A185950 -id:A185950 - cp's OEIS frontend

A033989 Write 0,1,2,... in a clockwise spiral on a square lattice, writing each digit at a separate lattice point, starting with 0 at the origin and 1 at x=0, y=-1; sequence gives the numbers on the negative x-axis.

0, 3, 1, 1, 3, 2, 7, 9, 1, 1, 6, 9, 4, 7, 9, 1, 2, 1, 2, 1, 6, 7, 4, 3, 6, 1, 2, 9, 5, 1, 1, 0, 9, 3, 1, 3, 6, 6, 1, 8, 6, 9, 2, 5, 0, 2, 2, 4, 6, 6, 2, 5, 6, 0, 3, 8, 9, 5, 3, 3, 6, 9, 4, 0, 5, 4, 4, 9, 8, 0, 5, 0, 4, 5, 5, 3, 3, 1, 6, 8, 5, 8, 6, 5, 1, 4, 7, 4, 9, 1, 8, 5, 1, 9, 9, 8, 6, 6, 9, 1, 1, 6, 4, 8, 1

Offset: 0

N. J. A. Sloane

			  2---3---2---4---2---5---2
  |                       |
  2   1---3---1---4---1   6
  |   |               |   |
  2   2   4---5---6   5   2
  |   |   |       |   |   |
  1   1   3   0   7   1   7
  |   |   |   |   |   |   |
  2   1   2---1   8   6   2
  |   |           |   |   |
  0   1---0---1---9   1   8
  |                   |   |
  2---9---1---8---1---7   2
We begin with the 0 and wrap the numbers 1 2 3 4 ... around it. Then the sequence is obtained by reading leftwards, starting from the initial 0. - _Andrew Woods_, May 20 2012

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Sequences based on the same spiral: A033953, A033988, A033990. Spiral without zero: A033952.

Other sequences from spirals: A001107, A002939, A007742, A033951, A033954, A033991, A002943, A033996, A033988.

a033989 0 = 0
a033989 n = a033307 $ a185950 n  -- Reinhard Zumkeller, Aug 14 2013

nmax = 1000; A033307 = Flatten[IntegerDigits /@ Range[0, nmax^2+10*nmax]]; a[n_] := If[n==0, 0, A033307[[4n^2-n+1]]]; Table[a[n], {n, 0, nmax}] (* Jean-François Alcover, Apr 23 2017, after Andrew Woods *)

a(n) = A033307(4*n^2-n-1) for n > 0. - Andrew Woods, May 20 2012

More terms from Andrew J. Gacek (andrew(AT)dgi.net)

Edited by Charles R Greathouse IV, Nov 01 2009

A156140 Accumulation of Stern's diatomic series: a(0)=-1, a(1)=0, and a(n+1) = (2e(n)+1)*a(n) - a(n-1) for n > 1, where e(n) is the highest power of 2 dividing n.

Original entry on oeis.org

-1, 0, 1, 3, 2, 7, 5, 8, 3, 13, 10, 17, 7, 18, 11, 15, 4, 21, 17, 30, 13, 35, 22, 31, 9, 32, 23, 37, 14, 33, 19, 24, 5, 31, 26, 47, 21, 58, 37, 53, 16, 59, 43, 70, 27, 65, 38, 49, 11, 50, 39, 67, 28, 73, 45, 62, 17, 57, 40, 63, 23, 52, 29, 35, 6, 43, 37, 68, 31, 87, 56, 81, 25, 94, 69

