This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
After 10 the next term is 23 and not 11. Any number containing 0 or 1 would occur only after all the digits from 2 to 9 have occurred once.
counts:= Array(0..9):
cp:= Array(0..9):
counts[0]:= 1:
A[0]:= 0:
for n from 1 to 70 do
for x from A[n-1]+1 do
L:= convert(x,base,10);
ArrayTools:-Copy(counts,cp);
for t in L do cp[t]:= cp[t]+1 od:
if max(cp) - min(cp) <= 1 then
A[n]:= x;
ArrayTools:-Copy(cp,counts);
break
fi
od
od:
seq(A[i],i=0..70); # Robert Israel, Sep 03 2015
The first eight rows of the array (and the last one) are: 0 1 2 3 4 5 6 7 8 9 10 23 45 67 89 12 30 46 57 98 13 20 47 58 69 14 25 36 78 90 15 24 37 80 96 16 27 34 5089 17 26 35 4089 ... 9876543210
a(11) = 10. The first ten terms contain the digits 0 to 9 so a(11) begins the second block of ten digits so is free to choose any distinct digits that can form the lowest number not yet seen. The lowest such number is 10. a(12) = 23 as a(11) used the digits 0 and 1, so the next lowest distinct number that can be created from the available digits 2 to 9 is 23. a(16) = 12 as a(15) ended the second block of ten digits so a(16) is free to choose any distinct digits that can form the lowest number not yet seen. The lowest such number is 12. a(25) = 890. This term ends the fourth block of ten unique digits and the only available digits to use are 8 and 9. But both 89 and 98 have previously appeared so the next lowest number possible is 890. This term uses the 0 digit in the fifth block of ten digits. This is the first term that differs from A120125. a(64) = 607. This is the first time where a term is determined by the requirement that the next term must exist. As a(64) is written the three available digits are 0, 6 and 7. The number 67 has been used but 76 is available. But setting a(64) to 76 would force the next term to start with 0. As we do not allow 0 as a leading digit this would mean a(65) would not exist. Thus 76 cannot be chosen, and the next lowest number that can be created from 0, 6 and 7 that avoids leaving 0 as the only available digit is 607.
See Links section.
a(1) = 0 = 0_2, a(2) = 1 = 1_2 (the only way to use two numbers in one pair). a(3) = 2 = 10_2 (the next smallest unused number to contain 1 then 0 and fill the next pair). a(4) = 4 = 100_2 (the next smallest unused number to contain 1 then 0, which fills the next pair, and then a 0 in the second-next pair; note that 3 = 11_2 can never be a term). a(5) = 5 = 101_2 (the next smallest unused number to contain a 1 to fill the pair started by a(4) and then 0 and 1 to fill the next pair). a(6) = 9 = 1001_2 (the next smallest unused number to contain two pairs both of which contain 0 and 1 and fill the next two pairs; note that 6 = 110_2 and 7 = 111_2 would fill the next pair with two 1's while 8 = 1000_2 would fill the second-next pair with two 0's). Neither 7 nor 8 can ever be terms.
\\ See Links section.
# Script (R, using base only) to build values in the sequence defined # in A248651: # a(n) is the smallest number greater than a(n-1) such that no # digit appears in the listing of all terms more than one time # more than any other digit in the listing, with a(1) = 1. digits <- c(0:9) #vector of the digits digitCounts <- rep(0L,10) #vector for digit count tracking A <- c() #vector of entries A[1] <- 1L #first entry failed <- 1L #failed entry digitCounts[2] = 1L #impact of first entry on digitCounts # Build a function that turns a number into a vector of its # component digits digits <- function(x) { if(length(x) > 1 ) { lapply(x, digits) } else { n <- floor(log10(x))+1L rev( x %/% 10^seq(0, length.out=n) %% 10 ) } } # Engine for testing digits, runs VERY SLOW, but checks every integer. # I welcome improvements, note that it respects incrementing # more than one digit at a time if that makes sense. while (length(A) < 300) { i <- max(failed, max(A)) + 1 newCounts <- digitCounts digitList <- digits(i) for (j in digitList) { newCounts[j + 1] <- newCounts[j + 1] + 1 } if (max(newCounts) - min(newCounts) > 1) { failed <- i } else { digitCounts <- newCounts A <- c(A,as.integer(i)) } } # Jeb Adams, Jul 17 2020
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