The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A106231

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A106231

Least j > 1 such that j^2 = (4*n^2 + 2)*(k^2) + (4*n^2 + 2)*k + 1.
(history; published version)

#16 by Susanna Cuyler at Tue Apr 07 08:48:44 EDT 2020

STATUS

proposed

approved

#15 by Jinyuan Wang at Tue Apr 07 04:06:50 EDT 2020

STATUS

editing

proposed

#14 by Jinyuan Wang at Tue Apr 07 04:06:29 EDT 2020

COMMENTS

For n=1 , j(1,1) = 1, j(2,1) = 10*j(1,1) + 1 , then j(i,1) = 10*j(i-1,1) - j(i-3).

For n>1 , j(1,n) = 1, j(2,n) = 4*n^3 - 4*n^2 + 2*n - 1, j(3,n) = 4*n^3 + 4*n^2 + 2*n+1 , j(4,n) = (8*n^2+2)*j(2,n) + 1 then j(i,n) = (8*n^2+2)*j(i-2) - j(i-4,n).

FORMULA

For n=a(1 j) = 11, for n > 1 ja(n) = 4*n^3 - 4*n^2 + 2*n - 1, for n > 1, k sequence = A106232.

G.f.: x*(10*x^4-39*x^3+67*x^2-25*x+11) / (x-1)^4. [_- _Colin Barker_, Mar 06 2013]

PROG

(PARI) a(n) = if(n==1, 11, 4*n^3-4*n^2+2*n-1); \\ Jinyuan Wang, Apr 07 2020

STATUS

approved

editing

#13 by Charles R Greathouse IV at Sat Jun 17 02:59:25 EDT 2017

LINKS

<a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

Discussion

Sat Jun 17

02:59

OEIS Server: https://oeis.org/edit/global/2662

#12 by Charles R Greathouse IV at Sat Jun 13 00:51:48 EDT 2015

LINKS

<a href="/index/Rec">Index to sequences with entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

Discussion

Sat Jun 13

00:51

OEIS Server: https://oeis.org/edit/global/2439

#11 by Charles R Greathouse IV at Fri Jun 12 15:25:52 EDT 2015

LINKS

<a href="/index/Rea#recLCCRec">Index to sequences with linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

Discussion

Fri Jun 12

15:25

OEIS Server: https://oeis.org/edit/global/2436

#10 by Joerg Arndt at Thu Mar 07 11:24:23 EST 2013

STATUS

proposed

approved

#9 by Michel Marcus at Thu Mar 07 11:08:31 EST 2013

STATUS

editing

proposed

#8 by Michel Marcus at Thu Mar 07 11:08:23 EST 2013

COMMENTS

For j there is always a recurrence.

For n=1 j(1,1)=1, j(2,1)=10*j(1,1)+1 then j(i,1)=10*j(i-1,1)-j(i-3).

For j there is alway a recurrence for n=1 j(1,1)=1, j(2,1)=10*j(1,1)+1 then j(i,1)=10*j(i-1,1)-j(i-3) for n>1 j(1,n)=1, j(2,n)=4*n^3-4*n^2+2*n-1, j(3,n)=4*n^3+4*n^2+2*n+1 j(4,n)=(8*n^2+2)*j(2,n)+1 then j(i,n)=(8*n^2+2)*j(i-2)-j(i-4,n).

FORMULA

For n=1 j=11, for n > 1 j(n) = 4*n^3 - 4*n^2 + 2*n - 1 , k sequence = A106232.

STATUS

approved

editing

#7 by Joerg Arndt at Wed Mar 06 13:08:41 EST 2013

STATUS

proposed

approved