1,2

If n+a(n) has exactly a(n) 1's in binary, then a(n+1) = a(n)+1, but if n+a(n) has less than a(n) 1's, then a(n+1) = a(n)-1. a(n) is the number of terms needed to represent n as a sum of numbers of the form 2^k-1. - Jeffrey Shallit (shallit(AT)graceland.math.uwaterloo.ca) Is a(n) = A080468(n+1)+1?

Compute a(n) by repeatedly subtracting the largest number 2^k-1<=n until zero is reached. The number of times a term was subtracted gives a(n). Examples: 5 = 3 + 1 + 1 ==> a(5) = 3 6 = 3 + 3 ==> a(6) = 2 Conjecture: replace all zeros in A079559 by -1, then the a(n) are obtained as cumulative sums (equivalent to the generating function given). - Joerg Arndt (arndt(AT)jjj.de), Jun 12 2006

Leroy Quet, Home Page (listed in lieu of email address)

Joerg Arndt, Fxtbook

David Wasserman, Quet Transform

a(2^k-1) = 1. For 2^k <= n <= 2^(k+1)-2, a(n) = a(n-2^k+1)+1.

Conjectured OGF: frac{2\,(1-x)\,prod_{n=1}^{infty}{(1+x^{2^n-1})} - 1}{(1-x)^2} = 1 + 2 x + x^2 + 2 x^3 + 3 x^4 + 2 x^5 + x^6 + 2 x^7 + 3 x^8 + 2 x^9 + ... - Joerg Arndt (arndt(AT)jjj.de), Jun 12 2006

a(4) = 2 because 4+2 (110) has two 1's, but 4+1 (101) has more than one 1.

Conjecture (Joerg Arndt):

a(n) is the number of bits in the binary words of sequence A108918

......A108918.A108918..n..=..n.=.(sum.of.term.2^k-1)

........00001.1.....00001.=..1.=..1

........00011.2.....00010.=..2.=..1.+.1

........00010.1.....00011.=..3.=..3

........00101.2.....00100.=..4.=..3.+.1

........00111.3.....00101.=..5.=..3.+.1.+.1

........00110.2.....00110.=..6.=..3.+.3

........00100.1.....00111.=..7.=..7

........01001.2.....01000.=..8.=..7.+.1

........01011.3.....01001.=..9.=..7.+.1.+.1

........01010.2.....01010.=.10.=..7.+.3

........01101.3.....01011.=.11.=..7.+.3.+.1

........01111.4.....01100.=.12.=..7.+.3.+.1.+.1

........01110.3.....01101.=.13.=..7.+.3.+.3

........01100.2.....01110.=.14.=..7.+.7

........01000.1.....01111.=.15.=.15

(PARI) TInverse(v) = local(l, w, used, start, x); l = length(v); w = vector(l); used = vector(l); start = 1; for (i = 1, l, while (start <= l && used[start], start++); x = start; for (j = 2, v[i], x++; while (x <= l && used[x], x++)); if (x > l, return (vector(i - 1, k, w[k])), w[i] = x; used[x] = 1)); w; PInverse(v) = local(l, w); l = length(v); w = vector(l); for (i = 1, l, if (v[i] <= l, w[v[i]] = i)); w; T(v) = local(l, w, c); l = length(v); w = vector(l); for (n = 1, l, if (v[n], c = 0; for (m = 1, n - 1, if (v[m] < v[n], c++)); w[n] = v[n] - c, return (vector(n - 1, i, w[i])))); w; Q(v) = T(PInverse(TInverse(v))); v = vector(150); for (n = 1, 150, m = n; x = 1; while (!(m%2), m\=2; x *= 2); v[n] = x); Q(v)

Cf. A006519; A080468.

Sequence in context: A081771 A066856 A089280 this_sequence A088696 A004738 A043554

Adjacent sequences: A100658 A100659 A100660 this_sequence A100662 A100663 A100664

easy,nonn

David Wasserman (dwasserm(AT)earthlink.net), Jan 14 2005