A037293 -id:A037293 - cp's OEIS frontend

A058077 Binomial coefficients formed from consecutive primes: a(n) = binomial( prime(n+1), prime(n) ).

3, 10, 21, 330, 78, 2380, 171, 8855, 475020, 465, 2324784, 101270, 903, 178365, 22957480, 45057474, 1830, 99795696, 971635, 2628, 277962685, 1837620, 581106988, 144520208820, 4082925, 5253, 5160610, 5886, 6438740

Offset: 1

Labos Elemer, Nov 13 2000

nonn

Conjecture: for each value of n > 1, if a(n+1) has the same number of digits as a(n) and a(n+1) > a(n), then prime(n+2) - prime(n+1) = prime(n+1) - prime(n). This conjecture has been verified for all n < 3*10^7. - Ahmad J. Masad, Oct 08 2019

			n=6: a(6) = C(p(7),p(6)) = C(17,13) = 57120/24 = 2380.

Michel Marcus, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Colored logarithmic scatterplot of the first 100000 terms (where the color is function of A001223(n))

Cf. A000040, A001223.

Cf. A037293, A066526, A080911, A277341.

Table[Binomial[Prime[n+1],Prime[n]],{n,1,20}] (* Vaclav Kotesovec, Nov 13 2014 *)

a(n) = binomial(A000040(n+1), A001223(n)).

Offset corrected by Vaclav Kotesovec, Nov 13 2014

A201461 Triangle read by rows: n-th row (n>=0) gives coefficients of the polynomial ((x+1)^(2^n) + (x-1)^(2^n))/2.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 28, 70, 28, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 496, 35960, 906192, 10518300, 64512240, 225792840, 471435600, 601080390, 471435600, 225792840, 64512240, 10518300, 906192, 35960, 496, 1

Offset: 0

N. J. A. Sloane, Dec 01 2011

Wanted: reference for the fact that these polynomials are irreducible. Washington, Cyclotomic Fields, perhaps?
The algorithm r(n) = (1/2)*(r(n-1) + A/r(n-1)), starting with r(0) = A, used for approximating sqrt(A), which is known as the Babylonian method or Hero's method after the first-century Greek mathematician Hero of Alexandria and which can be derived from Newton's method, generates fractions beginning with (A+1)/2, (A^2 + 6*A + 1)/(4*(A+1)), (A^4 + 28*A^3 + 70*A^2 + 28*A + 1)/(8*(A+1)*(A^2 + 6*A + 1)), ... This is p(n,sqrt(A))/(2^n*Product_{k=1..n-1} p(k,sqrt(A))) with the given polynomial p(n,x) = ((x+1)^(2^n) + (x-1)^(2^n))/2. - Martin Renner, Jan 11 2017
The quadratic coefficient of this polynomial is A006516(n), the even-indexed coefficients are binomial(2^n,2*k) or A086645(2^(n-1),k) for 0 <= k <= 2^(n-1), in each row the maximum central coefficient for n>=2 is A037293(n) or A000984(2^(n-1)). - Martin Renner, Jan 14 2017
T(n,k) and A281122 are a bisection of row 2^n of Pascal's triangle A007318. - Martin Renner, Jan 15 2017
For nonnegative real x, sqrt(x) = (2*x/(1 + x)) * (2*(1 + x)^2/(1 + 6*x + x^2)) * (2*(1 + 6*x + x^2)^2/(1 + 28*x + 70*x^2 + 28*x^3 + x^4)) * .... See Bauer. - Peter Bala, Jan 18 2022

			The first few polynomials are:
1,
x^2 + 1,
x^4 + 6*x^2 + 1,
x^8 + 28*x^6 + 70*x^4 + 28*x^2 + 1,
x^16 + 120*x^14 + 1820*x^12 + 8008*x^10 + 12870*x^8 + 8008*x^6 + 1820*x^4 + 120*x^2 + 1.
The triangle of coefficients begins:
[0] [1]
[1] [1, 0, 1]
[2] [1, 0, 6, 0, 1]
[3] [1, 0, 28, 0, 70, 0, 28, 0, 1]
[4] [1, 0, 120, 0, 1820, 0, 8008, 0, 12870, 0, 8008, 0, 1820, 0, 120, 0, 1]
The triangle of nonzero coefficients begins:
[0] 1
[1] 1, 1
[2] 1, 6, 1
[3] 1, 28, 70, 28, 1
[4] 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1
[5] 1, 496, 35960, 906192, 10518300, 64512240, 225792840, 471435600, 601080390, 471435600, 225792840, 64512240, 10518300, 906192, 35960, 496, 1
...

