Pietro Majer - cp's OEIS frontend

User: Pietro Majer

Pietro Majer has authored 6 sequences.

A158118 Number of solutions of +-1+-2^3+-3^3..+-n^3=0.

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 4, 2, 0, 0, 4, 124, 0, 0, 536, 712, 0, 0, 4574, 2260, 0, 0, 10634, 73758, 0, 0, 406032, 638830, 0, 0, 4249160, 3263500, 0, 0, 21907736, 82561050, 0, 0, 485798436, 945916970, 0, 0, 5968541478, 6839493576, 0, 0

Offset: 1

Pietro Majer, Mar 12 2009

nonn

Constant term in the expansion of (x + 1/x)(x^8 + 1/x^8)..(x^n^3 + 1/x^n^3).
a(n) = 0 for any n=1 (mod 4) or n=2 (mod 4).
The expansion above and the integral representation formula below are due to Andrica & Tomescu. The asymptotic formula is a conjecture; see Andrica & Ionascu. - Jonathan Sondow, Nov 06 2013

			Example: For n=12 the a(12) = 2 solutions are:
+1+8-27+64-125-216-343+512+729-1000-1331+1728=0,
-1-8+27-64+125+216+343-512-729+1000+1331-1728=0.

Ray Chandler, Table of n, a(n) for n = 1..130
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides.
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Equals twice A113263.

Cf. A063865, A158092, A019568. - Pietro Majer, Mar 15 2009

N:=60: p:=1: a:=[]: for n from 1 to N do p:=expand(p*( x^(n^3) + x^(-n^3) )): a:=[op(a), coeff(p,x,0)]: od:a;

a(n) = 2 * A113263(n).
Integral representation: a(n)=((2^n)/Pi)*int_0^Pi prod_{k=1}^n cos(x*k^3) dx.
Asymptotic formula: a(n)=(2^n)*sqrt(14/(Pi*n^7))*(1+o(1)) as n-->infty; n=-1 or 0 (mod 4).

A158465 Number of solutions to +-1+-2^4+-3^4+-4^4...+-n^4=0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 16, 18, 0, 0, 32, 100, 0, 0, 424, 510, 0, 0, 2792, 5988, 0, 0, 29058, 45106, 0, 0, 276828, 473854, 0, 0, 2455340, 4777436, 0, 0, 27466324, 46429640, 0, 0, 280395282, 526489336, 0, 0, 3193589950, 5661226928, 0, 0

Offset: 1

Pietro Majer, Mar 19 2009

nonn

Constant term in the expansion of (x + 1/x)(x^16 + 1/x^16)..(x^n^4 + 1/x^n^4).
a(n)=0 for any n=1 (mod 4) or n=2 (mod 4).
Andrica & Tomescu give a more general integral formula than the one below. The asymptotic formula below is a conjecture by Andrica & Ionascu; it remains unproven. - Jonathan Sondow, Nov 11 2013

			For n=16 the a(16) = 2 solutions are +1 +16 +81 +256 -625 -1296 -2401 +4096 +6561 +10000 +14641 +20736 -28561 -38416 -50625 +65536 = 0 and the opposite.

D. Andrica and E. J. Ionascu, Variations on a result of Erdős and SurányiINTEGERS 2013 slides.
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Cf. A063865, A158092, A158118, A158380, A019568.

A111253(n) = a(n)/2. - Alois P. Heinz, Oct 31 2011

N:=32: p:=1 a:=[]: for n from 32 to N do p:=expand
(p*(x^(n^4)+x^(-n^4))): a:=[op(a), coeff(p,x,0)]: od:a;

Integral representation: a(n) = ((2^n)/Pi)*int_0^pi prod_{k=1}^n cos(x*k^4) dx.

Asymptotic formula: a(n) = (2^n)*sqrt(18/(Pi*n^9))*(1+o(1)) as n->infinity; n=-1 or 0 (mod 4).

a(35)-a(58) from Alois P. Heinz, Oct 31 2011

A167028 Number of terms in the expansion of the determinant of a skew-symmetric matrix of order n.

