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.
a(19) = 1 + 2 + 3 + 7 + 8 + 52 + 53 + 288 + 1209 + 5247 + 5248 + 71395 + 71396 + 375779 + 6957533 + 52310862 + 52310863 + 1152622553 + 1152622554.
Table[Length[IntegerPartitions[1, All, 1/Range[n]]], {n, 1, 20}] (* Ben Branman, Apr 21 2012 *)
a037268 n = a037268_list !! (n-1)
a037268_list = filter ((== 1) . a168046) $
takeWhile (<= 999999999) a214959_list
-- Reinhard Zumkeller, Aug 02 2012
A037268 := proc(n) option remember: local d,k: if(n=1)then return 1: fi: for k from procname(n-1)+1 do d:=convert(k,base,10): if(not member(0,d) and add(1/d[j],j=1..nops(d))=1)then return k: fi: od: end: seq(A037268(n),n=1..50); # Nathaniel Johnston, May 28 2011
Select[Range[10000],Total[1/(IntegerDigits[#]/.(0->1))]==1&] (* Harvey P. Dale, Jul 23 2025 *)
lista(nn) = {for (n=1, nn, d = digits(n); if (vecmin(d) && (sum(k=1, #d, 1/d[k])==1), print1(n, ", ")););} \\ Michel Marcus, Jul 06 2015
from fractions import Fraction def ok(n): sn = str(n) return False if '0' in sn else sum(Fraction(1, int(d)) for d in sn) == 1 def aupto(limit): return [m for m in range(1, limit+1) if ok(m)] print(aupto(8824)) # Michael S. Branicky, Jan 22 2021
a(6)=2 since there are two fractions 1=1/1 and 1=1/2+1/3+1/6.
a(4) = 4 since there are four fractions 1=1/2+1/4+1/4, 1=1/4+1/2+1/4, 1=1/4+1/4+1/2 and 1=1/4+1/4+1/4+1/4.
a(4) = 7 because we have 16 * (1/16) = 12 * (1/16) + 1/4 = 8 * (1/16) + 2 * (1/4) = 4 * (1/16) + 3 * (1/4) = 9 * (1/9) = 4 * (1/4) = 1.
From _David A. Corneth_, Nov 24 2024: (Start)
To find a(12) we can rewrite the problem as "Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, 1/4^2, 1/5^2, 1/6^2, 1/8^2, 1/9^2, 1/10^2, 1/12^2} + |{7, 11}|". The lcm of (1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 8^2, 9^2, 10^2, 12^2) is 129600. So this comes a partition problem of (number of partitions of 129600 into parts 129600, 32400, 14400, 8100, 5184, 3600, 2025, 1600, 1296, 900) + |{7, 11}|. (End)
a(4) = 11 because we have 64 * (1/64) = 56 * (1/64) + 1/8 = 48 * (1/64) + 2 * (1/8) = 40 * (1/64) + 3 * (1/8) = 32 * (1/64) + 4 * (1/8) = 24 * (1/64) + 5 * (1/8) = 16 * (1/64) + 6 * (1/8) = 8 * (1/64) + 7 * (1/8) = 27 * (1/27) = 8 * (1/8) = 1.
a(2) = 5 because we have [1/2, 1/2, 1/2, 1/2], [1/2, 1/2, 1], [1/2, 1, 1/2], [1, 1/2, 1/2] and [1, 1].
b:= proc(n, r) option remember; `if`(r=0, 1,
add(`if`(r*j<1, 0, b(n, r-1/j)), j=1..n))
end:
a:= n-> b(n$2):
seq(a(n), n=0..10); # Alois P. Heinz, Dec 12 2024
from functools import lru_cache from fractions import Fraction def A378842(n): @lru_cache(maxsize=None) def f(r): return 1 if r==0 else sum(f(r-Fraction(1,j)) for j in range(int(Fraction(1,r))+(r.numerator!=1),n+1)) return f(n) # Chai Wah Wu, Dec 14 2024
a(5) = 7 because we have 9 * (1/9) = 6 * (1/9) + 1/3 = 3 * (1/9) + 2 * (1/3) = 7 * (1/7) = 5 * (1/5) = 3 * (1/3) = 1.
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