Jeffrey R. Goodwin - cp's OEIS frontend

User: Jeffrey R. Goodwin

Jeffrey R. Goodwin has authored 2 sequences.

A193912 Partial sums of A193911.

1, 4, 11, 25, 50, 93, 162, 272, 439, 694, 1069, 1627, 2432, 3611, 5292, 7730, 11181, 16156, 23167, 33237, 47390, 67673, 96134, 136868, 193971, 275634, 390049, 553599, 782668, 1110023, 1568432, 2223430, 3140553, 4450872, 6285459, 8906457, 12576010, 17818405

Offset: 1

Jeffrey R. Goodwin, Aug 08 2011

			We have A193911(1)=1, A193911(2)=3, and A193911(3)=7. Thus a(1)=1, a(2)=4, and a(3)=11.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-8,7,3,-6,2).

LinearRecurrence[{3,0,-8,7,3,-6,2},{1,4,11,25,50,93,162},40] (* Harvey P. Dale, Sep 09 2015 *)
CoefficientList[Series[(1 + x - x^2)/((1 - x)^4*(1 + x)*(1 - 2*x^2)), {x, 0, 50}], x] (* G. C. Greubel, Feb 25 2017 *)

my(x='x+O('x^50)); Vec((1+x-x^2)/((1-x)^4*(1+x)*(1-2*x^2))) \\ G. C. Greubel, Feb 25 2017

a(n) = Sum_{i=1..n} 1/8*(2^(i/2+2)*((10-7*sqrt(2))*(-1)^(i) + 10 + 7*sqrt(2))-(-1)^(i)-2*i*(i+12)-79).
G.f.: x*(1+x-x^2)/((1-x)^4*(1+x)*(1-2*x^2)). - Alexander R. Povolotsky, Aug 12 2011
a(n) = (1/32)*( (-1/2)^n + 32*(41*sqrt(2)-58)*(sqrt(2)-2)^n - 32*(58+41*sqrt(2))*(-2-sqrt(2))^n ).

A193911 Sums of the diagonals of the matrix formed by listing the h-Stohr sequences in increasing order.

Original entry on oeis.org

1, 3, 7, 14, 25, 43, 69, 110, 167, 255, 375, 558, 805, 1179, 1681, 2438, 3451, 4975, 7011, 10070, 14153, 20283, 28461, 40734, 57103, 81663, 114415, 163550, 229069, 327355, 458409, 654998, 917123, 1310319, 1834587, 2620998, 3669553, 5242395, 7339525, 10485230

Offset: 1

Jeffrey R. Goodwin, Aug 08 2011

			Portion of the first three rows:
A033627, 2-Stohr  1  2  4  7
A026474, 3-Stohr  1  2  4  8
A051039, 4-Stohr  1  2  4  8
Thus a(1)=1, a(2)=2+1=3, and a(3)=4+2+1=7.

Vincenzo Librandi, Table of n, a(n) for n = 1..2000
Eric Weisstein's World of Mathematics, Stöhr Sequence.
Index entries for linear recurrences with constant coefficients, signature (2,2,-6,1,4,-2).

A193911=t={0,1}; Do[AppendTo[t,t[[-2]]+t[[-1]]]; AppendTo[t,2*t[[-2]]],{n,41}]; Drop[Nest[Accumulate,t,2],1] (* Vladimir Joseph Stephan Orlovsky, Jan 27 2012 *)
LinearRecurrence[{2,2,-6,1,4,-2},{1,3,7,14,25,43},40] (* Harvey P. Dale, Jun 20 2015 *)

All h-Stohr sequences have formula: h terms 1,2,..,2^(n-1),..,2^(h-1) and then continue (2^h-1)(n-h)+1. - Henry Bottomley, Feb 04 2000
So we get the sums from the piecewise function:
for odd n>=1, a(n)=2^((n+1)/2)-n+((n+1)/2)-2+Sum_{i=0..((n+1)/2)-1}(2*i+1)*(2^(((n+1)/2)-i) -1);
for even n>=2, a(n)=2^((n/2)+2)-n-4+Sum_{i=0..(n/2)-1}(2*i+1)*(2^((n/2)-i) -1). - Jeffrey R. Goodwin, Aug 09 2011
Let odd m>=3, then a(n)=a(m)-A000295(((m+1)/2)+1), where n>=2 is even. - Jeffrey R. Goodwin, Aug 09 2011
Let even m>=2, then a(n)=a(m)-A077802(m/2)=a(m)-A095151(m/2), where n>=1 is odd. - Jeffrey R. Goodwin, Aug 09 2011
From Alexander R. Povolotsky, Aug 09 2011: (Start)
G.f.: x*(1 + x - x^2)/((-1 + x)^3*(-1 - x + 2*x^2 + 2*x^3)).
a(n+4) = -2*a(n)+3*a(n+2)+n+5.
a(n) = 1/8*(2^(n/2+2)*((10-7*sqrt(2))*(-1)^n+10+7*sqrt(2))-(-1)^n-2*n*(n+12)-79). (End)

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User: Jeffrey R. Goodwin

A193912 Partial sums of A193911.

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