Original entry on oeis.org

0, 1, 0, 6, 0, 120, 0, 5250, 0, 395010, 0, 45197460, 0, 7299452160, 0, 1580682203100, 0, 441926274289500, 0, 154940341854097800, 0, 66565404923242024800, 0, 34389901168124209507800, 0, 21034386936107260971255000, 0, 15032296693671903309613950000, 0, 12411582569784462888618434640000, 0

Offset: 1

Pietro Majer, Oct 27 2009

If n is odd a(n)=0.

Essentially a duplicate of A002370. - N. J. A. Sloane, Oct 27 2009

			Example: the determinant of a skew symmetric matrix of order n=4 is
det(A)=A(1,2)A(1,2)A(3,4)A(3,4) + 2A(1,2)A(2,3)A(1,4)A(3,4) -2A(1,2)A(2,4)A(1,3)A(3,4)+ A(1,3)A(1,3)A(2,4)A(2,4)-2A(1,3)A(2,4)A(1,4)A(2,3)+A(1,4)A(1,4)A(2,3)A(2,3).

Cf. A002370, A167028, A002135.

for n from 1 to 20 do a[n]:=n!coeftayl( (1-x^2)^(-1/4)*exp(x^2/4),x=0,n) od;

Rest[CoefficientList[Series[(1-x^2)^(-1/4)*E^(x^2/4), {x, 0, 20}], x] * Range[0, 20]!] (* Vaclav Kotesovec, Feb 15 2015 *)

Exponential generating function: (1-x^2)^(-1/4) exp(x^2/4).

Asymptotics (for even n): a(n)= (n!/Pi)exp( (-3log(n)+1+log(2))/4 ) GAMMA(3/4) (1+O(1/n)). [corrected by Vaclav Kotesovec, Feb 15 2015]. More elegant form is a(n) ~ n! * 2^(1/4) * exp(1/4) * GAMMA(3/4) / (Pi * n^(3/4)).

A167029 Difference between the number of positive and negative terms in the expansion of a skew symmetric matrix of order n.

Original entry on oeis.org

1, 0, 2, 0, 8, 0, 18, 0, 578, 0, -15460, 0, 1012512, 0, -81237604, 0, 8572174172, 0, -1139408178984, 0, 186543348044576, 0, -36888247922732008, 0, 8669441321229610968, 0, -2388740252077518073072, 0, 762715125987833507921408, 0, -279382350611903941569174000, 0

Offset: 1

Pietro Majer, Oct 27 2009

For even n, a(n)=0.

Cf. A167028.

Rest[Rest[CoefficientList[Series[Sqrt[Cosh[x]]*E^(x^2/4), {x, 0, 20}], x] * Range[0, 20]!]] (* Vaclav Kotesovec, Feb 15 2015 *)

E.g.f. (for offset 2): sqrt(cosh(x))*exp(x^2/4).
Asymptotics (for even n): a(n)=exp(Pi^2/16)*(2^(n-2))*(n!)*(Pi^(-n))*n^(3/4)*(1+O(1/n)) [This formula is wrong. - Vaclav Kotesovec, Feb 15 2015]
If n is odd |a(n)| ~ exp(-Pi^2/16) * 2^(n+1/2) * n! / (sqrt(n) * Pi^(n+1)). - Vaclav Kotesovec, Feb 15 2015

More terms from Vaclav Kotesovec, Feb 15 2015

A158380 Number of solutions to +-1 +- 3 +- 6 +- ... +- n(n+1)/2 = 0.

