A160096 - OEIS

%I #52 Jun 03 2023 23:57:55

%S 1,2,2,2,1,1,0,0,0,0,0,1,1,1,2,2,2,2,2,2,2,1,1,1,1,0,0,0,0,0,0,0,0,0,

%T 1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,

%U 0,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2

%N Partial sums of A010815 starting with offset 1, and signed (+ + - - + + ...).

%C INVERT transform of the sequence = A137682: (1, 3, 7, 17, 40, 96, 228, ...).

%C From _Mats Granvik_, Jan 01 2015: (Start)

%C (1) The positive integers are the row sums of the infinite lower triangular matrix "t" starting:

%C   1, 0, 0, 0, 0, 0, 0, ...

%C   1, 1, 0, 0, 0, 0, 0, ...

%C   1, 1, 1, 0, 0, 0, 0, ...

%C   1, 1, 1, 1, 0, 0, 0, ...

%C   1, 1, 1, 1, 1, 0, 0, ...

%C   1, 1, 1, 1, 1, 1, 0, ...

%C   1, 1, 1, 1, 1, 1, 1, ...

%C   ...

%C which satisfies the recurrence:

%C t(n, 1) = 1; t(n, k) = Sum_{i=1..n-1} t(n-i, k-1) - Sum_{i=1..n-1} t(n-i, k) if n >= k, otherwise 0;

%C (2) This sequence a(n), in turn, is the row sums of the infinite lower triangular matrix "t" starting:

%C   1, 0, 0, 0, 0, 0, 0, ...

%C   1, 1, 0, 0, 0, 0, 0, ...

%C   1, 0, 1, 0, 0, 0, 0, ...

%C   1, 0, 0, 1, 0, 0, 0, ...

%C   1, 0,-1, 0, 1, 0, 0, ...

%C   1, 0, 0,-1, 0, 1, 0, ...

%C   1, 0, 0,-1,-1, 0, 1, ...

%C   ...

%C which satisfies the recurrence:

%C t(n, 1) = 1; t(n, k) = Sum_{i=1..k-1} t(n-i, k-1) - Sum_{i=1..n-1} t(n-i, k) if n >= k, otherwise 0;

%C (3) The partition numbers are the row sums of the infinite lower triangular matrix "t" starting:

%C   1, 0, 0, 0, 0, 0, 0, ...

%C   1, 1, 0, 0, 0, 0, 0, ...

%C   1, 1, 1, 0, 0, 0, 0, ...

%C   1, 2, 1, 1, 0, 0, 0, ...

%C   1, 2, 2, 1, 1, 0, 0, ...

%C   1, 3, 3, 2, 1, 1, 0, ...

%C   1, 3, 4, 3, 2, 1, 1, ...

%C   ...

%C which satisfies the recurrence:

%C t(n, 1) = 1; t(n, k) = Sum_{i=1..n-1} t(n-i, k-1) - Sum_{i=1..k-1} t(n-i, k) if n >= k, otherwise 0;

%C (4) The number of divisors of "n" is the row sums of the infinite lower triangular matrix "t" starting:

%C   1, 0, 0, 0, 0, 0, 0, ...

%C   1, 1, 0, 0, 0, 0, 0, ...

%C   1, 0, 1, 0, 0, 0, 0, ...

%C   1, 1, 0, 1, 0, 0, 0, ...

%C   1, 0, 0, 0, 1, 0, 0, ...

%C   1, 1, 1, 0, 0, 1, 0, ...

%C   1, 0, 0, 0, 0, 0, 1, ...

%C   ...

%C which satisfies the recurrence:

%C t(n, 1) = 1; t(n, k) = Sum_{i=1..k-1} t(n-i, k-1) - Sum_{i=1..k-1} t(n-i, k) if n >= k, otherwise 0.

%C In the four cases of recurrences only the summation indices within the sums change, from (1) "n-1" and "n-1" to (2) "k-1" and "n-1" to (3) "n-1" and "k-1" to (4) "k-1" and "k-1".

%C (End)

%F Partial sums of Euler's q series (signed), starting from offset 1 = (1, 1, 0, 0, -1, 0, -1, 0, 0, 0, 0, 1, 0, 0, 1, ...).

%F G.f.: (1 - f(-x)) / (1 - x) where f(-x) is the g.f. of A010815. - _Michael Somos_, Jan 02 2015

%F Partial sums of A257628. - _Georg Fischer_, May 29 2023

%e The series begins (1, 2, 2, 2, 1, 1, 0, ...) since the signed q-series = (1, 1, 0, 0, -1, 0, ...).

%e G.f. = x + 2*x^2 + 2*x^3 + 2*x^4 + x^5 + x^6 + x^12 + x^13 + x^14 + ...

%t (* A160096 as row sums of recursively defined table *)

%t Clear[t]; nn = 90; t[n_, 1] = 1; t[n_, k_] := t[n, k] = If[n >= k, Sum[t[n - i, k - 1], {i, 1, k - 1}] - Sum[t[n - i, k], {i, 1, n - 1}], 0]; PartialSumsOfEulerqSeries = Table[Sum[t[n, k], {k, 1, n}], {n, 1, nn}] (* _Mats Granvik_, Jan 01 2015 *)

%t a[ n_] := SeriesCoefficient[ (1 - QPochhammer[ x]) / (1 - x), {x, 0, n}]; (* _Michael Somos_, Jan 02 2015 *)

%t CoefficientList[Series[q*(1/(1 - q)^(2)*QHypergeometricPFQ[{q, q}, {q^2, q}, q, q^2]), {q, 0, 89}], q] (* _Mats Granvik_, Jan 09 2015 *)

%o (PARI) {a(n) = if( n<0, 0, polcoeff( (1 - eta(x + x * O(x^n))) / (1 - x), n))}; /* _Michael Somos_, Jan 02 2015 */

%Y Cf. A010815, A257628.

%Y Cf. (1) A000027, (2) A160096, (3) A000041, (4) A000005.

%K nonn

%O 1,2

%A _Gary W. Adamson_, May 01 2009

%E More terms from _Mats Granvik_, Jan 01 2015