Indranil Ghosh, Rows 0..11, flattened
F. L. Bauer, Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.

Cf. A000984, A006516, A007318, A037293, A086645, A281122.

Flatten[Table[Binomial[2^n,2k],{n,0,6},{k,0,2^(n-1)}]] (* Indranil Ghosh, Feb 22 2017 *)

row(n) = my(v = Vec(((x+1)^(2^n)+(x-1)^(2^n))/2)); vector(#v\2 + 1, k, v[2*k-1]); \\ Michel Marcus, Jan 14 2017

T(n,k)=binomial(2^n,2*k);
for(n=0,5,for(k=0,2^(n-1),print1(T(n,k),", "));print()); \\ Joerg Arndt, Jan 15 2017

def A201461_polynomial(n): return expand(((x+1)^(2^n) + (x-1)^(2^n))/2)
for n in range(6): print(A201461_polynomial(n))
for n in range(6): print(A201461_polynomial(n).list()) # coefficients
for n in range(6): # depunched (not a mathematical operation)
    if n == 0: print([1])
    else: print(A201461_polynomial(n).list()[::2]) # Peter Luschny, Jan 11 2021

T(n,k) = binomial(2^n,2*k). - Joerg Arndt, Jan 15 2017

A340263 T(n, k) = [x^k] ((1-x)^(2^n) + 2^(-n)((2^n-1)(x-1)^(2^n) + (x+1)^(2^n)))/2. Irregular triangle read by rows, for n >= 0 and 0 <= k <= 2^n.

Original entry on oeis.org

1, 1, -1, 1, 1, -3, 6, -3, 1, 1, -7, 28, -49, 70, -49, 28, -7, 1, 1, -15, 120, -525, 1820, -4095, 8008, -10725, 12870, -10725, 8008, -4095, 1820, -525, 120, -15, 1

Offset: 0

Peter Luschny, Jan 06 2021

Conjecture: for n >= 1 the polynomials are irreducible.

			Polynomials begin:
[0] 1;
[1] x^2 - x + 1;
[2] x^4 - 3*x^3 + 6*x^2 - 3*x + 1;
[3] x^8 - 7*x^7 + 28*x^6 - 49*x^5 + 70*x^4 - 49*x^3 + 28*x^2 - 7*x + 1;
Triangle begins:
[0] [1]
[1] [1, -1, 1]
[2] [1, -3, 6, -3, 1]
[3] [1, -7, 28, -49, 70, -49, 28, -7, 1]
[4] [1, -15, 120, -525, 1820, -4095, 8008, -10725, 12870, -10725, 8008, -4095, 1820, -525, 120, -15, 1]

Peter Luschny, Table of n, a(n) for n = 0..1031

Row sums are 2^(2^n - n - 1) = A016031(n-1).
Central terms of the rows are A037293(n) for n >= 2.
Cf. A340312.

A340263_row := proc(n) local a, b;
if n = 0 then return [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);
expand(b(n) - (2^n-1)*a(n-1));
[seq(coeff(%, x, j), j = 0..2^n)] end:
for n from 0 to 5 do A340263_row(n) od;
# Alternatively:
CoeffList := p -> [op(PolynomialTools:-CoefficientList(p, x))]:
Tpoly := n -> ((1-x)^(2^n) + 2^(-n)*((2^n-1)*(x-1)^(2^n) + (x + 1)^(2^n)))/2:
seq(print(CoeffList(Tpoly(n))), n=0..5); # Peter Luschny, Feb 03 2021

def A340263():
    a, b, c = 1, 1, 1
    yield [1]
    while True:
        c *= 2
        a *= b
        b = sum(binomial(c, 2 * k) * x ^ (2 * k) for k in range(c + 1))
        yield ((b - (c - 1) * x * a)).list()
A340263_row = A340263()
for _ in range(6):
    print(next(A340263_row))

Let p_n(x) = b(n) - (2^n-1)*a(n-1), b(n) = Sum_{k=0..2^n} binomial(2^n, 2*k)* x^(2*k), and a(n) = x*Product_{k=0..n} b(k). Then T(n, k) = [x^k] p_n(x).

Shorter name by Peter Luschny, Feb 03 2021

A000721 Number of NP-equivalence classes of balanced Boolean functions of n variables.

