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Total number of bases and exponents in Quetian Superfactorization of n, excluding the unity-exponents at the tips of branches.
+10
15
0, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 4, 1, 2, 2, 3, 1, 3, 1, 3, 3, 2, 1, 4, 2, 3, 2, 3, 1, 3, 2, 3, 2, 2, 1, 4, 1, 2, 3, 3, 2, 3, 1, 3, 2, 3, 1, 4, 1, 2, 3, 3, 2, 3, 1, 4, 3, 2, 1, 4, 2, 2, 2, 3, 1, 4, 2, 3, 2, 2, 2, 3, 1, 3, 3, 4, 1, 3
COMMENTS
Quetian Superfactorization proceeds by factoring a natural number to its unique prime-exponent factorization (p1^e1 * p2^e2 * ... pj^ej) and then factoring recursively each of the (nonzero) exponents in similar manner, until unity-exponents are finally encountered.
FORMULA
Additive with a(p^e) = 1 + a(e).
Other identities. For all n >= 1:
EXAMPLE
a(64) = 3, as 64 = 2^6 = 2^(2^1*3^1) and there are three non-1 nodes in that superfactorization. Similarly, for 360 = 2^(3^1) * 3^(2^1) * 5^1 we get a(360) = 5. a(65536) = a(2^(2^(2^(2^1)))) = 4.
MAPLE
a:= proc(n) option remember; `if`(n=1, 0,
add(1+a(i[2]), i=ifactors(n)[2]))
end:
PROG
(Scheme, with memoization-macro definec)
(PARI)
A067029(n) = if(n<2, 0, factor(n)[1, 2]);
CROSSREFS
Cf. A276230 (gives first k such that a(k) = n, i.e., this sequence is a left inverse of A276230). After n=1 differs from A038548 for the first time at n=24, where A038548(24)=4, while a(24)=3.
AUTHOR
Antti Karttunen, May 09 2005 based on Leroy Quet's message ('Super-Factoring' An Integer) posted to SeqFan-mailing list on Dec 06 2003.
Additive function a(n) defined by the recursive formula a(1)=1 and a(p^k)=a(k) for any prime p.
+10
13
1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 3, 2, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 1, 3, 1, 2, 3
COMMENTS
That is, if i, j, k, ... are relatively prime, then a(i*j*k*...) = a(i) + a(j) + a(k) + ... - N. J. A. Sloane, Nov 20 2007 Starts almost the same as A001221 (the number of distinct primes dividing n): the first twelve terms which are different are a(1), a(64), a(192), a(320), a(448), a(576), a(704), a(729), a(832), a(960), a(1024) and a(1088), since the first non-unitary values of n are a(6) and(10). - Henry Bottomley, Sep 23 2002
EXAMPLE
a(30) = a(5^1 * 3^1 * 2^1) = a(1) + a(1) + a(1) = 3.
MAPLE
a:= proc(n) option remember; `if`(n=1, 1,
add(a(i[2]), i=ifactors(n)[2]))
end:
MATHEMATICA
a[1] = 1; a[n_] := a[n] = Plus @@ a /@ FactorInteger[n][[All, 2]]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Sep 19 2012 *)
PROG
(Haskell)
a064372 1 = 1
a064372 n = sum $ map a064372 $ a124010_row n
Exponent-recursed cross-domain bijection from N to GF(2)[X]. Variant of A091202 and A106442.
+10
10
0, 1, 2, 3, 4, 7, 6, 11, 8, 5, 14, 13, 12, 19, 22, 9, 16, 25, 10, 31, 28, 29, 26, 37, 24, 21, 38, 15, 44, 41, 18, 47, 128, 23, 50, 49, 20, 55, 62, 53, 56, 59, 58, 61, 52, 27, 74, 67, 48, 69, 42, 43, 76, 73, 30, 35, 88, 33, 82, 87, 36, 91, 94, 39, 64, 121, 46, 97, 100, 111
FORMULA
a(0)=0, a(1)=1, a(p_i) = A014580(i) for primes p_i with index i and for composites n = p_i^e_i * p_j^e_j * p_k^e_k * ..., a(n) = A048723(a(p_i), a(e_i)) X A048723(a(p_j), a(e_j)) X A048723(a(p_k), a(e_k)) X ..., where X stands for carryless multiplication of GF(2)[X] polynomials ( A048720) and A048723(n, y) raises the n-th GF(2)[X] polynomial to the y:th power.
