# [seqfan] Re: offset of A000254 wrong?

Richard Mathar mathar at strw.leidenuniv.nl
Tue Aug 4 15:01:02 CEST 2009

```In response to
http://list.seqfan.eu/pipermail/seqfan/2009-August/002040.html

ja> From: Joerg Arndt <arndt at jjj.de>
ja> To: seqfan at list.seqfan.eu
ja> Subject: [seqfan]  offset of A000254 wrong?
ja>
ja> Stirling numbers of first kind s(n,2):
ja> 0, 1, 3, 11, 50, 274, 1764, ...
ja>
ja> offset is 0, but should be 1:

I agree, this should be consistent with the other standard sources,
like Abramowitz-Stegun or the other columns s(n,i), i=3,4,.. in the OEIS.

i) This needs a change in the recurrence of A000254:

%N A000254 Stirling numbers of first kind s(n,2): a(n+1) = n*a(n)+(n-1)!

ii) Adaptation of the factor in the Cloitre comment:

%C A000254 Let P(n,X)=(X+1)(X+2)(X+3)...(X+n-1); then a(n) is the coefficient of X;
or a(n)=P'(n,0) - Benoit Cloitre (benoit7848c(AT)orange.fr), May
09 2002

%C A000254 a(n+1)=number of cycles in all permutations of [n]. Example: a(4)=11 because the permutations (1)(2)(3), (1)(23), (12)(3), (13)(2), (132), (123) have 11 cycles alltogether. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 12 2004

%C A000254 a(n+1) is divisible by n for all composite n >= 6. a(2n+1) is divisible by (2n+1): see A114450 - Leroy Quet May 20 2007

The Lajos maple program should be replaced anyway by the intrinsic function:

%p A000254 A000254 := proc(n) abs(combinat[stirling1](n,2)) ; end: seq(A000254(n),n=1..20) ; # R. J. Mathar, Aug 04 2009

%C A000254 For n >= 3 the determinant of the n-2 X n-2 matrix M(i,j) = i + 2 for i = j and 1 otherwise (i,j = 1..n-2). E.g. for n = 4 the determinant of [(3, 1), (1, 4)]. See 53rd Putnam Examination, 1992, Problem B5.  - Franz Vrabec (franz.vrabec(AT)planetuniqa.at), Jan 13 2008, Mar 26 2008

and perhaps other changes in A000254..

iii) An index change in the comment of A136125:

%I A136125
%C A136125 Row sums are the factorials (A000142). T(n,1)=n!/2 for n>=2; Sum(k*T(n, k),k=1..n)=s(n,2)=A000254(n+1) (Stirling numbers of the first kind).

iv) An index change in a formula in A158442

%I A158442
%F A158442 Row sums: (n+2)*A000254(n+1).

v) A change in a formula in A114450:

%F A114450 a(n) = A000254(2n+1)/(2n+1).

vi) An index change in a formula in A138771:

%C A138771 T(n,1)=A000254(n).

vii) An index change in a formula in A161128:

%C A161128 a(n)=A000254(n+1) - A007489(n).

viii) Removal of (a superfluous duplication) in A074246:

%F A074246 First column is A000254 (Stirling numbers of first kind),

ix) Change of A079303:

%F A079303 a(n)=A000254(A000040(n))/(2*A000040(n)^2) .

x) Change of the comment in A074845:

%C A074845 It appears that terms >6 are simply given by : composite n such that n^2 doesn't divide A000254(n+1) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 09 2004

There is no change to the comment in A097422 (which was wrong up to now
and becomes correct).

xi) change of the index in a comment in A109822:

%C A109822 T(n,n)=n!. Sum of row n is the signless Stirling number of the first
kind s(n+1,2)(A000254). T(n,k)=A096747(n,k) for 1<=k<=n.

xii) Change of an index in a  comment in A130679:

... The row sums of Q are (n+2)*A000254(n+1).

xiii) in A052881 change of an index

.... = A000027(n)*A000254(n) = ...

ixx) in A001706 three indices need a change:

%F A001706 a(n-2)=(1/2)*sum(i=0, n, C(n, i)*A000254(i+1)*A000254(n-i+1)) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 09 2004

Another problem is with the e.g.f. A001706, which currently refers to
offset 2 and misses a pair of parenthesis in the denominator. I suspect
that the better offset in A001706 is 2, not 0, but this is a separate problem,
probably to be attacked also in A001719.

xx) in A056612 shift the Vrabec index:

%F A056612 a(n)=A000142(n)/A002805(n)=A000254(n+1)/A001008(n). - Franz Vrabec (franz.vrabec(AT)planetuniqa.at), Sep 13 2005

xxi) in A105954 remove all these "starting at.." comments. They are always
irritating because one cannot be sure whether these indices refer to a(n) or
the particular referenced sequence, and because a(n) and the referenced
sequence (guaranteed!) start at their offset.

xxii) both indices in A067176 need a shift
%F A067176 a(n, k) =(n!/k!)*sum_{k<j<=n}1/j =(A000254(n+1)-A000254(k+1)*A008279(n, n-k))/ A000142(k) =a(n-1, k)*n+(n-1)!/k! =(a(n, k-1)-n!/k!)/k.

xxiii) in A056792 increase the index:

%F A056792 n>0 a(n)=n-valuation(A000254(n+1), 2) - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 09 2004

xxiv) in A025527 shift the index:
%F A025527 a(n)=A000142(n)/A003418(n)=A000254(n+1)/A025529(n). - Franz Vrabec (franz.vrabec(AT)planetuniqa.at), Sep 13 2005

xxxv) in A052517 shift the index (which was wrong anyway):
%Y A052517 Equals 2 * A000254(n), n>0.

xxxvi) in A001705 shift an index
%F A001705 a(n) = a(n-1)*(n+1)+n! = A000254(n+2)-A000142(n+1) = A067176(n+1, 1)
- Henry Bottomley (se16(AT)btinternet.com), Jan 09 2002

xxxvii) in A120299 shift the offset (I think the Mma start at 2 remains correct):
%N A120299 Largest prime factor of Stirling numbers of first kind s(n,2) A000254[n]
%O A120299 3,1

xxxviii) in A091828 shift an index (the alternative is to shift the offset, but
this may lead to avalanche effects elsewhere ..)
%N A091828 a(n)=n-2*valuation(A000254(n+1),3).

xxxix) an index shift in A091827
%N A091827 Least k such that k^n divides A000254(k+1).

xv) A091826 needs an index shift:
%N A091826 a(n)=(1+A000254(n+1)-n)/n as n runs through the primes.

The same work is then to be done on behalf of A081048, the signed version....

ja> Also I fail to see why the seq. contains a link to a page
ja> titled "Prism Of Spirals: Images, visions, and hallucinations".