Original entry on oeis.org

1, 0, 0, 0, 2, 0, 2, 2, 4, 0, 12, 16, 26, 0, 66, 104, 210, 0, 620, 970, 1748, 0, 5948, 10480, 18976, 0, 60836, 111430, 209460, 0, 704934, 1284836, 2387758, 0, 8331820, 15525814, 28987902, 0, 101242982, 190267598, 358969426, 0, 1275032260, 2404124188, 4547419694

Offset: 0

Pietro Majer, Mar 17 2009

Equivalently, number of partitions of the set of the first n triangular numbers {t(1),...,t(n)} into two classes with equal sums.
Constant term in the expansion of (x + 1/x)(x^3 + 1/x^3)...(x^t(n) + 1/x^t(n)).
a(n) = 0 for all n == 1 (mod 4).
Andrica & Tomescu give a more general integral formula than the one below. - Jonathan Sondow, Nov 11 2013

			For n=6 the 2 solutions are +1-3+6-10-15+21 = 0 and -1+3-6+10+15-21 = 0.

Ray Chandler, Table of n, a(n) for n = 0..633
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.

Cf. A058498, A063865, A158092, A158118.

N:=70: p:=1: a:=[]: for n from 0 to N do
p:=expand(p*(x^(n*(n+1)/2)+x^(-n*(n+1)/2))):
a:=[op(a), coeff(p, x, 0)]: od:a;
# second Maple program:
b:= proc(n, i) option remember; (m-> `if`(n>m, 0,
      `if`(n=m, 1, b(abs(n-i*(i+1)/2), i-1)+
      b(n+i*(i+1)/2, i-1))))((2+(3+i)*i)*i/6)
    end:
a:= n-> `if`(irem(n, 4)=1, 0, b(0, n)):
seq(a(n), n=0..50);  # Alois P. Heinz, Sep 17 2017

a[n_] := With[{t = Table[k(k+1)/2, {k, 1, n}]}, Coefficient[Times @@ (x^t + 1/x^t), x, 0]];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 0, 50}] (* Jean-François Alcover, Jun 16 2018 *)

t(k) = k*(k+1)/2;
a(n) = polcoeff(prod(k=1, n, (x^t(k)+ 1/x^t(k))), 0); \\ Michel Marcus, May 19 2015

a(n) = (2^n/Pi) * Integral_{x=0..Pi} cos(x)*cos(3x)*...*cos(n(n+1)x/2) dx.
a(n) ~ 2^(n+1)*sqrt(10/Pi)*n^(-5/2)*(1+o(1)) as n --> infinity, n !== 1 (mod 4).
a(n) = 2 * A058498(n) for n > 0. - Alois P. Heinz, Nov 01 2011

a(0) = 1 prepended by Joerg Arndt, Sep 17 2017

Example corrected by Ilya Gutkovskiy, Feb 02 2022

A158092 Number of solutions to +- 1 +- 2^2 +- 3^2 +- 4^2 +- ... +- n^2 = 0.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 2, 10, 0, 0, 86, 114, 0, 0, 478, 860, 0, 0, 5808, 10838, 0, 0, 55626, 100426, 0, 0, 696164, 1298600, 0, 0, 7826992, 14574366, 0, 0, 100061106, 187392994, 0, 0, 1223587084, 2322159814, 0, 0, 16019866270, 30353305134, 0, 0

Offset: 1

Pietro Majer, Mar 12 2009

nonn

Twice A083527.
Number of partitions of the half of the n-th-square-pyramidal number into parts that are distinct square numbers in the range 1 to n^2. Example: a(7)=2 since, squarePyramidal(7)=140 and 70=1+4+16+49=9+25+36. - Hieronymus Fischer, Oct 20 2010
Erdős & Surányi prove that this sequence is unbounded. More generally, there are infinitely many ways to write a given number k as such a sum. - Charles R Greathouse IV, Nov 05 2012
The expansion and integral representation formulas below are due to Andrica & Tomescu. The asymptotic formula is a conjecture; see Andrica & Ionascu. - Jonathan Sondow, Nov 11 2013

			For n=8 the a(8)=2 solutions are: +1-4-9+16-25+36+49-64=0 and -1+4+9-16+25-36-49+64=0.