EXAMPLE
a(5) = 7, as 5 is the 3rd prime and the third irreducible GF(2)[X] polynomial x^2+x+1 is encoded as A014580(3) = 7. a(32) = a(2^5) = A048723( A014580(1),a(5)) = A048723(2,7) = 128. a(48) = a(3 * 2^4) = 3 X A048723(2,a(4)) = 3 X A048723(2,4) = 3 X 16 = 48.
Exponent-recursed cross-domain bijection from GF(2)[X] to N. Variant of A091203 and A106443.
+10
10
0, 1, 2, 3, 4, 9, 6, 5, 8, 15, 18, 7, 12, 11, 10, 27, 16, 81, 30, 13, 36, 25, 14, 33, 24, 17, 22, 45, 20, 21, 54, 19, 512, 57, 162, 55, 60, 23, 26, 63, 72, 29, 50, 51, 28, 135, 66, 31, 48, 35, 34, 19683, 44, 39, 90, 37, 40, 99, 42, 41, 108, 43, 38, 75, 64, 225, 114, 47
COMMENTS
This map from the multiplicative domain of GF(2)[X] to that of N preserves 'superfactorized' structures, e.g. A106493(n) = A106490(a(n)), A106494(n) = A106491(a(n)), A106495(n) = A064372(a(n)). Shares with A091203 and A106443 the property that maps A014580(n) to A000040(n). Differs from the plain variant A091203 for the first time at n=32, where A091203(32)=32, while a(32)=512. Differs from the variant A106443 for the first time at n=48, where A106443(48)=768, while a(48)=48. Differs from a yet deeper variant A106447 for the first time at n=13, where A106447(13)=23, while a(13)=11.
FORMULA
a(0)=0, a(1)=1. For irreducible GF(2)[X] polynomials ir_i with index i (i.e. A014580(i)), a(ir_i) = A000040(i) and for composite polynomials n = A048723(ir_i, e_i) X A048723(ir_j, e_j) X A048723(ir_k, e_k) X ..., a(n) = a(ir_i)^a(e_i) * a(ir_j)^a(e_j) * a(ir_k)^a(e_k) * ... = A000040(i)^a(e_i) * A000040(j)^a(e_j) * A000040(k)^a(e_k), where X stands for carryless multiplication of GF(2)[X] polynomials ( A048720) and A048723(n, y) raises the n-th GF(2)[X] polynomial to the y:th power, while * is the ordinary multiplication and ^ is the ordinary exponentiation.
EXAMPLE
a(5) = 9, as 5 encodes the GF(2)[X] polynomial x^2+1, which is the square of the second irreducible GF(2)[X] polynomial x+1 (encoded as 3) and the square of the second prime is 3^2=9. a(32) = a( A048723(2,5)) = 2^a(5) = 2^9 = 512. a(48) = a(3 X A048723(2,4)) = 3 * 2^a(4) = 3 * 2^4 = 3 * 16 = 48.
Total number of bases and exponents in GF(2)[X] Superfactorization of n, including the unity-exponents at the tips of branches.
