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.
str10:=proc(n) if n<>0 then RETURN(convert(n,base,10)) else RETURN([0]) fi end:ds:=proc(s) local j: RETURN(add(s[j]*10^(j-1),j=1..nops(s))): end: ap:=1: for n from 1 to 12 do m:=ds([seq(op(str10(i*n)),i=str10(ap))]): printf("%d, ",m):ap:=m od: # C. Ronaldo
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1, (l-> parse(cat(seq(
l[-i]*n, i=1..nops(l)))))(convert(a(n-1), base, 10)))
end:
seq(a(n), n=0..12); # Alois P. Heinz, Dec 12 2020
nxt[{n_,a_}]:={n+1,FromDigits[Flatten[IntegerDigits/@((n+1)*IntegerDigits[ a])]]}; Transpose[NestList[nxt,{1,1},10]][[2]] (* Harvey P. Dale, Mar 26 2015 *)
Example for n = 3: Take the number 569. Sum the permutations of its digits: 569 + 596 + 659 + 695 + 956 + 965 = 4440. Add all its digits: 5 + 6 + 9 = 20. Divide: 4440 / 20 = 222. General proof for n = 3: Number: abc where a,b,c are distinct. The sum of the permutations is 200*(a+b+c) + 20*(a+b+c) + 2*(a+b+c) = 222*(a+b+c), so a(3) = 222.
Table[(n-1)! (10^n-1)/9,{n,20}] (* Harvey P. Dale, Mar 15 2024 *)
a(n) = (10^n - 1) / 9 * (n-1)! \\ David A. Corneth, Jan 13 2019
f = lambda n:+(n==0) or n*f(n-1)
def seq(n):
if n==0: return
l = []
for i in range(1, n + 1):
# following line with a string repeat
# s = int('1'*i)
s = 0
for j in range(i):
s += 10 ** j
l += [s*f(i-1)]
return l
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