A115637 In the binary expansion of n+2, transform 0->1 and 1->0 then interpret as base 4.

1, 0, 5, 4, 1, 0, 21, 20, 17, 16, 5, 4, 1, 0, 85, 84, 81, 80, 69, 68, 65, 64, 21, 20, 17, 16, 5, 4, 1, 0, 341, 340, 337, 336, 325, 324, 321, 320, 277, 276, 273, 272, 261, 260, 257, 256, 85, 84, 81, 80, 69, 68, 65, 64, 21, 20, 17, 16, 5, 4, 1, 0, 1365, 1364, 1361, 1360, 1349

Offset: 0

Author

Paul Barry, Jan 27 2006

Keywords

• easy
• nonn
• look

Comments

Row sums of number triangle A115636. Partial sums of A115638.

Old name was "A divide and conquer sequence".

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..16384 (first 1025 terms from Antti Karttunen)
• Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003, see equation 2.4 with a(n) = a_{n+2} for case alpha=4, c=1, d=0.

Crossrefs

Cf. A000695, A035327, A115633, A115636, A115638 (first differences), A374625.

Programs

Maple

b:= n-> 1-(n mod 2)+`if`(n<2, 0, b(iquo(n, 2))*4):
a:= n-> b(n+2):
seq(a(n), n=0..66);  # Alois P. Heinz, Jul 16 2024

Mathematica

A115637[n_] := FromDigits[1 - IntegerDigits[n + 2, 2], 4];
Array[A115637, 100, 0] (* Paolo Xausa, Jul 16 2024 *)

PARI

up_to = 1024;
A115633array(n, k) = (((-1)^n)*if(n==k,1, if((k+k+2)==n, -4, if((k+1)==n, -(1+(-1)^k)/2, 0))));
A115637list(up_to) = { my(mA115633=matrix(up_to,up_to,n,k,A115633array(n-1,k-1)), mA115636 = matsolve(mA115633,matid(up_to)), v = vector(up_to)); for(n=1,up_to,v[n] = vecsum(mA115636[n,])); (v); };
v115637 = A115637list(up_to+1);
A115637(n) = v115637[1+n]; \\ Antti Karttunen, Nov 02 2018

PARI

a(n) = fromdigits([!b |b<-binary(n+2)], 4); \\ Kevin Ryde, Jul 15 2024

Python

def A115637(n): return int(bin((~(n+2))^(-1<<(n+2).bit_length()))[2:],4) # Chai Wah Wu, Jul 17 2024

Formula

G.f.: (1/(1-x))*Sum_{k>=0} 4^k*x^(2^(k+1)-2)/(1+x^(2^k)); the g.f. G(x) satisfies G(x) - 4(1+x)*x^2*G(x^2) = 1/(1-x^2).

a(n) = A000695(A035327(n+2)). - Kevin Ryde, Jul 15 2024

Extensions

New name from Kevin Ryde, Jul 15 2024