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%I #69 Oct 27 2023 22:07:52
%S 0,1,2,3,10,11,12,13,20,21,22,30,31,100,101,102,103,110,111,112,113,
%T 120,121,122,130,200,201,202,210,211,212,220,221,300,301,310,311,1000,
%U 1001,1002,1003,1010,1011,1012,1013,1020,1021,1022,1030,1031,1100,1101,1102
%N Skinny numbers: numbers n such that there are no carries when n is squared by "long multiplication".
%C There are several equivalent formulations. Suppose the decimal expansion of n is n = Sum_{i = 0..k } d_i 10^i, where 0 <= d_i <= 9 for i = 0..k.
%C Then n is skinny if and only if:
%C (i) e_i = Sum_{ j = 0..i } d_j d_{i-j} <= 9 for i = 0 .. 2k-1;
%C (ii) if P_n(X) = Sum_{i = 0..k } d_i X^i (so P_n(10) = n) then P_{n^2}(X) = P_n(X)^2;
%C (iii) R(n^2) = R(n)^2, where R(n) means the digit reversal of n;
%C (iv) (sum of digits of n)^2 = sum of digits of n^2.
%C This entry is a merging and reworking of earlier entries from _Asher Auel_, May 17 2001 and _Amarnath Murthy_, Aug 15 2005. Thanks to _Andrew S. Plewe_ for suggesting that these sequences might be identical.
%C Also, numbers n in base 10 whose base 10 expansion of n^2 is the same as if n were interpreted in some base b>10 and n^2 also calculated in that base. - Andrew Silberman (sandrew(AT)math.upenn.edu), Oct 30 2006
%C From _David Applegate_ and _N. J. A. Sloane_, Jun 14 2007: (Start)
%C The decimal expansion of a skinny number n may contain only 0's, 1's, 2's and 3's.
%C There may be at most one 3 and if there is a 3 then there can be no 2's. (Hence of course if there are any 2's then there can be no 3's.)
%C There is no limit to the number of 1's and 2's - consider for example Sum_{i=0..m} 10^{2^i} and 2*Sum_{i=0..m} 10^{2^i}.
%C These are necessary conditions, but are not sufficient (e.g., 131 is not skinny). (End)
%C There are fifty-five skinny numbers without a 0 digit, the greatest being a(5203) = 111111111. - _Jason Zimba_, Jul 05 2020
%H T. D. Noe, <a href="/A061909/b061909.txt">Table of n, a(n) for n = 1..15276</a> (terms less than 10^10)
%H <a href="/index/Sq#sqrev">Index entry for sequences related to reversing digits of squares</a>
%H <a href="/index/Ca#CARRYLESS">Index entries for sequences related to carryless arithmetic</a>
%F a(n) >> n^2.0959..., where the exponent is log 10/log 3. - _Charles R Greathouse IV_, Sep 21 2012
%e 12 is a member as 12^2 = 144, digit reversal of 144 is 441 = 21^2.
%e 13 is a member as 13 squared is 169 and sqrt(961) = 31.
%e 113 is a member as 113^2 = 12769, reversal(12769) = 96721 = 311^2.
%e (Sum of digits of 13)^2 = 4^2 = 16 and sum of digits of 13^2 = sum of digits of 169 = 16.
%e 10^k is in the sequence for all k >= 0, since reversal((10^k)^2) = 1 = (reversal(10^k))^2. - _Ryan Propper_, Sep 09 2005
%p rev:=proc(n) local nn, nnn: nn:=convert(n,base,10): add(nn[nops(nn)+1-j]*10^(j-1),j=1..nops(nn)) end: a:=proc(n) if sqrt(rev(n^2))=rev(n) then n else fi end: seq(a(n),n=1..1200); # _Emeric Deutsch_, Mar 31 2007
%p f := []: for n from 1 to 1000 do if (convert(convert(n,base,10),`+`))^2 = convert(convert(n^2,base,10),`+`) then f := [op(f), n] fi; od; f; # _Asher Auel_
%t r[n_] := FromDigits[Reverse[IntegerDigits[n]]]; Do[If[r[n]^2 == r[n^2], Print[n]], {n, 1, 10^4}] (* _Ryan Propper_, Sep 09 2005 *)
%t Select[Range[0,1200],IntegerReverse[#^2]==IntegerReverse[#]^2&] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Aug 02 2017 *)
%o (Haskell)
%o a061909 n = a061909_list !! (n-1)
%o a061909_list = filter (\x -> a004086 (x^2) == (a004086 x)^2) [0..]
%o -- _Reinhard Zumkeller_, Jul 08 2011
%o (PARI) is(n)=sumdigits(n)^2==sumdigits(n^2) \\ _Charles R Greathouse IV_, Jun 21 2017
%o (Python)
%o from itertools import count, islice, product
%o def sd(n): return sum(map(int, str(n)))
%o def ok(n): return sd(n**2) == sd(n)**2
%o def agen(): # generator of terms
%o yield from [0, 1, 2, 3]
%o for d in count(2):
%o for f in "123":
%o rset = "01" if f == "3" else "012" if f == "2" else "0123"
%o for r in product(rset, repeat=d-1):
%o t = int(f+"".join(r))
%o if ok(t): yield t
%o print(list(islice(agen(), 53))) # _Michael S. Branicky_, Dec 23 2022
%Y A085305 is a subsequence.
%Y Cf. A007953, A004159, A061903, A061910, A129967, A129968, A129969, A129970, A129971, A123977, A159953, A169939.
%Y The primes in this sequence are given by A085306.
%Y Numbers n such that A067552(n) = 0.
%K base,easy,nonn,nice,look
%O 1,3
%A _N. J. A. Sloane_, Jun 14 2007
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