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%I #9 Sep 05 2019 05:03:46
%S 0,1,1,2,3,2,3,7,8,3,4,9,26,9,4,5,13,32,27,10,5,6,15,50,33,28,11,6,7,
%T 25,56,51,34,29,30,7,8,27,122,57,52,35,126,31,8,9,31,128,123,58,53,
%U 150,127,32,9,10,33,146,129,124,59,246,151,128,33,10,11,37,152,147,130,125,270
%N Square array A(x,y) = the number whose factorial expansion A007623 is that of x and y concatenated; zero expanded as empty string; read by ascending antidiagonals: A(0,0), A(1,0), A(0,1), A(2,0), A(1,1), A(0,2), ...
%F A(x,y) = A322001(concat(A007623(x), A007623(y))), where A322001 is a left inverse of A007623. - _M. F. Hasler_, Nov 27 2018
%e From _M. F. Hasler_, Nov 27 2018:
%e The array starts:
%e 0 1 2 3 4 5 6 ...
%e 1 3 8 9 10 11 30 ...
%e 2 7 26 27 28 29 ...
%e 3 9 32 33 34 ...
%e 4 13 50 51 ...
%e (...) (End)
%e A(4,3) = 51 which has a factorial expansion '2011' (2*24+0*6+1*2+1*1), a concatenation of factorial expansions of 4, '20' and of 3, '11'. Similarly, A(3,4) = 34 which has a factorial expansion '1120' (1*24+1*6+2*2+0*1). See A085217 for the corresponding factorial expansions.
%o (MIT Scheme) (define (A085215bi x y) (let loop ((x x) (y y) (i 2) (j (1+ (A084558 y)))) (cond ((zero? x) y) (else (loop (floor->exact (/ x i)) (+ (* (A000142 j) (modulo x i)) y) (1+ i) (1+ j))))))
%o (define (A085215 n) (A085215bi (A025581 n) (A002262 n)))
%o (define (A085216 n) (A085215bi (A002262 n) (A025581 n)))
%o (PARI) A085215(x,y)=A322001(eval(Str(A007623(x),A007623(y)))) \\ _M. F. Hasler_, Nov 27 2018
%Y Transpose: A085216. Variant: A085219. Can be used to compute A085201. Cf. A000142, A007623, A084558, A025581, A002262.
%K nonn,tabl
%O 0,4
%A _Antti Karttunen_, Jun 23 2003
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