A191685 Eighth diagonal a(n) = s(n,n-7) of the unsigned Stirling numbers of the first kind with n>7.

5040, 109584, 1172700, 8409500, 45995730, 206070150, 790943153, 2681453775, 8207628000, 23057159840, 60202693980, 147560703732, 342252511900, 756111184500, 1599718388730, 3256091103430, 6400590336096, 12191224980000, 22563937825000, 40681506808800

Offset: 8

Author

Paul Weisenhorn, Jun 11 2011

Keywords

• nonn
• easy

Comments

The Maple programs under I generate the sequence. The Maple program under II generates explicit formulas for a(n+1) = s(n+1,n+1-c) with c>=1 and n>=c.

Examples

			c=1; a(n+1) = binomial(n+1,2)
c=2; a(n+1) = binomial(n+1,3)*(2+3*n)/4
c=3; a(n+1) = binomial(n+1,4)*(n+n^2)/2
c=4; a(n+1) = binomial(n+1,5)*(-8-10*n+15*n^2 +15*n^3)/48
c=5; a(n+1) = binomial(n+1,6)*(-6*n-7*n^2+2*n^3+ 3*n^4)/16
c=6; a(n+1) = binomial(n+1,7)*(96+140*n-224*n^2-315*n^3+63*n^5)/576
c=7; a(n+1) = binomial(n+1,8)*(80*n+114*n^2-23*n^3-75*n^4-9*n^5+9*n^6)/144
c=8; a(n+1) = binomial(n+1,9)*(-1152-1936*n+2820*n^2+
  5320*n^3+735*n^4-1575*n^5-315*n^6+135*n^7)/3840
c=9; a(n+1) = binomial(n+1,10)*(-1008*n-1676*n^2 +100*n^3+1295*n^4+392*n^5-210*n^6-60*n^7 +15*n^8)/768
		

References

• K. Seidel, Variation der Binomialkoeffizienten, Bild
• der Wissenschaft, 6 (1980), pp. 127-128.

Links

• T. D. Noe, Table of n, a(n) for n = 8..1000

Crossrefs

Cf. A130534, A000012 (c=0; 1st diagonal), A000217 (c=1; 2nd diagonal), A000914 (c=2; 3rd diagonal), A001303 (c=3; 4th diagonal), A000915 (c=4; 5th diagonal), A053567 (c=5; 6th diagonal), A112002 (c=6; 7th diagonal), A191685 (c=7; 8th diagonal).

Programs

Maple

I: programs generate the sequence:
with(combinat): c:=7; a:= proc(n) a(n):=abs(stirling1(n,n-c)); end: seq(a(n), n=c+1..28);
for n from 7 to 27 do a(n+1) := binomial(n+1,8)*(80*n+ 114*n^2- 23*n^3- 75*n^4- 9*n^5+ 9*n^6)/144 end do: seq(a(n),n=8..28);
II: program generates explicit formulas for a(n+1) =  s(n+1,n+1-c):
k[1,0]:=1: v:=1:
for c from 2 to 10 do
  c1:=c-1: c2:=c-2: p0:=0:
  for j from 0 to c2 do p0:=p0+k[c1,j]*m^j: end do:
  f:=expand(2*c*(m+1)*p0/v):
  p1:=0: p2:=0:
  for j from 0 to c1 do
    p1:=p1+k[c,j]*(m+1)^j:
    p2:=p2+k[c,j]*m^j:
  end do:
  g:=collect((m+2)*p1-(m-c1)*p2-f,m):
  kh[0]:=rem(g,m,m): Mk:=[kh[0]]: Mv:=[k[c,0]]:
  for j from 1 to c1 do
    kh[j]:=coeff(g,m^j):
    Mk:=[op(Mk),kh[j]]: Mv:=[k[c,j],op(Mv)]:
  end do:
  sol:=solve(Mk,Mv):
  v:=1:
  for j from 1 to c do
    k[c,c-j]:=eval(k[c,c-j],sol[1,j]):
    nen[j]:=denom(k[c,c-j]):
    v:=ilcm(v,nen[j]):
  end do:
  for j from 0 to c1 do k[c,j]:=k[c,j]*v:
    printf("%8d",k[c,j]): end do:
  p3:=0:
  for j from 0 to c1 do p3:=p3+k[c,j]*n^j: end do:
  s[n+1,n+1-c]:=binomial(n+1,c+1)*(c+1)*p3/(2^c*k[c,c1]):
end do:
for c from 2 to 10 do print("%a\n",s[n+1,n+1-c]):
end do:

Formula

a(n+1) = A130534(T(n,n-7)) = s(n+1,n+1-7)

a(n+1) = binomial(n+1,8)*(80*n+114*n^2-23*n^3-75*n^4-9*n^5+9*n^6)/144