+10
6
1, 2, 2, 3, 3, 4, 2, 3, 4, 5, 2, 5, 2, 4, 3, 4, 4, 6, 2, 6, 3, 4, 4, 5, 2, 4, 5, 5, 4, 5, 2, 4, 4, 6, 4, 7, 2, 4, 5, 6, 2, 5, 4, 5, 5, 6, 2, 6, 4, 4, 4, 5, 4, 7, 2, 5, 5, 6, 2, 6, 2, 4, 5, 5, 6, 6, 2, 7, 3, 6, 4, 7, 2, 4, 5, 5, 4, 7, 4, 7, 3, 4, 6, 6, 5, 6, 2, 5, 4, 7, 2, 7, 4, 4, 5, 6, 2, 6, 5, 5, 6, 6
EXAMPLE
a(64) = 5, as 64 = A048723(2,6) = A048723(2,( A048723(2,1) X A048723(3,1))) and there are five nodes in that superfactorization. Similarly, for 27 = 5x7 = A048723(3, A048723(2,1)) X A048273(7,1) we get a(27) = 5. The operation X stands for GF(2)[X] multiplication defined in A048720, while A048723(n,y) raises the n-th GF(2)[X] polynomial to the y:th power.
Total sum of bases and exponents in Quetian Superfactorization of n, excluding the unity-exponents at the tips of branches.
+10
4
0, 2, 3, 4, 5, 5, 7, 5, 5, 7, 11, 7, 13, 9, 8, 6, 17, 7, 19, 9, 10, 13, 23, 8, 7, 15, 6, 11, 29, 10, 31, 7, 14, 19, 12, 9, 37, 21, 16, 10, 41, 12, 43, 15, 10, 25, 47, 9, 9, 9, 20, 17, 53, 8, 16, 12, 22, 31, 59, 12, 61, 33, 12, 7, 18, 16, 67, 21, 26, 14, 71, 10, 73, 39, 10, 23, 18
FORMULA
Additive with a(p^e) = p + a(e).
EXAMPLE
a(64) = 7, as 64 = 2^6 = 2^(2^1*3^1) and 2+2+3=7. Similarly, for 360 = 2^(3^1) * 3^(2^1) * 5^1 we get a(360) = 2+3+3+2+5 = 15. See comments at A106490.
MAPLE
a:= proc(n) option remember;
add(i[1]+a(i[2]), i=ifactors(n)[2])
end:
MATHEMATICA
a[1] = 0; a[n_] := a[n] = #[[1]] + a[#[[2]]]& /@ FactorInteger[n] // Total; Array[a, 100] (* Jean-François Alcover, Mar 03 2016 *)
AUTHOR
Antti Karttunen, May 09 2005 based on Leroy Quet's message ('Super-Factoring' An Integer) posted to SeqFan-mailing list on Dec 06 2003.
A "cosh transform" of binomial(2n,n-1).
+10
0
0, 1, 4, 18, 80, 365, 1692, 7945, 37648, 179595, 861020, 4143832, 20004096, 96810779, 469508340, 2281123530, 11100465216, 54093131147, 263929559436, 1289217255934, 6303934406640, 30853639964847, 151139139048084
COMMENTS
Mean of binomial and inverse binomial transform of A001791.
FORMULA
E.g.f.: cosh(x)exp(2x)I_1(2x), where I_1 is Bessel function; a(n)=sum{k=0..floor(n/2), binomial(n, 2k)binomial(2(n-2k), n-2k+1)}.
Conjecture: -(n+1)*(n-2)*a(n) +4*n*(3*n-7)*a(n-1) +(-49*n^2+193*n-148)*a(n-2) +8*(9*n-19)*(n-4)*a(n-3) +(5*n^2+115*n-418)*a(n-4) -12*(7*n-19)*(n-4)*a(n-5) +45*(n-4)*(n-5)*a(n-6)=0. - R. J. Mathar, Feb 20 2015 Conjecture: -(n-2)*(n+1)*(4*n^2-16*n+19)*a(n) +8*n*(n-2)*(4*n^2-14*n+13)*a(n-1) +2*(-28*n^4+168*n^3-375*n^2+401*n-180)*a(n-2) -8*(n-2)*(4*n^3-18*n^2+27*n-10)*a(n-3) +15*(n-2)*(n-3)*(4*n^2-8*n+7)*a(n-4)=0. - R. J. Mathar, Feb 20 2015
MAPLE
add(binomial(n, 2*k)*binomial(2*(n-2*k), n-2*k+1), k=0..floor(n/2)) ;
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