A265894 a(n) = A099563(A001813(n)); the most significant digit in factorial base representation of (2n)! / n!.
1, 1, 2, 1, 2, 6, 1, 4, 1, 2, 7, 1, 3, 10, 1, 3, 11, 1, 3, 10, 1, 3, 8, 25, 2, 6, 19, 1, 4, 13, 38, 2, 7, 23, 1, 4, 13, 39, 2, 6, 20, 1, 3, 9, 29, 1, 4, 13, 40, 1, 5, 16, 51, 2, 6, 20, 62, 2, 7, 23, 70, 2, 8, 25, 77, 2, 8, 25, 79, 2, 8, 25, 78, 2, 7, 23, 73, 2, 6, 21, 66, 1, 6, 18, 57, 1, 4, 15, 47, 1, 3, 12, 38, 118, 3, 9
Offset: 0
Examples
The terms A001813(0) .. A001813(8) in factorial base representation (A007623) look as: 1, 10, 200, 10000, 220000, 6000000, 174000000, 4760000000, 110000000000, ... Taking the first digit (actually: a place holder value) of each gives the terms a(0) .. a(8) of this sequence: 1, 1, 2, 1, 2, 6, 1, 4, 1, ...
Links
Crossrefs
Programs
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Mathematica
factBaseIntDs[n_] := Module[{m, i, len, dList, currDigit}, i = 1; While[n > i!, i++ ]; m = n; len = i; dList = Table[0, {len}]; Do[ currDigit = 0; While[m >= j!, m = m - j!; currDigit++ ]; dList[[len - j + 1]] = currDigit, {j, i, 1, -1}]; If[dList[[1]] == 0, dList = Drop[dList, 1]]; dList]; (* taken from A007623, Alonso del Arte, May 03 2006 *) f[n_] := factBaseIntDs[(2 n)!/n!][[1]]; Array[f, 96, 0] (* Robert G. Wilson v, Dec 25 2015 *)
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PARI
allocatemem((2^31)); \\ Enough? A099563(n) = { my(i=2,dig=0); until(0==n, dig = n % i; n = (n - dig)/i; i++); return(dig); }; A265894 = n -> A099563((2*n)! / n!);
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Scheme
(define (A265894 n) (A099563 (A001813 n)))
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Scheme
(define (A265894 n) (A265890bi (+ 1 n) n)) ;; Code for A265890bi given in A265890.
Comments