Original entry on oeis.org

1, 2, 6, 74, 169112, 39785643746726, 37126652766640082937217814348006, 558874591495497577231218517843968898077072559983411918227348931497772

Offset: 1

N. J. A. Sloane

These are distinct groups of Boolean functions of n variables that output an equal number of 0s and 1s across all possible inputs (balanced), and cannot be transformed into each other by negating or permuting inputs. - Aniruddha Biswas, Nov 12 2024

			For n = 2 the a(2) = 2. Solutions are f(x,y)=x and f(x,y)=x⊕y; are two 2-variable NP-inequivalent balanced Boolean functions. - _Aniruddha Biswas_, Nov 12 2024

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Herman Jamke, Table of n, a(n) for n = 1..10
B. Elspas, Self-complementary symmetry types of Boolean functions, IEEE Trans. Electron. Computers, 9 (1960), 264-266.
B. Elspas, Self-complementary symmetry types of Boolean functions, IEEE Transactions on Electronic Computers 2, no. EC-9 (1960): 264-266. [Annotated scanned copy]
E. M. Palmer and R. W. Robinson, Enumeration of self-dual configurations, Pacific J. Math., 110 (1984), 203-221. [From Herman Jamke (hermanjamke(AT)fastmail.fm), Aug 21 2010]
D. Zeilberger, First 7 terms of the sequence of weight-enumerators enumerating equivalence classes of Boolean functions under permutation of variable and negation . [From Herman Jamke (hermanjamke(AT)fastmail.fm), Aug 21 2010]
D. Zeilberger, First 7 terms of the sequence of weight-enumerators enumerating equivalence classes of Boolean functions under permutation of variable and negation [Local copy]
Index entries for sequences related to Boolean functions

Cf. A037293, A000370.

a(n) is asymptotic to binomial(2^n, 2^(n-1)) / ( n! * 2^n ) as n -> oo. - Aniruddha Biswas, Dec 16 2024

More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Aug 21 2010, Sep 05 2010

A069954 a(n) = binomial(2^(n+1), 2^n)/2 = binomial(2^(n+1) - 1, 2^n) = binomial(2^(n+1) - 1, 2^n-1).

Original entry on oeis.org

1, 3, 35, 6435, 300540195, 916312070471295267, 11975573020964041433067793888190275875, 2884329411724603169044874178931143443870105850987581016304218283632259375395

Offset: 0

Benoit Cloitre, Apr 27 2002

Terms are always odd. a(1) = A061548(2), a(2) = A061548(3), a(3) = A061548(5), a(4) = A061548(9), a(5) = A061548(17), ... Hence it seems that a(n) = A061548(A000051(n)).

C(2*k, k)/2 = C(2*k-1, k) = C(2*k-1, k-1) is odd if and only if k = 2^n. - Michael Somos, Mar 12 2014

			C(2,1)/2 = C(1,0) = C(1,1) = 1. C(4,2)/2 = C(3,1) = C(3,2) = 3. C(8,4)/2 = C(7,3) = C(7,4) = 35. - _Michael Somos_, Mar 12 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..10

Cf. A000051, A000108, A000984, A001790, A061548, A037293.

[Binomial(2^(n+1)-1, 2^n-1): n in [0..10]]; // Vincenzo Librandi, Mar 14 2014

Table[Binomial[2^(n+1) -1, 2^n -1], {n, 0, 10}] (* Vincenzo Librandi, Mar 14 2014 *)

[binomial(2^(n+1) -1, 2^n) for n in (0..9)] # G. C. Greubel, Aug 16 2022

From Harry Richman, May 18 2023: (Start)
a(n) = A001790(2^n).
a(n) = 1/2 * A000984(2^n).
a(n) = 1/2 * (2^n + 1) * A000108(2^n).
log log a(n) ~ (log 2) * (n + 1) + log log 2 + O(n / 2^n). (End)
a(n) = A037293(n+1) / 2. - Tilman Piesk, Oct 11 2024

a(0) = 1 added by Michael Somos, Mar 12 2014

A080911 a(n) = binomial(n!,(n-1)!).

Original entry on oeis.org

1, 2, 15, 134596, 10872202353646160680764975

Offset: 1

Labos Elemer, Apr 01 2003

nonn

Amiram Eldar, Table of n, a(n) for n = 1..7

Cf. A000142, A037293, A066526, A058077.

Table[Binomial[n!, (n-1)!], {n, 1, 7}] (* Vaclav Kotesovec, Nov 13 2014 *)

A345135 Number of ordered rooted binary trees with n leaves and with minimal Sackin tree balance index.

