A178748 Total number of '1' bits in the terms of 'rows' of A178746.

1, 2, 7, 14, 37, 80, 187, 410, 913, 1988, 4327, 9326, 20029, 42776, 91027, 192962, 407785, 859244, 1805887, 3786518, 7922581, 16544192, 34486507, 71769194, 149130817, 309446420, 641262487, 1327264190, 2744006893, 5666970728, 11691855427, 24099538706, 49630733209

Offset

0,2

Comments

Sum of adjacent terms equals the difference of adjacent terms in A127981. - David Scambler, Jun 10 2010

Links

Andrew Howroyd, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,3,-4,-4).

Formula

G.f: (1/2)*x^3 - 1/4 + (x^4 + x^3 - (3/4)*x^2 - (1/2)*x + 1/4)*F(x) = 0. [From GUESSS]

From David Scambler, Jun 10 2010: (Start)

a(n) = (2^n*(3*n+8) + (3*n+1)*(-1)^n)/9.

a(n) + a(n-1) = A127981(n+1) - A127981(n).

(End)

From Colin Barker, Mar 04 2020: (Start)

G.f.: (1 - 2*x^3) / ((1 + x)^2*(1 - 2*x)^2).

a(n) = 2*a(n-1) + 3*a(n-2) - 4*a(n-3) - 4*a(n-4) for n>3.

(End)

Example

a(0) = bitcount(1) = 1.
a(1) = bitcount(3) = 2.
a(2) = bitcount(6) + bitcount(6) + bitcount(7) = 2 + 2 + 3 = 7.

Mathematica

LinearRecurrence[{2, 3, -4, -4}, {1, 2, 7, 14}, 40] (* Harvey P. Dale, Aug 27 2021 *)

Prog

(PARI) seq(n)={my(a=vector(n+1), f=0, p=0, k=1, s=0); while(k<=#a, my(b=bitxor(p+1, p)); f=bitxor(f, b); p=bitxor(p, bitand(b, f)); if(p>2^k, a[k]=s; k++; s=0); s+=hammingweight(p)); a} \\ Andrew Howroyd, Mar 03 2020
(PARI) a(n) = {(2^n*(3*n+8) + (3*n+1)*(-1)^n)/9} \\ Andrew Howroyd, Mar 03 2020
(PARI) Vec((1 - 2*x^3) / ((1 + x)^2*(1 - 2*x)^2) + O(x^30)) \\ Colin Barker, Mar 04 2020

Crossrefs

Cf. A178747 (sum of terms in rows of A178746).

Cf. A127981.

Sequence in context: A191389 A191319 A018497 * A194590 A107373 A336579

Adjacent sequences: A178745 A178746 A178747 * A178749 A178750 A178751

Keyword

nonn,easy

Author

David Scambler, Jun 09 2010

Extensions

Terms a(16) and beyond from Andrew Howroyd, Mar 03 2020

Status

approved