A101701
Numbers n such that n = sum of the reversals of divisors of n.
Original entry on oeis.org
1, 207321, 890827, 7591023, 18368601, 4885292403
Offset: 1
18368601 is in the sequence because divisors of 18368601 are 1, 3, 6122867, 18368601 and 18368601 = 1 + 3 + 7682216 + 10686381.
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Do[h = Divisors[n]; l = Length[h]; If[n == Sum[ FromDigits[Reverse[IntegerDigits[h[[k]]]]], {k, l}], Print[n]], {n, 370000000}]
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from sympy import divisors
A101701_list = [n for n in range(1,10**6) if n == sum([int(d) for d in (str(x)[::-1] for x in divisors(n))])]
# Chai Wah Wu, Dec 06 2014
A244251
Numbers k such that if m = (sum of the reverses of the aliquot parts of k) then k = (sum of the reverses of the aliquot parts of m).
Original entry on oeis.org
6, 98, 145, 244, 285, 133857
Offset: 1
Aliquot parts of 98 are 1, 2, 7, 14, 49 and the sum of their reverses is 1 + 2 + 7 + 41 + 94 = 145. Aliquot parts of 145 are 1, 5, 29 and the sum of their reverses is 1 + 5 + 92 = 98.
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with(numtheory): T:=proc(w) local x,y,z; x:=0; y:=w;
for z from 1 to ilog10(w)+1 do x:=10*x+(y mod 10); y:=trunc(y/10); od; x; end:
P:=proc(q) local a,b,k,n; for n from 1 to q do
a:=sort([op(divisors(n))]); b:=add(T(a[k]),k=1..nops(a)-1); a:=sort([op(divisors(b))]); b:=add(T(a[k]),k=1..nops(a)-1);
if b=n then print(n); fi; od; end: P(10^12);
A254009
Numbers that divide the sum of the reverse of their aliquot parts (A069250).
Original entry on oeis.org
1, 6, 244, 285, 944, 1242, 3738, 22644, 37686, 58950, 85512, 124944, 130410, 133857, 235644, 3202101, 5367582, 5663697, 45165231, 141635817, 214939686, 736140702, 2395863144, 4992033177, 28406362140, 30364415451
Offset: 1
Aliquot parts of 944 are 1, 2, 4, 8, 16, 59, 118, 236, 472 and the sum of their reverse is 1 + 2 + 4 + 8 +61 + 95 + 811 + 632 + 274 = 1888. Finally, 1888 / 944 = 2.
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with(numtheory):
T:=proc(w) local x,y,z; x:=w; y:=0;
for z from 1 to ilog10(x)+1 do y:=10*y+(x mod 10); x:=trunc(x/10); od; y; end:
P:=proc(q) local a,b,k; global n;
for n from 1 to q do a:=sort([op(divisors(n))]); b:=add(T(a[k]),k=1..nops(a)-1);
if type(b/n,integer) then print(n); fi; od; end: P(10^9);
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isok(n) = (sumdiv(n, d, (d != n)* eval(concat(Vecrev(Str(d))))) % n) == 0; \\ Michel Marcus, Feb 27 2015
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