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%I #33 Sep 09 2017 19:22:16
%S 1,2,1,2,5,2,5,10,1,10,25,10,5,50,5,10,11,10,25,250,5,250,125,50,11,
%T 250,25,250,55,50,55,110,1,110,275,250,55,6250,125,1250,121,1250,625,
%U 31250,55,6250,1375,550,11,2750,275,6250,605,6250,1375,13750,11,2750,3025,2750,55,6050,55,110,17,110,275,30250,55,68750,15125,13750,121
%N Prime factorization representation of Stern polynomials B(n,x) with only the even powers of x present: a(n) = A247503(A260443(n)).
%C a(n) = Prime factorization representation of Stern polynomials B(n,x) where the coefficients of odd powers of x are replaced by zeros. In other words, only the constant term and other terms with even powers of x are present. See the examples.
%C Proof that A001222(a(1+n)) matches _Ralf Stephan_'s formula for A000360(n): Consider functions A001222(a(n)) and A001222(A284554(n)) (= A284556(n)). They can be reduced to the following mutual recurrence pair: b(0) = 0, b(1) = 1, b(2n) = c(n), b(2n+1) = b(n) + b(n+1) and c(0) = c(1) = 0, c(2n) = b(n), c(2n+1) = c(n) + c(n+1). From the definitions it follows that the difference b(n) - c(n) for even n is b(2n) - c(2n) = -(b(n) - c(n)), and for odd n, b(2n+1) - c(2n+1) = (b(n)+b(n+1))-(c(n)+c(n+1)) = (b(n)-c(n)) + (b(n+1)-c(n+1)). Then by induction, if we assume that for 3n, 3n+1, 3n+2, ..., 6n, the value of difference b(n)-c(n) is always [0, +1, -1; repeated], it follows that from 6n to 12n the differences are [0, +1, -1; 0, +1, -1; repeated], which proves that b(n) - c(n) = A102283(n). As an implication, recurrence b can be defined without referring to c as: b(0) = 0, b(1) = 1, b(2n) = b(n) - A102283(n), b(2n+1) = b(n)+b(n+1), and this is equal to _Ralf Stephan_'s Oct 05 2003 formula for A000360, but shifted once right, with prepended zero.
%H Antti Karttunen, <a href="/A284553/b284553.txt">Table of n, a(n) for n = 0..8192</a>
%H S. Klavzar, U. Milutinovic and C. Petr, <a href="http://dx.doi.org/10.1016/j.aam.2006.01.003">Stern polynomials</a>, Adv. Appl. Math. 39 (2007) 86-95.
%F a(0) = 1, a(1) = 2, a(2n) = A003961(A284554(n)), a(2n+1) = a(n)*a(n+1).
%F Other identities. For all n >= 0:
%F a(n) = A247503(A260443(n)).
%F a(n) = A260443(n) / A284554(n).
%F a(n) = A064989(A284554(2n)).
%F A001222(a(1+n)) = A000360(n). [Proof in Comments section.]
%e n A260443(n) Stern With odd powers
%e prime factorization polynomial of x cleared -> a(n)
%e ------------------------------------------------------------------------
%e 0 1 (empty) B_0(x) = 0 0 | 1
%e 1 2 p_1 B_1(x) = 1 1 | 2
%e 2 3 p_2 B_2(x) = x 0 | 1
%e 3 6 p_2 * p_1 B_3(x) = x + 1 1 | 2
%e 4 5 p_3 B_4(x) = x^2 x^2 | 5
%e 5 18 p_2^2 * p_1 B_5(x) = 2x + 1 1 | 2
%e 6 15 p_3 * p_2 B_6(x) = x^2 + x x^2 | 5
%e 7 30 p_3 * p_2 * p_1 B_7(x) = x^2 + x + 1 x^2 + 1 | 10
%e 8 7 p_4 B_8(x) = x^3 0 | 1
%e 9 90 p_3 * p_2^2 * p_1 B_9(x) = x^2 + 2x + 1 x^2 + 1 | 10
%e 10 75 p_3^2 * p_2 B_10(x) = 2x^2 + x 2x^2 | 25
%t a[n_] := a[n] = Which[n < 2, n + 1, EvenQ@ n, Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &@ a[n/2], True, a[#] a[# + 1] &[(n - 1)/2]]; Table[Times @@ (FactorInteger[#] /. {p_, e_} /; e > 0 :> (p^Mod[PrimePi@ p, 2])^e) &@ a@ n, {n, 0, 72}] (* _Michael De Vlieger_, Apr 05 2017 *)
%o (PARI)
%o A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From _Michel Marcus_
%o A260443(n) = if(n<2, n+1, if(n%2, A260443(n\2)*A260443(n\2+1), A003961(A260443(n\2)))); \\ Cf. _Charles R Greathouse IV_'s code for "ps" in A186891 and A277013.
%o A247503(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 2] *= (primepi(f[i, 1]) % 2); ); factorback(f); } \\ After _Michel Marcus_
%o A284553(n) = A247503(A260443(n));
%o (Scheme) (define (A284553 n) (A247503 (A260443 n)))
%Y Cf. A000360, A001222, A003961, A064989, A102283, A247503, A260443, A284554, A284556, A284563 (odd bisection).
%K nonn
%O 0,2
%A _Antti Karttunen_, Mar 29 2017
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