# [seqfan] Re: Sequence to A141530

Klaus Brockhaus klaus-brockhaus at t-online.de
Thu Feb 12 14:10:25 CET 2009

```"vincenzo.librandi at tin.it" wrote:
>
> I think that
> a(n)=4*n^3+6*n^2-1
> is equal to A141530

Not exactly. Offset is 0, therefore a(n) = 4*n^3-6*n^2+1.
PARI:
? for(n=0,36,print1(4*n^3-6*n^2+1,","))
1,-1,9,55,161,351,649,1079,1665,2431,3401,4599,6049,7775,9801,12151,
14849,17919,21385,25271,29601,34399,39689,45495,51841,58751,66249,
74359,83105,92511,102601,113399,124929,137215,150281,164151,178849,

Klaus

> 0          -1
>            1           9
>            2          55
>            3         161
>            4         351
>            5
> 649
>            6        1079
>            7        1665
>
> 8        2431
>            9        3401
>           10        4599
>           11        6049
>           12        7775
>           13
> 9801
>           14       12151
>           15       14849
>
> 16       17919
>           17       21385
>           18       25271
>           19       29601
>           20       34399
>           21
> 39689
>           22       45495
>           23       51841
>
> 24       58751
>           25       66249
>           26       74359
>           27       83105
>           28       92511
>           29
> 102601
>           30      113399
>
> Regards,
> Vincenzo Librandi
>
> %I A141530
> %S A141530 1,1,9,55,161,351,649,1079,1665
> %V A141530 1,
> -1,9,55,161,351,649,1079,1665
> %N A141530 Third from a recurrences
> family concerning numerators of a(i,j) square
>                defined
> by j!*a(i,j)=Integral (from i to i+1) u*(u-1)*(-2)* .. *(u-j+1)
>                du.

[......]

> %K A141530 sign,uned
> %O A141530
> 0,3
> %A A141530 Paul Curtz (bpcrtz(AT)free.fr), Aug 12 2008

```