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Suppose n=(p1^e1)(p2^e2)... where p1,p2,... are the prime numbers and e1,e2,... are nonnegative integers. Then we can define Pn(x) = e1 + (e2)*x + (e3)*(x^2) + (e4)*(x^3) + ... + (ek)*(x^(k-1)) + ... The sequence is the table T(x,n)=Pn(x) read by antidiagonals.
3

%I #11 Aug 02 2015 17:51:46

%S 0,0,1,0,1,1,0,1,2,2,0,1,3,2,1,0,1,4,2,4,2,0,1,5,2,9,3,1,0,1,6,2,16,4,

%T 8,3,0,1,7,2,25,5,27,3,2,0,1,8,2,36,6,64,3,4,2,0,1,9,2,49,7,125,3,6,5,

%U 1,0,1,10,2,64,8,216,3,8,10,16,3,0,1,11,2,81,9,343,3,10,17,81,4,1,0,1,12,2

%N Suppose n=(p1^e1)(p2^e2)... where p1,p2,... are the prime numbers and e1,e2,... are nonnegative integers. Then we can define Pn(x) = e1 + (e2)*x + (e3)*(x^2) + (e4)*(x^3) + ... + (ek)*(x^(k-1)) + ... The sequence is the table T(x,n)=Pn(x) read by antidiagonals.

%C This square array is the transpose of A104244, see comments there.

%e a(13)=3 because 3=(p1^0)(p2^1)(p3^0)..., so P3(x)=x. Hence a(13) = T(3,3) = P3(3) = 3.

%e The top left corner of the array:

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

%e 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

%e 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

%e 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

%e 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

%e 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13

%e 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728

%e 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

%e 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

%e 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145

%e 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, 14641, 20736

%e 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

%e ...

%o (Scheme)

%o (define (A104245 n) (A104244bi (A004736 n) (A002260 n))) ;; A104244bi given in A104244.

%o ;; _Antti Karttunen_, Jul 29 2015

%Y Transpose: A104244.

%Y Main diagonal: A090883.

%K easy,nonn,tabl

%O 1,9

%A _Olaf Voß_, Feb 26 2005

%E Starting offset changed from 0 to 1 by _Antti Karttunen_, Jul 29 2015