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If n = 2^a * 3^b * 5^c * 7^d * ... then a(n) = a + 10 * b + 100 * c + 1000 * d + ... .
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%I #49 Mar 17 2024 12:13:23

%S 0,1,10,2,100,11,1000,3,20,101,10000,12,100000,1001,110,4,1000000,21,

%T 10000000,102,1010,10001,100000000,13,200,100001,30,1002,1000000000,

%U 111,10000000000,5,10010,1000001,1100,22,100000000000,10000001,100010

%N If n = 2^a * 3^b * 5^c * 7^d * ... then a(n) = a + 10 * b + 100 * c + 1000 * d + ... .

%C Are there any other numbers besides n=12 for which n=a(n) ? - _Ctibor O. Zizka_, Oct 08 2008

%C The sequence is a morphism from (N*,*) into (N,+), cf. formula. Up to n=1023, the digit sum A007953(a(n)) equals Omega(n) = A001222(n). This holds whenever A051903(n)<10. Restricted to these n, the sequence is also injective. However, when n is a multiple of 2^10, 3^10, 5^10 etc, then a(n) is equal to some a(m) with m<n. - _M. F. Hasler_, Nov 16 2008

%C This has been called the "Exponential Prime Power Representation" of n by W. Nissen in a post to the sci.math newsgroup (where probably some more sophisticated convention for representing digits > 10 would be used). - _M. F. Hasler_, Jul 03 2016

%H Robert Israel, <a href="/A054841/b054841.txt">Table of n, a(n) for n = 1..10000</a> (n=1..1023 from Michael De Vlieger)

%H Evans A Criswell, <a href="http://www.hsvmovies.com/static_subpages/personal/math/puzzle_sequence.html">A Sequence Puzzle</a> (Posted to rec.puzzles Jan 01 1997)

%H Walter Nissen, <a href="http://upforthecount.com/math/epprep.html">Exponential Prime Power Representation</a>, sci.math newsgroup, May 23 1995.

%F a(m*n) = a(m) + a(n) for all m,n > 0. A007953(a(n))=A001222(n) for all n such that A051903(n) < 10. - _M. F. Hasler_, Nov 16 2008

%F a(n) = sum(10^(A049084(A027748(k))-1) * A124010(k): k = 1..A001221(n)). - _Reinhard Zumkeller_, Aug 03 2015

%F a(A054842(n)) = n for all n >= 0. - _Antti Karttunen_, Aug 29 2016

%F a(n) = Sum_{i>0} e_i*10^(i-1) when n = Product_{i>0} prime(i)^e_i. - _M. F. Hasler_, Mar 14 2018

%e a(25) = 200 because 25 = 5^2 * 3^0 * 2^0.

%e a(1024) = 10 = a(3), because 1024 = 2^10; but this two-digit multiplicity overflows into the 10^1 position, which encodes for powers of three.

%p A:= n -> add(t[2]*10^(numtheory:-pi(t[1])-1),t= ifactors(n)[2]);

%p seq(A(n), n=1..1000); # _Robert Israel_, Jul 24 2014

%t a054841[n_Integer] := Catch[FromDigits[IntegerDigits[Apply[Plus,

%t Which[n == 0, Throw["undefined"],

%t n == 1, 0,

%t Max[Last /@ FactorInteger @ n] > 9, Throw["overflow"],

%t True, Power[10, PrimePi[Abs[#]] - 1]] & /@

%t Flatten[ConstantArray @@@ FactorInteger[n]]]]]] (* _Michael De Vlieger_, Jul 24 2014 *)

%o (PARI) A054841(n)=sum(i=1,#n=factor(n)~,10^primepi(n[1,i])*n[2,i])/10 \\ _M. F. Hasler_, Nov 16 2008

%o (Haskell)

%o a054841 1 = 0

%o a054841 n = sum $ zipWith (*)

%o (map ((10 ^) . subtract 1 . a049084) $ a027748_row n)

%o (map fromIntegral $ a124010_row n)

%o -- _Reinhard Zumkeller_, Aug 03 2015

%o (Python)

%o from sympy import factorint, primepi

%o def a(n): return sum(e*10**(primepi(p)-1) for p, e in factorint(n).items())

%o print([a(n) for n in range(1, 41)]) # _Michael S. Branicky_, Mar 17 2024

%Y Row 10 of A104244.

%Y Left inverse of A054842.

%Y Cf. A001222, A048675, A090880, A090881, A090882, A276075, A276085 (analogous constructions for other bases), A090883, A090884, A049084, A027748, A124010, A056239.

%K base,nonn

%O 1,3

%A _Henry Bottomley_, Apr 11 2000