Alois P. Heinz and Ray Chandler, Table of n, a(n) for n = 1..500 (first 240 terms from Alois P. Heinz)
D. Andrica and E. J. Ionascu, Variations on a result of Erdős and Surányi, INTEGERS 2013 slides.
Dorin Andrica and Ioan Tomescu, On an Integer Sequence Related to a Product of Trigonometric Functions, and Its Combinatorial Relevance, J. Integer Sequences, 5 (2002), Article 02.2.4.
P. Erdős and J. Surányi, Egy additív számelméleti probléma (in Hungarian; Russian and German summaries), Mat. Lapok 10 (1959), pp. 284-290.

Cf. A063865, A158118, A019568, A083527, A231015.

From Pietro Majer, Mar 15 2009: (Start)
N:=60: p:=1: a:=[]: for n from 1 to N do p:=expand(p*(x^(n^2)+x^(-n^2))):
a:=[op(a), coeff(p, x, 0)]: od:a; (End)
# second Maple program:
b:= proc(n, i) option remember; local m; m:= (1+(3+2*i)*i)*i/6;
      `if`(n>m, 0, `if`(n=m, 1, b(abs(n-i^2), i-1) +b(n+i^2, i-1)))
    end:
a:= n-> `if`(irem(n-1, 4)<2, 0, 2*b(n^2, n-1)):
seq(a(n), n=1..60);  # Alois P. Heinz, Nov 05 2012

b[n_, i_] := b[n, i] = With[{m = (1+(3+2*i)*i)*i/6}, If[n>m, 0, If[n == m, 1, b[ Abs[n-i^2], i-1] + b[n+i^2, i-1]]]]; a[n_] := If[Mod[n-1, 4]<2, 0, 2*b[n^2, n-1]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Mar 13 2015, after Alois P. Heinz *)

a(n)=2*sum(i=0,2^(n-1)-1,sum(j=1,n-1,(-1)^bittest(i,j-1)*j^2)==n^2) \\ Charles R Greathouse IV, Nov 05 2012

from itertools import count, islice
from collections import Counter
def A158092_gen(): # generator of terms
    ccount = Counter({0:1})
    for i in count(1):
        bcount = Counter()
        for a in ccount:
            bcount[a+(j:=i**2)] += ccount[a]
            bcount[a-j] += ccount[a]
        ccount = bcount
        yield(ccount[0])
A158092_list = list(islice(A158092_gen(),20)) # Chai Wah Wu, Jan 29 2024

Constant term in the expansion of (x + 1/x)(x^4 + 1/x^4)..(x^n^2 + 1/x^n^2).
a(n)=0 for any n == 1 or 2 (mod 4).
Integral representation: a(n)=((2^n)/pi)*int_0^pi prod_{k=1}^n cos(x*k^2) dx
Asymptotic formula: a(n) = (2^n)*sqrt(10/(pi*n^5))*(1+o(1)) as n-->infty; n == -1 or 0 (mod 4).
a(n) = 2 * A083527(n). - T. D. Noe, Mar 12 2009
min{n : a(n) > 0} = A231015(0) = 7. - Jonathan Sondow, Nov 06 2013

a(51)-a(56) from R. H. Hardin, Mar 12 2009

Edited by N. J. A. Sloane, Sep 15 2009

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User: Pietro Majer

A158118 Number of solutions of +-1+-2^3+-3^3..+-n^3=0.

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A158465 Number of solutions to +-1+-2^4+-3^4+-4^4...+-n^4=0.

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A167028 Number of terms in the expansion of the determinant of a skew-symmetric matrix of order n.

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A167029 Difference between the number of positive and negative terms in the expansion of a skew symmetric matrix of order n.

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A158380 Number of solutions to +-1 +- 3 +- 6 +- ... +- n(n+1)/2 = 0.

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A158092 Number of solutions to +- 1 +- 2^2 +- 3^2 +- 4^2 +- ... +- n^2 = 0.

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