Original entry on oeis.org

1, 1, 1, 2, 1, 4, 6, 4, 1, 8, 28, 56, 70, 56, 28, 8, 1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 11440, 8008, 4368, 1820, 560, 120, 16, 1, 32, 496, 4960, 35960, 201376, 906192, 3365856, 10518300, 28048800, 64512240, 129024480, 225792840, 347373600, 471435600

Offset: 0

Mareike Fischer, Jun 09 2021

Ordered rooted binary trees are trees with two descendants per inner node where left and right are distinguished.
The Sackin tree balance index is also known as total external path length.
a(0) = 1 by convention.

			a(1) = a(2) = 1 because there are only the trees (o) and (o,o) which get counted. a(3) = 2 because the trees ((o,o),o) and (o,(o,o)) get counted. a(4) = 1 because only the tree ((o,o),(o,o)) is counted. Note that the other possible rooted binary ordered trees with four leaves, namely the different orderings of (((o,o),o),o), are not Sackin minimal. a(5) = 4 because the following trees get counted: (((o,o),o),(o,o)), ((o,(o,o)),(o,o)), ((o,o),((o,o),o)), ((o,o),(o,(o,o))).

Alois P. Heinz, Table of n, a(n) for n = 0..4096
Mareike Fischer, Extremal Values of the Sackin Tree Balance Index. Ann. Comb. 25, 515-541 (2021), Theorem 7.

Cf. A000079, A000108, A011782, A037293, A051179, A053644, A299037.

a:= n-> (b-> binomial(b, n-b))(2^ilog2(n)):
seq(a(n), n=0..46);  # Alois P. Heinz, Jun 09 2021

a[0] := 1; a[n_] := Module[{k = 2^(BitLength[n] - 1)}, Binomial[k, n - k]];
Table[a[n], {n, 0, 46}]

a(n) = binomial(2^(ceiling(log_2(n))-1),n-2^(ceiling(log_2(n))-1)).
a(n) = binomial(A053644(n),n-A053644(n)).
a(2^n) = 1.
a(2^n-1) = A011782(n).
a(2^n+1) = A000079(n).
From Alois P. Heinz, Jun 09 2021: (Start)
max({ a(k) | k = 2^n..2^(n+1) }) = A037293(n).
Sum_{i=2^n..2^(n+1)-1} a(i) = 2^(2^n) - 1 = A051179(n). (End)
Conjecture: Sum_{n>=0} a(n)*x^n = 1 + x + Sum_{n>=0} x^(2^n) * ((1 + x)^(2^n) - 1). - Paul D. Hanna, Jul 18 2024

A211600 a(n) = (binomial(p^n, p^(n-1)) - binomial(p^(n-1), p^(n-2))) / p^(3n-3) for p = 2.

Original entry on oeis.org

1, 25, 146745, 55927250376633, 91366371314728099305354933301689, 2750710880016902131123422793322699970110063817946068739768171777481145

Offset: 3

Alexander Adamchuk, Apr 16 2012

Consider the difference between two binomials f(p,k) = binomial(p^k, p^(k-1)) - binomial(p^(k-1), p^(k-2)).
A theorem from the A. I. Shirshov paper (in Russian) states:
p^(3k - 3) divides f(p,k) for prime p = 2 and k > 2.
p^(3k - 2) divides f(p,k) for prime p = 3 and k > 1.
p^(3k - 1) divides f(p,k) for prime p > 3 and k > 1.

			a(3) = 1 is the difference between central binomials C(8,4) - C(4,2) = 70 - 6 = 64 divided by 2^(3*2 - 3) = 64.

D. B. Fuks and Serge Tabachnikov, Mathematical Omnibus: Thirty Lectures on Classic Mathematics, American Mathematical Society, 2007. Lecture 2. Arithmetical Properties of Binomial Coefficients, pages 27-44

D. B. Fuks and M. B. Fuks, Arithmetics of binomial coefficients, Kvant 6 (1970), 17-25. (in Russian)
A. I. Shirshov, On one property of binomial coefficients, Kvant 10 (1971), 16-20. (in Russian)

Cf. A037293, A211601, A211602.

A211600:=n->(binomial(2^n, 2^(n - 1)) - binomial(2^(n - 1), 2^(n - 2))) / 2^(3*n - 3): seq(A211600(n), n=3..9); # Wesley Ivan Hurt, Apr 25 2017

p = 2; Table[(Binomial[p^n, p^(n - 1)] - Binomial[p^(n - 1), p^(n - 2)]) / 2^(3n - 3), {n, 3, 9}]

a(n) = (binomial(2^n, 2^(n-1)) - binomial(2^(n-1), 2^(n-2))) / 2^(3*n-3).

a(n) = (A037293(n) - A037293(n-1)) / 2^(3*n - 3).

A112884 Number of bits required to represent binomial(2^n, 2^(n-1)).

