2011
a(n) = (2^n*(2^(n+1)-1)*(2^((n+1)+1)-1)*(2^(n+3)-1))/21. - Harvey P. Dale, Jun 18 2011; offset corrected by Charles R Greathouse IV, Feb 10 2017
E.g.f.: exp(2*x)*(64*exp(14*x) - 56*exp(6*x) + 14*exp(2*x) - 1)/21. - Stefano Spezia, Jun 23 2022

A020528 3rd Bernoulli polynomial evaluated at powers of 2 (multiplied by 6).

Original entry on oeis.org

0, 18, 252, 2520, 22320, 187488, 1536192, 12435840, 100074240, 802948608, 6433016832, 51501864960, 412165877760, 3297930928128, 26385863196672, 211096568954880, 1688811205754880, 13510644263682048
Offset: 0

Views

Author

Simon Plouffe

Keywords

Links

Crossrefs

Cf. A016290, A020527.

Programs

  • Maple
    seq(6*bernoulli(3,2^i),i=0..24);
  • Mathematica
    6*BernoulliB[3, 2^Range[0, 30]] (* Paolo Xausa, Sep 16 2024 *)

Formula

a(n) = 3*2^(n+1)*(2^(n+1)-1)*(2^(n+1)-2)/4. - Ralf Stephan, Apr 06 2004
From R. J. Mathar, May 22 2013: (Start)
a(n) = 18*A016290(n-1).
G.f.: -18*x/((8*x-1)*(2*x-1)*(4*x-1)). (End)
a(n) = 14*a(n-1) - 56*a(n-2) + 64*a(n-3). - Wesley Ivan Hurt, Mar 02 2023

A140047 Triangle, read by rows: T(n,k) = (1/2)*Sum_{j=0..2^n-1} j^k for k=0..n-1, n>=1; related to the Prouhet-Tarry-Escott problem.

Original entry on oeis.org

1, 2, 3, 4, 14, 70, 8, 60, 620, 7200, 16, 248, 5208, 123008, 3098760, 32, 1008, 42672, 2032128, 103223568, 5461682688, 64, 4064, 345440, 33032192, 3369214496, 357969864704, 39119789090720, 128, 16320, 2779840, 532684800, 108880217152
Offset: 1

Views

Author

Paul D. Hanna, May 12 2008

Keywords

Examples

			Triangle begins:
1;
2, 3;
4, 14, 70;
8, 60, 620, 7200;
16, 248, 5208, 123008, 3098760;
32, 1008, 42672, 2032128, 103223568, 5461682688;
64, 4064, 345440, 33032192, 3369214496, 357969864704, 39119789090720; ...
For n=3, since A010060(k) = 0 at k={0,3,5,6}, then
T(3,k) = 0^k + 3^k + 5^k + 6^k for k=0..2;
and since A010060(k) = 1 at k={1,2,4,7}, we also have
T(3,k) = 1^k + 2^k + 4^k + 7^k for k=0..2.
For n=4, since A010060(k) = 0 at k={0,3,5,6,9,10,12,15}, then
T(4,k) = 0^k + 3^k + 5^k + 6^k + 9^k + 10^k + 12^k + 15^k for k=0..3;
and since A010060(k) = 1 at k={1,2,4,7,8,11,13,14}, we also have
T(4,k) = 1^k + 2^k + 4^k + 7^k + 8^k + 11^k + 13^k + 14^k for k=0..3.

Crossrefs

Cf. A140048 (main diagonal), A010060, A016290.

Programs

  • PARI
    {T(n,k)=(1/2)*sum(j=0,2^n-1,j^k)}
  • PARI
    {T(n,k)=local(Tnk=0);for(j=0,2^n-1,if(subst(Pol(binary(j)),x,1)%2==0,Tnk+=j^k));Tnk}

Formula

T(n,k) = Sum_{j=0..2^n-1, A010060(j)=0 } j^k for k=0..n-1, n>=1; also,
T(n,k) = Sum_{j=0..2^n-1, A010060(j)=1 } j^k for k=0..n-1, n>=1;
where A010060 is the Thue-Morse sequence (identity due to Prouhet).
T(n,0) = 2^n; T(n,1) = 4^n - 2^(n-1); T(n,2) = A016290(n)/2;
T(n,n-1) = A140048(n).
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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