Offset: 0

Arie Werksma (Werksma(AT)Tiscali.nl), Feb 04 2009

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Cf. A002487, A007814.
From Yosu Yurramendi, Mar 09 2018: (Start)
a(2^m +  0) = A000027(m),   m >= 0.
a(2^m +  1) = A002061(m+2), m >= 1.
a(2^m +  2) = A002522(m),   m >= 2.
a(2^m +  3) = A033816(m-1), m >= 2.
a(2^m +  4) = A002061(m),   m >= 2.
a(2^m +  5) = A141631(m),   m >= 3.
a(2^m +  6) = A084849(m-1), m >= 3.
a(2^m +  7) = A056108(m-1), m >= 3.
a(2^m +  8) = A000290(m-1), m >= 3.
a(2^m +  9) = A185950(m-1), m >= 4.
a(2^m + 10) = A144390(m-1), m >= 4.
a(2^m + 12) = A014106(m-2), m >= 4.
a(2^m + 16) = A028387(m-3), m >= 4.
a(2^m + 18) = A250657(m-4), m >= 5.
a(2^m + 20) = A140677(m-3), m >= 5.
a(2^m + 32) = A028872(m-2), m >= 5.
a(2^m -  1) = A005563(m-1), m >= 0.
a(2^m -  2) = A028387(m-2), m >= 2.
a(2^m -  3) = A033537(m-2), m >= 2.
a(2^m -  4) = A008865(m-1), m >= 3.
a(2^m -  7) = A140678(m-3), m >= 3.
a(2^m -  8) = A014209(m-3), m >= 4.
a(2^m - 16) = A028875(m-2), m >= 5.
a(2^m - 32) = A108195(m-5), m >= 6.
(End)

A156140 := proc(n)
    option remember ;
    if n <= 1 then
        n-1 ;
    else
        (2*A007814(n-1)+1)*procname(n-1)-procname(n-2) ;
    end if;
end proc:
seq(A156140(n),n=0..80) ; # R. J. Mathar, Mar 14 2009

Fold[Append[#1, (2 IntegerExponent[#2, 2] + 1) #1[[-1]] - #1[[-2]] ] &, {-1, 0}, Range[73]] (* Michael De Vlieger, Mar 09 2018 *)

first(n)=my(v=vector(n+1)); v[1]=-1; v[2]=0; for(k=1,n-1,v[k+2]=(2*valuation(k,2)+1)*v[k+1] - v[k]); v \\ Charles R Greathouse IV, Apr 05 2016

fusc(n)=my(a=1, b=0); while(n>0, if(bitand(n, 1), b+=a, a+=b); n>>=1); b
a(n)=my(m=1,s,t); if(n==0, return(-1)); while(n%2==0, s+=fusc(n>>=1)); while(n>1, t=logint(n,2); n-=2^t; s+=m*fusc(n)*(t^2+t+1); m*=-t); m*(n-1) + s \\ Charles R Greathouse IV, Dec 13 2016

a <- c(0,1)
maxlevel <- 6 # by choice
for(m in 1:maxlevel) {
  a[2^(m+1)] <- m + 1
  for(k in 1:(2^m-1)) {
    r <- m - floor(log2(k)) - 1
    a[2^r*(2*k+1)] <- a[2^r*(2*k)] + a[2^r*(2*k+2)]
}}
a
# Yosu Yurramendi, May 08 2018

Let b(n) = A002487(n), Stern's diatomic series.
a(n+1)*b(n) - a(n)*b(n+1) = 1 for n >= 0.
a(2n+1) = a(n) + a(n+1) + b(n) + b(n+1) for n >= 0.
a(2n) = a(n) + b(n) for n >= 0.
a(2^n + k) = -n*a(k) + (n^2 + n + 1)*b(k) for 0 <= k <= 2^n.
b(2^n + k) = -a(k) + (n + 1)*b(k) for 0 <= k <= 2^n.
a(2^m + k) = b(2^m+k)*m + b(k), m >= 0, 0 <= k < 2^m. - Yosu Yurramendi, Mar 09 2018
a(2^(m+1)+2^m+1) = 2*m+1, m >= 0. - Yosu Yurramendi, Mar 09 2018
From Yosu Yurramendi, May 08 2018: (Start)
a(2^m) = m, m >= 0.
a(2^r*(2*k+1)) = a(2^r*(2*k)) + a(2^r*(2*k+2)), r = m - floor(log_2(k)) - 1, m > 0, 1 <= k < 2^m.
(End)

cp's OEIS Frontend

A033989 Write 0,1,2,... in a clockwise spiral on a square lattice, writing each digit at a separate lattice point, starting with 0 at the origin and 1 at x=0, y=-1; sequence gives the numbers on the negative x-axis.

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