A129587 -id:A129587 - cp's OEIS frontend

A130679 a(n) = (n+1+(-1)^n)*A024167(n), related to alternating harmonic sums.

1, 4, 15, 84, 470, 3552, 26796, 255840, 2435184, 28114560, 323405280, 4380445440, 59105255040, 918796677120, 14228252640000, 249644312064000, 4363865549568000, 85297521899520000, 1661265370695168000

Offset: 1

Paul Curtz, Jun 29 2007

nonn

Inspired by a formula in the reference, the study of the singular points of planar differential systems leads to 3 two-dimensional polynomial families, one ordinary (degenerate case, considered in one dimension, see A129326) and two odd (the second, considered in one dimension, see A129587).
The first is in one dimension P(2n-1,x)=(n+1+x^n)*sum_{i=0..n-1} x^i/(i+1), n>=1.
The table of coefficients of P() with 2n coefficients per row starts:
2, 1;
3, 3/2, 1, 1/2;
4, 2, 4/3, 1, 1/2, 1/3;.. .
Rows multiplied by n!, the table becomes Q():
2, 1;
6, 3, 2, 1;
24, 12, 8, 6, 3, 2;
120, 60, 40, 30, 24, 12, 8, 6;
720, 360, 240, 180, 144,...
The sequence gives the alternating row sums of this table Q, positive sign for coefficients in front of even and negative sign for coefficients in front of odd powers of x.
The row sums of Q are (n+2)*A000254(n)= 3, 12, 55, 300...
Adding the alternating and ordinary row sums yields the sequence 4, 16, 70, 384....
The sequence of sums of antidiagonals in the Q table starts 2, 6+1=7, 24+3=27, 120+12+1=134.

			a(1) = 2-1.
a(2) = 6-3+2-1.
a(3) = 24-12+8-6+3-2.

G. C. Greubel, Table of n, a(n) for n = 1..250
Jean-François Alcover, Roots of P(2n-1,x) lie close to a circle (example)
P. Curtz, Stabilité locale des systèmes quadratiques, Ann. Sc. Ec. Norm. Sup., vol. 13 no. 3 (1980) 293-302.

a[n_] := (1/2)*(n + (-1)^n + 1)*n!*((-1)^n*(HarmonicNumber[(n-1)/2] - HarmonicNumber[n/2]) + Log[4]); Table[a[n] // FullSimplify, {n, 1, 19}] (* Jean-François Alcover, Oct 03 2012 *)

a(n) = n!*(n+1+(-1)^n)*A058313(n)/A058312(n). - R. J. Mathar, Jul 28 2008

Edited and extended by R. J. Mathar, Jul 28 2008

A057058 Let R(i,j) be the rectangle with antidiagonals 1; 2,3; 4,5,6; ...; each k is an R(i(k),j(k)) and a(n)=i(A057027(n)).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 1, 4, 2, 3, 1, 5, 2, 4, 3, 1, 6, 2, 5, 3, 4, 1, 7, 2, 6, 3, 5, 4, 1, 8, 2, 7, 3, 6, 4, 5, 1, 9, 2, 8, 3, 7, 4, 6, 5, 1, 10, 2, 9, 3, 8, 4, 7, 5, 6, 1, 11, 2, 10, 3, 9, 4, 8, 5, 7, 6, 1, 12, 2, 11, 3, 10, 4, 9, 5, 8, 6, 7, 1, 13, 2, 12, 3, 11, 4

Offset: 1

Clark Kimberling, Jul 30 2000

nonn

Since A057027 is a permutation of the natural numbers, every natural number occurs infinitely many times in this sequence.
Consider the triangle TN := 1; 1, -2; 1, -3, 2; 1, -4, 2, -3; ... Antidiagonal sums give A129819(n+2). TN arises in studying the equation (E) dy/dx=Q(n,x,y)/P(n,x,y) involving saddle-points quantities, P and Q are bidimensional polynomials n=2,3,4.. . (E) leads also for instance to the one-dimension polynomials in A129326, A129587, A130679. - Paul Curtz, Aug 16 2008
First inverse function (numbers of rows) for pairing function A194982. - Boris Putievskiy, Jan 10 2013

Boris Putievskiy, Transformations Integer Sequences And Pairing Functions, arXiv:1212.2732 [math.CO], 2012.

Cf. A057059, A194982.

From Boris Putievskiy, Jan 10 2013: (Start)
a(n) = -((A002260(n)+1)/2)*((-1)^A002260(n)-1)/2+(A004736(n)+A002260(n)/2)*((-1)^A002260(n)+1)/2.
a(n) = -((i+1)/2)*((-1)^i-1)/2+(j+i/2)*((-1)^i+1)/2, where i = n-t*(t+1)/2, j = (t*t+3*t+4)/2-n, t = floor((-1+sqrt(8*n-7))/2). (End)

cp's OEIS Frontend

A130679 a(n) = (n+1+(-1)^n)*A024167(n), related to alternating harmonic sums.

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