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A224752
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a(1)=1; thereafter a(n) = smallest number m such that a(n-1)+m = (a(n-1) followed by the leading digit of m).
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10
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1, 10, 99, 899, 8099, 72898, 656088, 5904797, 53143178, 478288606, 4304597458, 38741377125, 348672394128, 3138051547155, 28242463924397, 254182175319575, 2287639577876177, 20588756200885595, 185298805807970356, 1667689252271733205, 15009203270445598846, 135082829434010389615
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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The sequence is infinite: a(n) always exists.
One could start with a(0) = 0, followed by the given a(n). - M. F. Hasler, Oct 10 2019
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REFERENCES
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Eric Angelini, Postings to the Sequence Fans Mailing List, Apr 13 2013
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LINKS
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E. Angelini, Magic Sums [Cached copy, with permission]
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EXAMPLE
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1+10=11 < 1+1 >
10+99=109 < 10+9 >
99+899=998 < 99+8 >
899+8099=8998 < 899+9 >
8099+72898=80997 < 8099+7 >
...
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MATHEMATICA
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leadingDigit[x_] := IntegerPart[N[x/10^IntegerPart[Log[10, x]]]];
successor[x_] :=(
y = 1;
If[leadingDigit[z = 9x+y] == y, z,
y = leadingDigit[9x];
If[leadingDigit[z = 9x+y] == y, z,
y += 1;
If[leadingDigit[z = 9x+y] == y, z,
Print["Bug"]]]]);
(* Gilles Esposito-Farèse, Apr 21 2013 *)
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PROG
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(PHP) <?
/*
calcul de la suite telle que la somme
s(n) + s(n+1) s'obtient en concaténant le
premier chiffre de s(n+1) après s(n).
Eric Angelini, 18/04/2013
*/
function leading_digit ($n) {
return (int) substr("$n", 0, 1) ;
}
function successor ($n) {
$p = 9*$n ;
for ( $a = 1 ; $a <= 9 ; $a++ ) {
if (leading_digit($p+$a) == $a) {
return ($p+$a) ;
}
}
die("nothing found for successor($n)") ;
}
$x = $_REQUEST["x"] ;
$n = $_REQUEST["n"] ;
if ( $n === "0" ) {
for ($i=1 ; $i<=$x ; $i++) {
echo "$i → ", successor($i), "<br>\n" ;
}
} else {
if (! $x) $x = 1 ;
if (! $n) $n = 15 ;
while ($n--) {
echo "$x, " ;
$x = successor($x) ;
}
}
?>
A224752_nxt(x, n=0)={x*=9; while(digits(x++)[1]!=n++, ); x} \\ (End)
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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