Q: OEIS duplicates?

[Maximilian Hasler]

>>  Are these OEIS duplicates:
>>  A138980 and A049561 (apart from the first value)?
>
>  NO, cf.
I meant to say : cf. A138982

>>  A138917 and A061242 ?
>
>  A061242                  Prime numbers == 8 (mod 9)
>  A138917                 Primes of the form 18n-1.
>
>  YES:
>  we have :
>  x=9n-1 <=> x = 18n-1 or x=18n+8
>  However, there is no prime of the latter form.

>> A138901 and A109074 and A094258 ?

YES:

A094258  	 	 a(1) = 1, a(n+1) = n*( a(1) + a(2) + ... + a(n) ).  	 	
A109074 		Number of integers which use each of 0-to-n decimal digits
exactly once.

With
%F A109074 a(0)=1 by definition, for n>0 a(n)=n*n!
let's call b(n)=A094258(n+1)=n*( b(0) + b(1) + ... + b(n-1) )
We have to show that b(n)/n=n!
By definition, b(n)/n = b(0) + b(1) + ... + b(n-1)
Assume that the hypothesis b(k)/k (=b(0)+...+b(k-1)) = k! is true for 0<k<n.
Then
b(0)+ ...+b(n-2)+b(n-1) = (n-1)! + (n-1)*(n-1)! =n*(n-1)! = n!
Thus, A109074=A094258, Q.E.D.


%N A138901 Number of monomials in determinant of n X n symbolic matrix
with only one zero entry.
We have:
%F A138901 a(2) = 1, a(n) = a(n-1)*(n-1)^2/(n-2). - Andrey Ryshevich

Again, this writes a(n)/(n-1) = (n-1)*a(n-1)/a(n-2) so that c(n)=a(n)/(n-1)
verifies c(n)=(n-1)c(n-1), thus c(n) = (n-1)! and thus a(n) = (n-1)(n-1)!
just like b() in the above.

Maximilian