Original entry on oeis.org

1, 2, 3, 7, 14, 30, 61, 125, 252, 508, 1019, 2043, 4090, 8186, 16377, 32761, 65528, 131064, 262135, 524279, 1048566, 2097142, 4194293, 8388597, 16777204, 33554420, 67108851, 134217715, 268435442, 536870898, 1073741809, 2147483633, 4294967280, 8589934576, 17179869167

Offset: 0

Matt Erbst (matt(AT)erbst.org), Oct 04 2005

			a(2) = 3 because binomial(2^2, 2^1) in binary = 110.

Alois P. Heinz, Table of n, a(n) for n = 0..3321
Sela Fried, The number of bits required to represent binomial(2^n, 2^(n-1)), 2024.
Sela Fried, Proofs of some Conjectures from the OEIS, arXiv:2410.07237 [math.NT], 6 October 2024.
Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,2).

a(n) represents the size of A037293 in binary - see also the central binomial coefficients: A001405.

Cf. A000079, A004526, A070939.

Table[IntegerLength[Binomial[2^n,2^(n-1)],2],{n,25}] (* or *)
CoefficientList[Series[(-2 x^3+3x-2)/((x-1)^2 (2x^2+x-1)), {x,0,25}], x] (* Harvey P. Dale, Apr 06 2011 *)

$LastFact = gmp_init('1'); for ($i = 2; $i !== 65536; $i *= 2) { $Fact = gmp_fact($i); $Result = gmp_div_q($Fact, gmp_pow($OldFact, 2)); $LastFact = $Fact; echo gmp_strval($Result, 2).'
'; }

Appears to be equal to 2^n - floor(n/2) = A000079(n) - A004526(n).
G.f.: (-3*x^3 + 2*x^2 + x - 1)/((x - 1)^2*(2*x^2 + x - 1)). - Conjectured by Harvey P. Dale, Apr 06 2011
a(n) = A070939(A037293(n)). - Alois P. Heinz, Feb 17 2024
The conjectured formula 2^n - floor(n/2) and consequent g.f. are true (see links). - Sela Fried, Oct 03 2024

a(0)=1 prepended and g.f. adapted by Alois P. Heinz, Oct 11 2024

A378302 Number of nondegenerate balanced Boolean functions of n variables.

Original entry on oeis.org

0, 2, 2, 58, 12618, 601016690, 1832624137336299922, 23951146041928082853307218802404658090, 5768658823449206338089748357862286887548602533639737369730665340966207267034

Offset: 0

Aniruddha Biswas, Nov 22 2024

A Boolean function is degenerate on some variable if its output does not depend on the variable, and it is said to be non-degenerate if it is not degenerate on any variable.

Aniruddha Biswas and Palash Sarkar, Counting unate and balanced monotone Boolean functions, arXiv:2304.14069 [math.CO], 2023.
Aniruddha Biswas and Palash Sarkar, Counting Unate and Monotone Boolean Functions Under Restrictions of Balancedness and Non-Degeneracy, J. Int. Seq. (2025) Vol. 28, Art. No. 25.3.4. See pp. 4, 6.

Cf. A037293, A000618.

a[n_]:=Sum[(-1)^(n-i)*Binomial[n,i]*Binomial[2^i,2^(i-1)],{i,n}]; Array[a,9,0] (* Stefano Spezia, Nov 24 2024 *)

from math import comb
def A378302(n): return sum(-comb(n,i)*comb(1<Chai Wah Wu, Dec 11 2024

a(n) = Sum_{i=1..n} (-1)^(n-i) * binomial(n,i) * binomial(2^i,2^(i-1)).

cp's OEIS Frontend

A058077 Binomial coefficients formed from consecutive primes: a(n) = binomial( prime(n+1), prime(n) ).

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A201461 Triangle read by rows: n-th row (n>=0) gives coefficients of the polynomial ((x+1)^(2^n) + (x-1)^(2^n))/2.

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A340263 T(n, k) = [x^k] ((1-x)^(2^n) + 2^(-n)*((2^n-1)*(x-1)^(2^n) + (x+1)^(2^n)))/2. Irregular triangle read by rows, for n >= 0 and 0 <= k <= 2^n.

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A000721 Number of NP-equivalence classes of balanced Boolean functions of n variables.

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A069954 a(n) = binomial(2^(n+1), 2^n)/2 = binomial(2^(n+1) - 1, 2^n) = binomial(2^(n+1) - 1, 2^n-1).

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A080911 a(n) = binomial(n!,(n-1)!).

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A345135 Number of ordered rooted binary trees with n leaves and with minimal Sackin tree balance index.

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A340263 T(n, k) = [x^k] ((1-x)^(2^n) + 2^(-n)((2^n-1)(x-1)^(2^n) + (x+1)^(2^n)))/2. Irregular triangle read by rows, for n >= 0 and 0 <= k <= 2^n.